on generalized (α,β)-derivations in prime rings

AlgebraColloquiumc© 2010 AMSS CAS

& SUZHOU UNIV

Algebra Colloquium 17 (Spec 1) (2010) 865–874

On Generalized (α, β)-Derivations in Prime Rings

Hidetoshi MarubayashiDepartment of Mathematics, Faculty of Engineering

Tokushima Bunri University, Japan

E-mail: [email protected]

Mohammad Ashraf Nadeem-ur Rehman Shakir Ali†Department of Mathematics, Aligarh Muslim University

Aligarh 202002, IndiaE-mail: [email protected] [email protected]

[email protected]

Received 4 January 2008Revised 18 November 2008

Communicated by Nanqing Ding

Abstract. Let R be a ring and α, β be endomorphisms of R. An additive mappingF : R → R is called a generalized (α, β)-derivation on R if there exists an (α, β)-derivationd : R → R such that F (xy) = F (x)α(y) + β(x)d(y) holds for all x, y ∈ R. In the presentpaper, we discuss the commutativity of a prime ring R admitting a generalized (α, β)-derivation F satisfying any one of the properties: (i) [F (x), x]α,β = 0, (ii) F ([x, y]) = 0,(iii) F (x ◦ y) = 0, (iv) F ([x, y]) = [x, y]α,β , (v) F (x ◦ y) = (x ◦ y)α,β , (vi) F (xy)−α(xy) ∈Z(R), (vii) F (x)F (y)− α(xy) ∈ Z(R) for all x, y in an appropriate subset of R.

2000 Mathematics Subject Classification: 16W25, 16N60, 16U80

Keywords: Lie ideals, prime rings, (α, β)-derivations, generalized (α, β)-derivations

1 Introduction

Let R be an associative ring with center Z(R). For x, y ∈ R, denote the commutatorxy− yx by [x, y] and the anti-commutator xy + yx by x ◦ y. Recall that a ring R isprime if for any a, b ∈ R, aRb = {0} implies a = 0 or b = 0. An additive subgroupU of R is said to be a Lie ideal of R if [U,R] ⊆ U . A Lie ideal U is said to be asquare-closed Lie ideal if u2 ∈ U for all u ∈ U . Let α and β be endomorphisms ofR. For any x, y ∈ R, set [x, y]α,β = xα(y)− β(y)x and (x ◦ y)α,β = xα(y) + β(y)x.Following [10], an additive mapping F : R → R is called a generalized derivationassociated with a derivation d if F (xy) = F (x)y + xd(y) holds for all x, y ∈ R. An

†Corresponding author.

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866 H. Marubayashi, M. Ashraf, N. Rehman, S. Ali

additive map d : R → R is called an (α, β)-derivation if d(xy) = d(x)α(y)+β(x)d(y)holds for all x, y ∈ R. For a fixed a, the map da : R → R given by da(x) = [a, x]α,β

for all x ∈ R is an (α, β)-derivation which is said to be an (α, β)-inner derivation.An additive mapping F : R → R is called a generalized (α, β)-inner derivationif F (x) = aα(x) + β(x)b holds for some fixed a, b ∈ R and for all x ∈ R. Asimple computation yields that if F is a generalized (α, β)-inner derivation, thenfor all x, y ∈ R, we have F (xy) = F (x)α(y) + β(x)d(−b)(y), where d(−b) is an(α, β)-inner derivation. With this viewpoint, an additive map F : R → R is calleda generalized (α, β)-derivation associated with an (α, β)-derivation d : R → R ifF (xy) = F (x)α(y) + β(x)d(y) holds for all x, y ∈ R. A (1, 1)-generalized derivationis simply called a generalized derivation, where 1 is the identity map on R.

The aim of the present paper is to extend many known results for generalized(α, β)-derivations. In fact, our results unify, extend and complement several well-known theorems previously obtained in [1, Theorems 2.1, 2.5–2.7], [3, Theorem3.1–3.4], [4, Theorem 1], [13, Theorem 3.1–3.8], [14, Theorem 2.1–2.4], etc.

2 Preliminary Results

Throughout the present paper, α and β will denote automorphisms of R. We shalluse without explicit mention the following basic identities:

[xy, z]α,β = x[y, z]α,β + [x, β(z)]y = x[y, α(z)] + [x, z]α,βy,

[x, yz]α,β = β(y)[x, z]α,β + [x, y]α,βα(z),(x ◦ (yz))α,β = (x ◦ y)α,βα(z)− β(y)[x, z]α,β = β(y)(x ◦ z)α,β + [x, y]α,βα(z),

((xy) ◦ z)α,β = x(y ◦ z)α,β − [x, β(z)]y = (x ◦ z)α,βy + x[y, α(z)].

The proof of the following Proposition 2.1 is rather elementary and is based onthe fact that a group cannot be written as the set-theoretic union of its two propersubgroups.

Proposition 2.1. Let R be a prime ring and S an additive subgroup of R. Letf : S → R and g : S → R be additive functions such that f(s)Rg(s) = {0} for alls ∈ S. Then either f(s) = 0 for all s ∈ S, or g(s) = 0 for all s ∈ S.

Lemma 2.2. [6, Lemma 1] Let R be a prime ring and I a nonzero left (or right)ideal of R. If d is an (α, β)-derivation on R such that d(I) = {0}, then d = 0.

The next two lemmas are essentially proved in [9, Lemma 4] and [13, Lemma2.6], respectively.

Lemma 2.3. If U * Z(R) is a Lie ideal of a 2-torsion free prime ring R anda, b ∈ R such that aUb = {0}, then a = 0 or b = 0.

Lemma 2.4. Let R be a 2-torsion free prime ring and U a nonzero Lie ideal of R.If U is commutative, then U ⊆ Z(R).

Lemma 2.5. Let R be a prime ring with charR 6= 2, and let U be a nonzerosquare-closed Lie ideal of R. If [u, v]α,β = 0 for all u, v ∈ U , then U ⊆ Z(R).

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On Generalized (α, β)-Derivations in Prime Rings 867

Proof. We have[u, v]α,β = 0 for all u, v ∈ U. (1)

Replacing v by [r, u] in (1), we get [u, [r, u]]α,β = 0 for all u ∈ U and r ∈ R. Again,replace r by rs to get [u, [rs, u]]α,β = 0 for all u ∈ U and r, s ∈ R. That is,

[u, [r, u]]α,βα(s) + β([r, u])[u, s]α,β + β(r)[u, [s, u]]α,β + [u, r]α,βα([s, u]) = 0

for all u ∈ U and r, s ∈ R. This implies that β([r, u])[u, s]α,β + [u, r]α,βα([s, u]) = 0for all u ∈ U and r, s ∈ R. Now replace r by v in the above expression and use (1)to get β([v, u])[u, s]α,β = 0 for all u, v ∈ U and s ∈ R. Again replacing s by sr andusing the above expression, we find that β([v, u])β(s)[u, r]α,β = 0 for all u, v ∈ Uand s ∈ R, that is, β([v, u])R[u, r]α,β = {0}. Thus, by Proposition 2.1, eitherβ([v, u]) = 0 for all u, v ∈ U , or [u, r]α,β = 0 for all u ∈ U and r ∈ R. In thefirst case, U ⊆ Z(R) by Lemma 2.4. In the second case, replace u by 2vu in theexpression [u, r]α,β = 0 to obtain [v, β(r)]u = 0 for all u, v ∈ U and r ∈ R. It followsby Lemma 2.3 that U ⊆ Z(R). ¤

Lemma 2.6. Let R be a 2-torsion free prime ring and U a nonzero square-closedLie ideal of R. Suppose that there exists a nonzero (α, β)-derivation d such thatd(u) = 0 for all u ∈ U . Then U ⊆ Z(R).

Proof. We have

d(u) = 0 for all u ∈ U. (2)

This yields that d([u, r]) = 0 for all u ∈ U and r ∈ R. Now an application of (2)yields that

[d(r), u]α,β = 0 for all u ∈ U, r ∈ R. (3)

Replace r by rv in (3), and use (3) and (2) to get d(r)α([v, u]) = 0 for all u, v ∈ Uand r ∈ R. Again replacing r by rs, we obtain d(r)Rα([u, v]) = {0} for all u, v ∈ Uand r ∈ R. Thus, the primeness of R implies either d(r) = 0 or α([u, v]) = 0. Sinced 6= 0, we find that α([u, v]) = 0 for all u, v ∈ U , i.e., [u, v] = 0 for all u, v ∈ U .Hence, by Lemma 2.4, we get the required result. ¤

Lemma 2.7. Let R be a prime ring and I be a nonzero right ideal of R. If Radmits a generalized (α, β)-derivation F such that {0} 6= F (I) ⊆ Z(R), then R iscommutative.

Proof. We have F (xr) = F (x)α(r) + β(x)d(r) ∈ Z(R) for all x ∈ I and r ∈ R.Therefore, [β(x)d(r), α(r)] = 0 = β(x)[d(r), α(r)] + [β(x), α(r)]d(r); and replac-ing x by yx gives [β(y), α(r)]β(x)d(r) = 0 for all x, y ∈ I and r ∈ R. Sincethe right annihilator of a nonzero right ideal is trivial, for a fixed r ∈ R, wehave [β(y), α(r)]β(x) = 0 for all x, y ∈ I, or d(r) = 0. Therefore, d = 0 or[β(y), α(r)]β(x) = 0 for all x, y ∈ I and r ∈ R. In the latter case, replacing r byrs gives [β(y), u]Rβ(x) = {0} for all x, y ∈ I and u ∈ R. Thus, β(I) ⊆ Z(R), soR is commutative. Now suppose d = 0. Choose x ∈ I such that F (x) 6= 0. We

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have F (xr) = F (x)α(r) ∈ Z(R); and since nonzero central elements are not zerodivisors, we get α(r) ∈ Z(R) so that R is commutative. ¤

Theorem 2.8. Let R be a 2-torsion free prime ring and U a nonzero square-closedLie ideal of R. If there exists a nonzero (α, β)-derivation d such that [d(u), u]α,β = 0for all u ∈ U , then U ⊆ Z(R).

Proof. Suppose on the contrary that U 6⊆ Z(R). Define B(· , ·) : R × R → R byB(u, v) = [d(u), v]α,β + [d(v), u]α,β for all u, v ∈ U ; and note that by linearizingthe condition [d(u), u]α,β = 0, we get B(u, v) = 0 for all u, v ∈ U . It is easilyverified that B(uv, w) = B(u,w)α(v) + β(u)B(v, w) + d(u)α([v, w]) + β([u,w])d(v)for all u, v, w ∈ U , so by taking w = u, we obtain d(u)α([v, u]) = 0 for all u, v ∈ U .Replacing v by 2vw yields d(u)α(v)α([w, u]) = 0 so that α−1(d(u))U [w, u] = {0}for all u,w ∈ U ; and applying Lemma 2.3 and the fact that (U,+) is not the unionof two of its proper subgroups shows that either d(U) = {0} or U is commutative.By Lemmas 2.4 and 2.6, either of these conditions yields a contradiction. ¤

Using the same techniques with necessary variations, we can prove the followingcorollary even without the characteristic assumption on the ring.

Corollary 2.9. Let R be a prime ring and I a nonzero ideal of R. If R admitsa nonzero (α, β)-derivation d such that [d(x), x]α,β = 0 for all x ∈ I, then R iscommutative.

3 Lie Ideals and Generalized (α, β)-Derivations

There is a close connection between the commutativity of a ring R and the existenceof certain specified derivations on R (cf. [2–5, 7, 8, 11–13]). In 2002, the secondand third authors [3] established that if a 2-torsion free prime ring R admits aderivation d such that d([u, v]) = [u, v] for all u, v ∈ U , where U is a Lie ideal of R,then U ⊆ Z(R). Further, the third author [13] extended the mentioned results forgeneralized derivations. In the present section, our aim is to extend these resultsfor generalized (α, β)-derivations in rings.

Theorem 3.1. Let R be a 2-torsion free prime ring and U a nonzero square-closed Lie ideal of R. If R admits a generalized (α, β)-derivation F associatedwith a nonzero (α, β)-derivation d such that [F (u), u]α,β = 0 for all u ∈ U , thenU ⊆ Z(R).

Proof. Suppose on the contrary that U * Z(R) and

[F (u), u]α,β = 0 for all u ∈ U. (4)

Linearizing (4), we obtain

[F (u), v]α,β + [F (v), u]α,β = 0 for all u, v ∈ U. (5)

Replacing v by 2vu in (5), we get [F (u), vu]α,β + [F (v)α(u) + β(v)d(u), u]α,β = 0,

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that is,

[F (u), v]α,βα(u) + β(v)[F (u), u]α,β + [F (v), u]α,βα(u)+ F (v)[α(u), α(u)] + β(v)[d(u), u]α,β + [β(v), β(u)]d(u) = 0. (6)

Now combining (4)–(6), we find that

β(v)[d(u), u]α,β + [β(v), β(u)]d(u) = 0 for all u, v ∈ U. (7)

Again, replace v by 2wv in (7) to get

[β(w), β(u)]β(v)d(u) = 0 for all u, v, w ∈ U. (8)

This implies that [w, u]Uβ−1(d(u)) = {0} for all u,w ∈ U . Thus, for each u ∈ U ,by Lemma 2.3, we have either [w, u] = 0 or β−1(d(u)) = 0. Now using similararguments as used in the proof of Theorem 2.8, we get the required result. ¤

Theorem 3.2. Let R be a 2-torsion free prime ring and U a nonzero square-closedLie ideal of R. If R admits a generalized (α, β)-derivation F associated with anonzero (α, β)-derivation d such that F ([u, v]) = 0 for all u, v ∈ U , then U ⊆ Z(R).

Proof. Suppose on the contrary that U 6⊆ Z(R) and F ([u, v]) = 0 for all u, v ∈ U .Replacing v by 2vu in the above expression and using the fact charR 6= 2, we findthat 0 = F ([u, vu]) = F ([u, v]u) = β([u, v])d(u). Again, replace v by 2wv to getβ([u,w]v)d(u) = 0 for all u, v, w ∈ U . This implies that [u,w]Uβ−1(d(u)) = {0}for all u,w ∈ U . Now applying similar techniques as used after (8) in the proof ofTheorem 3.1 yields the required result. ¤

Theorem 3.3. Let R be a 2-torsion free prime ring and U a nonzero square-closedLie ideal of R. Suppose that R admits a generalized (α, β)-derivation F associatedwith an (α, β)-derivation d such that

(i) F ([u, v]) = [u, v]α,β for all u, v ∈ U , or

(ii) F ([u, v]) = −[u, v]α,β for all u, v ∈ U .If F = 0 or d 6= 0, then U ⊆ Z(R).

Proof. (i) Let F be a generalized (α, β)-derivation of R such that

F ([u, v]) = [u, v]α,β for all u, v ∈ U. (9)

If F = 0, then [u, v]α,β = 0 for all u, v ∈ U . Thus, by Lemma 2.5, we get therequired result.

Henceforth, we shall assume d 6= 0. Suppose on the contrary that U 6⊆ Z(R).Replacing v by 2wv in (9) and using the fact charR 6= 2, we get F (w[u, v]+ [u,w]v)= [u,wv]α,β for all u, v, w ∈ U , that is,

F (w)α([u, v]) + β(w)d([u, v]) + F ([u,w])α(v) + β([u,w])d(v)= [u,w]α,βα(v) + β(w)[u, v]α,β

for all u, v, w ∈ U .

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Now by applying (9), we find that

F (w)α([u, v]) + β(w)d([u, v]) + β([u,w])d(v) = β(w)[u, v]α,β . (10)

Replace v by 2vu in (10) to get

F (w)α([u, v]u) + β(w)d([u, v]u) + β([u,w])(d(v)α(u) + β(v)d(u))= β(w)([u, v]α,βα(u) + β(v)[u, u]α,β).

This implies that(F (w)α([u, v]) + β(w)d([u, v]) + β([u,w])d(v)

)α(u)

+β(w)β([u, v])d(u) + β([u,w])β(v)d(u)= β(w)([u, v]α,βα(u) + β(w)β(v)[u, u]α,β)

for all u, v, w ∈ U . Now using (10), we obtain

β(w)β([u, v])d(u) + β([u,w])β(v)d(u) = β(w)β(v)[u, u]α,β . (11)

Again, replace w by 2w1w in (11) to get

β(w1)β(w)β([u, v])d(u) + β(w1[u,w] + [u,w1]w)β(v)d(u)= β(w1)β(w)β(v)[u, u]α,β

for all u, v, w,w1 ∈ U . An application of (11) gives β([u,w1])β(w)β(v)d(u) = 0 forall u, v, w,w1 ∈ U , and hence [u,w1]wUβ−1(d(u)) = {0} for all u,w, w1 ∈ U . Itfollows by Lemma 2.3 that d(U) = {0} or U is commutative, so by Lemmas 2.4 and2.6, we get a contradiction.

(ii) If F ([u, v]) = −[u, v]α,β for all u, v ∈ U , then −F satisfies the condition inpart (i), hence our result follows. ¤Theorem 3.4. Let R be a 2-torsion free prime ring and U a square-closed Lie idealof R. Suppose that R admits a generalized (α, β)-derivation F associated with an(α, β)-derivation d such that

(i) F (uv) = α(uv) for all u, v ∈ U , or

(ii) F (uv) = α(vu) for all u, v ∈ U .If F = 0 or d 6= 0, then U ⊆ Z(R).

Proof. (i) For any u, v ∈ U , we have F (uv−vu) = F (uv)−F (vu) = α(uv)−α(vu),and hence F ([u, v]) = α([u, v]). If F = 0, then α([u, v]) = 0 for all u, v ∈ U . Thus,[u, v] = 0 for all u, v ∈ U , and hence U ⊆ Z(R) by Lemma 2.4.

Henceforth, we shall assume d 6= 0. Suppose on the contrary that U * Z(R).For any u, v ∈ U , we have F ([u, v]) = α([u, v]). This can be rewritten as

F (u)α(v) + β(u)d(v)− F (v)α(u)− β(v)d(u) = α([u, v]). (12)

Replacing v by 2vu in (12) and using the fact charR 6= 2, we find that

F (u)α(v)α(u) + β(u)d(v)α(u) + β(u)β(v)d(u)−F (v)α(u)α(u)− β(v)d(u)α(u)− β(v)β(u)d(u)

= α([u, v])α(u)

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for all u, v ∈ U , and hence an application of (12) gives β([u, v])d(u) = 0 for allu, v ∈ U . Again, replace v by 2wv to get β([u,w])β(v)d(u) = 0 for all u, v, w ∈ Uand hence [u,w]Uβ−1(d(u)) = {0} for all u,w ∈ U . The last expression is same asthe equation (8) and hence the result follows.

(ii) Using similar techniques with necessary variations, the result follows. ¤

If the commutator is replaced by the anti-commutator in Theorems 3.2 and 3.3,then we see that the conclusions of these theorems also hold.

Theorem 3.5. Let R be a 2-torsion free prime ring and U a nonzero square-closedLie ideal of R. Suppose that R admits a generalized (α, β)-derivation F associatedwith an (α, β)-derivation d such that F (uov) = 0 for all u, v ∈ U . If d 6= 0, thenU ⊆ Z(R).

Proof. Suppose on the contrary that U 6⊆ Z(R). Replacing v by 2vu in our hy-pothesis, we obtain 0 = F (u ◦ vu) = F ((u ◦ v)u) = β(u ◦ v)d(u) for all u, v ∈ U .Now replace v by 2wv and use the above relation to get β([u,w]v)d(u) = 0 for allu, v, w ∈ U . This implies that [u,w]Uβ−1(d(u)) = {0} for all u,w ∈ U . Now anapplication of similar arguments as used after (8) in the proof of Theorem 3.1 yieldsthe required result. ¤Theorem 3.6. Let R be a 2-torsion free prime ring and U a square-closed Lie idealof R. Suppose that R admits a generalized (α, β)-derivation F associated with an(α, β)-derivation d such that

(i) F (u ◦ v) = (u ◦ v)α,β for all u, v ∈ U , or(ii) F (u ◦ v) = −(u ◦ v)α,β for all u, v ∈ U .

If F = 0 or d 6= 0, then U ⊆ Z(R).

Proof. (i) If F = 0, then we have

(u ◦ v)α,β = 0 for all u, v ∈ U. (13)

Replacing v by 2vw in (13), we get β(v)[u,w]α,β = 0 for all u, v, w ∈ U . Nowreplace v by [v, r] to get β([v, r])[u,w]α,β = 0. Again replacing r by rs in the aboveexpression, we find that β([v, r])R[u,w]α,β = {0} for all u, v, w ∈ U and r ∈ R.Thus, the primeness of R forces either β([v, r]) = 0 or [u,w]α,β = 0. Hence, ifβ([v, r]) = 0 for all v ∈ U and r ∈ R, then [v, r] = 0. This implies that U ⊆ Z(R).On the other hand, if [u,w]α,β = 0 for all u,w ∈ U , then by Lemma 2.5, we get therequired result.

Therefore, we shall assume d 6= 0. Suppose on the contrary that U * Z(R). Forany u, v ∈ U , we have F (u ◦ v) = (u ◦ v)α,β . This can be rewritten as

F (u)α(v) + β(u)d(v) + F (v)α(u) + β(v)d(u) = (u ◦ v)α,β . (14)

Replacing v by 2vu in (14), we find that

F (u)α(v)α(u) + β(u)d(v)α(u) + β(u)β(v)d(u)+F (v)α(u)α(u) + β(v)d(u)α(u) + β(v)β(u)d(u)

= (u ◦ v)α,βα(u)− β(v)[u, u]α,β

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for all u, v ∈ U . Thus, an application of (14) gives β(u ◦ v)d(u) + β(v)[u, u]α,β = 0for all u, v ∈ U . Again, replace v by 2wv to get β([u,w])β(v)d(u) = 0, that is,[u,w]Uβ−1(d(u)) = {0} for all u,w ∈ U . Notice that the arguments given in thelast paragraph of the proof of Theorem 3.1 are still valid in the present situation,and hence repeating the same process, we get the required result.

(ii) Use similar arguments as above. ¤Corollary 3.7. Let R be a prime ring and I a nonzero ideal of R. Suppose thatR admits a generalized (α, β)-derivation F associated with an (α, β)-derivation dsuch that any one of the following holds:

(i) [F (x), x]α,β = 0 for all x ∈ I.(ii) F ([x, y]) − [x, y]α,β = 0 for all x, y ∈ I, or F ([x, y]) + [x, y]α,β = 0 for all

x, y ∈ I.(iii) F (x ◦ y) − (x ◦ y)α,β = 0 for all x, y ∈ I, or F (x ◦ y) + (x ◦ y)α,β = 0 for all

x, y ∈ I.If F = 0 or d 6= 0, then R is commutative.

4 Ideals and Generalized (α, β)-Derivations

In the hypothesis of Theorem 3.4, if we choose the underlying subset as an idealinstead of a Lie ideal, then we can prove the following results even without thecharacteristic assumption on the ring.

Theorem 4.1. Let R be a prime ring and I a nonzero ideal of R. If R admits ageneralized (α, β)-derivation F 6= α such that F (xy)−α(xy) ∈ Z(R) for all x, y ∈ I,then R is commutative.

Proof. If F = 0, then α(I2) is a nonzero central ideal and thus R is commutative.Then suppose F 6= 0; and consider first the case d 6= 0. For any x, y ∈ I, we haveF (xy) − α(xy) ∈ Z(R). This can be rewritten as F (x)α(y) + β(x)d(y) − α(xy) ∈Z(R). Replacing y by yz, we obtain

F (x)α(y)α(z) + β(x)d(y)α(z) + β(x)β(y)d(z)− α(xy)α(z) ∈ Z(R)

for all x, y ∈ I. Thus, in particular, we have

[(F (x)α(y) + β(x)d(y)− α(xy))α(z) + β(xy)d(z), α(z)] = 0

for all x, y ∈ I. This gives [β(xy)d(z), α(z)] = 0 for all x, y, z ∈ I and hence

β(xy)[d(z), α(z)] + [β(xy), α(z)]d(z) = 0. (15)

For any r ∈ R, replacing x by rx in (15), we get [β(r), α(z)]β(x)β(y)d(z) = 0, so([β(r), α(z)])β(I)β(I)R(d(z)) = {0} for all x, z ∈ I and r ∈ R. By Proposition 2.1,we conclude either d(I) = {0} or ([β(r), α(z)])β(I)β(I) = {0} = [β(r), α(I)] for allr ∈ R. But d(I) 6= {0} by Lemma 2.2, hence α(I) ⊆ Z(R) and R is commutative.

The remaining case is F 6= 0 = d, in which case F (rs) = F (r)α(s) for allr, s ∈ R. For x, y, z,∈ I and r ∈ R, we have

F (rxy)− α(rx)α(y) = (F (r)− α(r))α(xy) ∈ Z(R) (16)

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and

(F (r)− α(r))α(xy)α(z) ∈ Z(R). (17)

We cannot have (F (r) − α(r))α(xy) = 0 for all r ∈ R and x, y ∈ I, for thatwould imply F = α. Therefore, (16) and (17) yield α(I) ⊆ Z(R); and hence R iscommutative. ¤

Theorem 4.2. Let R be a prime ring and I a nonzero ideal of R. If R admits ageneralized (α, β)-derivation F 6= α such that F (xy)+α(xy) ∈ Z(R) for all x, y ∈ I,then R is commutative.

Proof. If F (xy) + α(xy) ∈ Z(R) for all x, y ∈ I, the generalized (α, β)-derivation−F satisfies the condition (−F )(xy)− α(xy) ∈ Z(R) for all x, y ∈ R, and hence byTheorem 4.1, R is commutative. ¤

Theorem 4.3. Let R be a prime ring and I a nonzero ideal of R. If R admitsa generalized (α, β)-derivation F 6= α such that F (x)F (y) − α(xy) ∈ Z(R) for allx, y ∈ I, then R is commutative.

Proof. If F = 0, then α(xy) ∈ Z(R) for all x, y ∈ I. Using the same arguments aswe have used in the proof of Theorem 4.1, we get the required result. Hence, weshall assume d 6= 0. For any x, y ∈ I, we have F (x)F (y)−α(xy) ∈ Z(R). Replacingy by yr, we find that F (x)(F (y)α(r) + β(y)d(r))− α(xy)α(r) ∈ Z(R), that is,

(F (x)F (y)− α(xy))α(r) + F (x)β(y)d(r) ∈ Z(R) for all x, y ∈ I, r ∈ R. (18)

This implies that

[F (x)β(y)d(r), α(r)] = 0 for all x, y ∈ I, r ∈ R.

It can be rewritten as

F (x)[β(y)d(r), α(r)] + [F (x), α(r)]β(y)d(r) = 0 for all x, y ∈ I, r ∈ R.

Now replacing y by β−1(F (x))y in (18), we find that

[F (x), α(r)]F (x)β(y)d(r) = 0.

As in the proof of Theorem 4.1, we conclude that [F (x), α(r)]F (x) = 0 for allx ∈ I and r ∈ R; and replacing r by rs gives [F (x), α(r)]RF (x) = {0}. Therefore,F (I) ⊆ Z(R). If F (I) 6= {0}, the commutativity of R follows by Lemma 2.7. IfF (I) = {0}, R is commutative because α(I2) is a nonzero central ideal. ¤

By similar arguments as above with necessary variations, we can prove thefollowing:

Theorem 4.4. Let R be a prime ring and I a nonzero ideal of R. If R admitsa generalized (α, β)-derivation F 6= α such that F (x)F (y) + α(xy) ∈ Z(R) for allx, y ∈ I, then R is commutative.

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Acknowledgement. The authors are greatly indebted to the referee for his/her valuable

suggestions which have improved the paper immensely and shorten the proofs of various

theorems.

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on generalized (α,β)-derivations in prime rings

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