**Introduction**

In the last four decades, several authors discussed the commutativity of the prime rings and the semiprime rings that admitting automorphisms, derivations or generalized derivations that are centralizing or commuting on an appropriate subset of R (see [1], [2], [3], [4], [8] and [10]). Related concepts such as σ -prime rings and Lie structure of prime rings with generalized (α,β) -derivations are also discussed (see for example: [5], [6], [7] and [9]).

Let R be an associative ring with center Z(*R*) and involution *- For each *x, y* ∈ *R*, the symbol [*x, y* ] represents the commutator *x, y –* *y, x* and the symbol *x*_{ °} y stands for the skew-commutator *x, y* + *y,x* An additive mapping *x ↔ x* *on a ring *R* is called an involution if (*x**)* =* x* and (*xy*)* = *x*y* *for all *x, y* ∈ *R*. A ring *R* equipped with an involution * is said to be a *-prime ring if* a* *R* *b* = *a* *R* *b*^{*} = {0} implies *a* = 0 or *b* = 0 for any *a, b* ∈ *R* . Obviously, every prime ring equipped with involution * is* -prime. The converse, in general, is not true. The set of all symmetric and skew-symmetric elements of a *- ring will be denoted by *S*_{*} *(R)* An additive mapping *d : R → R* is called a derivation if d : (*xy*) = *d* (*x*)* y +* *d x *(*y) *for all *x, y* ∈ *R* . In particular R for a fixed *a, *∈ *R* the mapping * I*_{a} : *R → R *given by *I*_{a }(*x) = *[a, *x*]* * is a derivation called an inner derivation. Let *a* and β be endomorphisms of *R*. An additive map *d : R → R* is called an (*α, β*) -derivation if *d*(*x*)a(*y)+* *β(**x*)d(*y)** *for all *x, y* ∈ *R* . A (1, 1)-derivation is called simply a derivation, where 1 is the identity map on *R*. For a fixed *a, *∈ *R*, the map *d*_{a} : R→R given by *d*_{a}(x) = [a, x] for all *x* ∈ R is an (*α, β*) -derivation called an (*α, β*) -inner derivation. An additive mapping F: R →R is called a generalized (*α, β*)-inner derivation if *F(x)* = aa (x) + *β* *(x)b*, for some fixed *a, ∈ b R* and for all *x ∈ R*. A simple computation yields that if F is a generalized inner derivation, then for 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. An additive map F : R →R is called a generalized (*α, β*)-derivation associated with an -derivation d : R → R if F*(xy)* = F*(x) α (y)* + * (x)d(y)* for all *x, y ∈ R*, and is denoted by *(F, d).*

In this paper, we have investigated the commutativity of *R* such that *R* is a*-prime ring admitting generalized (*α, β*) -derivations *(F, d)* and *(G, g)* that satisfying certain properties for all *x, y∈ R*. Throughout this paper, the ordered pair *(F, d)* stands for a generalized (*α, β*)-derivation F associated with an (*α, β*)-derivation* d*.

**Preliminaries**

We shall use the following well known basic identities that hold for any x, y, z R and any automorphisms *α, β* on *R*:

[x y, z]

_{α,β} = x[y, z]

_{α,β}+ [x, β (z)]y = x[y, α (z)] + [x, z]

_{α,β} y;α

[x, y z]

_{α,β} =β (y)[x, z]

_{α,β}+ [x, y]

_{α,β} α(z);

(x (y z))_{α,β} = (x _{°} y)_{α,β} α(z)− β(y)[x _{°} z]_{α,β} = β(y)(x _{°} z)_{α,β} +[x, y]_{α,β} α(z);

((x y) z)_{α,β} = x(y _{° }z)_{α,β} − [x, β (z)]y = (x _{°} z)_{α,β} y + x[y, α (z)].

It is known that for a prime ring R and a nonzero element *a∈ Z(R)*, if a *b∈ Z(R*), then *b∈ Z(R)*. The following lemma is essential to prove our results:

**Lemma **

Let *R* be a 2-torsion free semiprime ring. If *α :R → R* is an automorphism on R, then for any 0 ≠ z *∈ * Z(R), we have (z)*∈ * Z(R).

**Proof**

We have 0 z Z(R), that is [z, r] = 0 for all r R and hence ([z, r]) = 0. Since is an automorphism on R, we have [ (z), (r)] = 0. Now, replacing r by (s), we get [ (z), s] = 0 for all s R. Therefore, (z) Z(R) as required.

**Theorem**

[4 : Theorem 3.5] Let R be a 2-torsion free prime ring and U a nonzero square-closed Lie ideal of R. Suppose that R admits a generalized -derivation (F, d) such that F(u∘ v) = 0 for all u, v U. If d ≠ 0, then UÍ Z(R).

**Results**

**Theorem**

Let R be a -prime ring and , be automorphisms on R. Suppose that R admits generalized -derivations (F, d) and (G, g) such that {0} g(Z(R)) Z(R) and [F(x),G(y)] − [x, y] Z(R), for all x, y R. If F = 0 (respectively G = 0) or d 0 (respectively g 0), then R is commutative where d and g are -derivations.

**Proof**

Let F and G be generalized -derivations on R. If F = 0 (or G = 0), then for any x, y R, we have

[x, y] Z(R). (3.1)

Replacing x by x (y) in (3.1) gives [x, y] (y) Z(R), which implies, for all x, y, c R,

[[x, y] (y), c] = [x, y] [ (y), c] = 0.

Again if we replace c by c (m) and use the last expression we get

[x, y] R[ (y), (m)] = {0}, for all x, y, m R. (3.2)

For any y S (R) R, we have

[x, y] R[ (y), (m)] = [x, y] R[ (y), (m)] = 0.

Thus, for any y S (R) R, the −primeness of R yields that either [x, y] = 0 or [ (y), (m)] = 0. Using the fact that y − y S (R) R, we get [x, y − y ] = 0 or ([y − y ,m]) = 0, for any x, y, m R. If [x, y − y ] = 0, then from equation (3.2), for all x, y, m R, we obtain [x, y] R[ (y), (m)] = [x, y] R[ (y), (m)]. If ([y − y ,m]) = 0, then from (3.2), we obtain

[x, y] R[ (y), (m)] = [x, y] R[ (y), (m)].

Consequently, for all y R either [x, y] = 0 or [ (y), (m)] = 0. Now, let A ={y R | [y, m] = 0} and B = {y R | [x, y] = 0}. Then A, B are both additive subgroups of R and R = A B. But a group cannot be a union of its two proper subgroups and hence either R = A or R = B. If A = R then we have [y, m] = 0 for all y, m R and hence R is commutative. If B = R, then we have

*[x, y] = 0, for all x, y R. (3.3)*

Replacing *x* by *xc* in (3.3) gives *[c, (y)]x = 0*, for all *x, c, y R*, that is *[c, (y)]Rx = 0*. For any *x S (R) R*, we have *[c, (y)]Rx = 0 = [c, (y)]Rx .* Thus, the −primeness of *R* yields that *[c, (y)] = 0* for all *c, y R*. As is an automorphism on *R*, we obtain that *R* is commutative.

** (ii)** Assume that *g 0 (or d 0)*. Then for all *x, y R*, we have

*[F(x),G(y)] − [x, y] Z(R). (3.4)*

Since *{0} g(Z(R)) Z(R)*, for any *z Z(R)* replacing *y* by *y z*, in (3.4) and using Lemma 2.1 give

*[F(x), (y)]g(z) − (y)[x, z] Z(R). (3.5)*

Again, replacing *y* by *my* in (3.5) gives

* (m){[F(x), (y)]g(z) − (y)[x, z] } + [F(x), (m)] (y)g(z) Z(R),*

or equivalently,

* [ (m){[F(x), (y)]g(z) − (y)[x, z] } + [F(x), (m)] (y)g(z), (m)] = 0,*

which implies * [[F(x), (m)] (y)g(z), (m)] = 0 *for all* x, y, m R.* Since *{0} g(Z(R)) Z(R)* and *R* is −prime, we get

* [F(x), (m)][ (y), (m)] + [[F(x), (m)], (m)] (y) = 0.*

Now, replacing *y* by *y x* in the last expression gives *[F(x), (m)] (y)[ (x), (m)] = 0*, which implies

[F(x), (m)]R[ (x), (m)] = {0} for all x, m R, (3.6)

as is an automorphism on R. Next, for all x R, let x S (R) R, then equation (3.6) yields [F(x), (m)]R[ (x), (m)] = 0 = [F(x), (m)]R([ (x), (m)]) . Hence, the −primeness of R implies that either [F(x), (m)] = 0 or [x, m] = 0. For any x R, we have x−x S (R) R. Thus, for some x R, [F(x−x ), (m)] = 0 or [x − x , m] = 0. If [F(x − x ), (m)] = 0, then for all x, m R we have [F(x), (m)]R[ (x), (m)] = ([F(x), (m)]) R[ (x), (m)], which means that either [F(x), (m)] = 0 or [x ,m] = 0. If [x − x ,m] = 0, then equation (3.6) yields [F(x), (m)]R[ (x), (m)] = [F(x), (m)]R([ (x), (m)]) which means that either [F(x), (m)] = 0 or [x ,m] = 0. Now, we let A = {x R | [x, m] = 0} and B = {x R | [F(x), (m)] = 0}. Then A and B are both additive subgroups of R whose union is R. Using Brauer’s trick we have either A = R or B = R. If A = R, then we have [x, m] = 0 for all x, m R which means that R is commutative. If B = R then [F(x), (m)] = 0 for all x, m R. Since is an automorphism on R, we get F(x) Z(R) for all x R. Thus, our hypothesis in (3.4), yields [x, y] Z(R). A similar argument as that after (3.1) assures that R is commutative. Therefore, for each case, we have R is commutative as required.

**Theorem**

Let R be a −prime ring and , be automorphisms on R. Suppose that R admits a generalized -derivation (F, d) satisfying any of the following conditions for all x, y R:

[F(x), x] = [y, x] ,

(F(x) x) = (y x) ,

(F(x) x) = [y, x] .

If F = 0 (or d 0), then R is commutative.

**Proof**

If F = 0, then [y, x] = 0 for all x, y R. Thus, using similar arguments as that after equation (3.3) gives that R is commutative. Henceforth, we shall assume that d 0. So, for all x, y R, we have

[F(x), x] = [y, x] . (3.7)

Replacing x by x + m in (3.7) gives

[F(x),m] + [F(m), x] = 0, for all x, m R. (3.8)

Now, we again replace m by mx in (3.8) to get

([m, x])d(x) + (m)[d(x), x] + (m)[F(x), x] = 0. (3.9)

Also, replacing m by w m in (3.9) gives ([w, x]) (m)d(x) = 0. Since is an automorphism on R, we get

([w, x])Rd(x) = {0} for all w, x R. (3.10)

If x S (R) R, then ([w, x])Rd(x) = ([w, x]) Rd(x). Thus, for some x S (R) R, the −primeness of R yields either ([w, x]) = 0 or d(x) = 0. But for any x R, x − x S (R) R. Thus, for some x R, either [w, x − x ] = 0 or d(x − x ) = 0. If [w, x − x ] = 0, then equation (3.10) follows that ([w, x])Rd(x) = 0 = ([w, x]) Rd(x). Hence the −primeness of R yields either ([w, x]) = 0 or d(x) = 0. If d(x−x ) = 0 then d(x) = (d(x)) for all x R. Consequently, for all x, w R, either ([w, x]) = 0 or d(x) = 0. Let A = {x R | d(x) = 0} and B = {x R | [w, x] = 0}. Then A and B are both additive subgroups of R whose union is R. Using Brauer’s trick we have either A = R or B = R. If A = R then d(x) = 0 for all x R, a contradiction. If B = R, then [w, x] = 0 for all x, w R and hence R is commutative.

**(ii)** If F = 0 then for all x, y R, we have

(x y) = 0. (3.11)

Replacing y by y m in (3.11), we get (y)[x, m] = 0 for all x, m R. Again, for any z R, replace y by y z in the last expression, to get y R ([x, m] ) = 0 = y R ([x, m] ). Applying the -primeness of R, we get ([x, m] ) = 0 for all x, m R. Since is an automorphism on R, we obtain [x ,m] = 0 for all x, m R. Now, using similar arguments as that follow equation (3.3), we get the required result. Therefore, we shall assume that d≠0 . Thus, for any x, y R we have

(F(x) x) = (y x) . (3.12)

Replacing x by x + m for any m R in (3.12) gives

(F(x) m) + (F(m) x) = 0. (3.13)

Again we replace m by mx in (3.13) to get

− (m)[F(x), x] + (m)(d(x) x) − [ (m), (x)]d(x) = 0. (3.14)

Now, replacing m by w m in (3.14) gives ([w, x]) (w)d(x) = 0 for all x, w R. Since is automorphism on R, we get

*([w, x])Rd(x) = {0}. (3.15)*

This is the same as equation (3.10), hence continuing in the same manner as above gives that R is commutative.

If *F = 0* then, we have *[x, y] = 0* for all *x, y R*. Hence we use similar arguments as that after (3.3) to get the required result. Therefore, we shall assume that *d 0*. So, for any x, y R, we have

*(F(x) x) = [y, x] . (3.16)*

For any *m* in *R*, replacing *x* by *x + m* in (3.16) gives

*(F(x) m) + (F(m) x) = 0. (3.17)*

Applying the same arguments that follow equation (3.13) yields the required result.

The following corollaries are immediate consequences of theorems 3.1 and 3.2 or by using the same techniques taking in our account theorem 2.1 for corollary 3.4:

**Corollary **

Let R be a 2-torsion free -prime ring and , be automorphisms on R. Suppose that R admits a generalized -derivation (F, d) such that

F([x, y]) − [F(x), y] = [d(y), x]

for all x, y R. If F = 0 (or d 0), then R is commutative.

**Corollary **

Let R be a 2-torsion free -prime ring and , be automorphisms on R. Suppose that R admits a generalized -derivation (F, d) such that

F([x, y]) − (F(x) y) = [d(y), x]

for all x, y R. If F = 0 (or d 0), then R is commutative.

**Corollary **

Let R be a -prime ring and , be automorphisms on R. Suppose that R admits generalized -derivations (F, d) and (G, g) such that

*F([x, y]) = [ (y),G(x)],*

*for all x, y R. If F = 0 (respectively G = 0) or d 0 (respectively g 0), then R is commutative.*

**Corollary **

*Let R be a -prime ring and , be automorphisms on R. Suppose that R admits generalized -derivations (F, d) and (G, g) such that *

*F(x y) = ( (y) G(x)),*

* for all x, y R. If F = 0 (respectively G = 0) or d 0 (respectively g 0), then R is commutative.*

**Acknowledgement**

This paper was financially supported by The Deanship of Scientific Research, Northern Border University, under the project no. 435-062-7. The authors would like to thank The Deanship of Scientific Research for their financial support and Professor Nadeem ur Rehman for many useful comments.

**References**

- Ashraf, M. and Rehman, N., On derivations and commutativity in prime rings, East West J. Math., 3(1)(2001), 87-91.
- Ashraf, M., Ali, A. and Ali, S., Some commutativity theorem for rings with generalized derivations, Southeast Asian Bull. Math., 32(2) (2007), 415-421.
- Bell, H. E. and Rehman, N., Generalized derivations with commutativity and anti-commutativity conditions, Math. J. Okayama Univ., 49(2007), 139-147.
- Marubayashi, H., Ashraf, M., Rehman, N. and Ali, S., On generalized -derivations in prime rings, Algebra Colloquim, 17(1) (2010), 865- 874.
- Oukhtite, L. and Salhi, S., On Commutativity of -prime rings, Glasnik Mathematicki, 41(1) (2006), 57-64.
- Oukhtite, L. and Salhi, S., On generalized derivations of -prime rings, African Diaspora J. Math., 5(1) (2006), 19-23.
- Rehman, N., Al-Omary, R. M. and Haetinger, C., On Lie structure of prime rings with generalized -derivations, Bol. Soc. Paran. Mat.,27(2) (2009), 43-52.
- Rehman, N., Al-Omary, R. M., On Commutativity of 2−torsion free −prime Rings with Generalized Derivations, Mathematica, 53 (76) (2011), 171-180.
- Rehman, N., Al-Omary, R. M. and Shuliang Huang, Lie ideals and generalized -derivations of −prime Rings, Africa Mathematica, Doi: 10.1007/s13370-012-0075-9 (2012).
- Rehman, N., Al-Omary, R. M. and Al-Shomrani M. M., Morita context and generalized -derivations, Bol. Soc. Paran. Mat., 31(1) (2013), 153-166.

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Hillary gave you dolts a clue. Remember when she pointed out how empty that suit was?Did anyone else bring that fact up?I believe they did. You voted for him anyway, and have precipitated the revival of good old conservatism. Congrats.Progressives sure are smart, ain't they?

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Hello. And Bye.

In front of me was the marital bed with sharon stripped naked

laid on their bed. I m sure you do, I thought.

Have Meadow send out a release on Twitter.

We walked out into the empty parking lot.

We decided to go and have a look in the downstairs playrooms.

While that paragraph is not going to win a writing award.

Jay Doe – November 30, 2016 22385 Views. I swung the car around as if

to run into them and they fucked off.

But i on the other hand don t operate or subscribe to that personally.

If he agreed to let me watch.

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