IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 (,)- Strongly Derivations Pairs on Rings I. A. Saed University of Technology Received in : 30 November 2010 Accepted in : 27 February 2011 Abstract Let R be an associative ring. In this paper we present the definition of (,)- Strongly derivation pair and Jordan (,)- strongly derivation pair on a ring R, and study the relation between them. Also, we study prime rings, semiprime rings, and rings that have commutator left nonzero divisior with (,)- strongly derivation pair, to obtain a (,)- derivation. Where ,: RR are two mappings of R. Keywords Prime ring, semiprime ring, (,)-derivation, (,)-Strongly derivation pair, Jordan (,)- Strongly derivation pair. §1 Basic Concepts Deinition 1.1: [1] A nonempty set R is said to be associative ring if in R there are defined two operations, denoted by + and . respectively, such that for all a, b, c in R: 1- a + b is in R 2- a + b = b + a 3- (a+b) + c = a + (b+c) 4- There is an element 0 in R such that a+0 = a (for every a in R) 5- There exists an element –a in R such that a + (-a)=0. 6- a . b is in R. 7- a. (b.c) = (a.b).c 8- a. (b+c) = a.b + a.c and (b+c).a = b.a + c.a Deinition 1.2: [1] A ring R is called prime ring if for any a, bR, a R b = {0}, implies that either a=0 or b=0. Definition 1.3:[1] A ring R is called semiprime ring if for any aR, aRa = {0}, implies that a=0. Remark 1.4:[1] Every prime ring is semiprime ring, but the converse in general is not true. The following example justifies this remark. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Example 1.5: [1] R = Z6 is a semiprime ring but is not prime. Let a  R such that aRa = {0}, implies that a 2 = 0, hence a=0, therefore R is a semiprime ring. But R is not prime, since 20 and 30 implies that 2R3 = {0}. Definition 1.6:[2] A ring R is said to be n-torsion free, where n≠0 is an integer if whenever n a=0, with aR, then a=0. Definition 1.7:[2] Let R be a ring. A Lie product[,] on R is defined as[x,y]=xy – yx, for all x,y  R. Definition 1.8:[2] Let R be a ring. An additive mapping d:RR is called a derivation if d(xy)= d(x)y + xd(y), for all x,y R and we say that d is a Jordan derivation if d(x 2)=d(x)x+xd(x), for all xR. Definition 1.9:[3] Let R be a ring. An additive mapping d:RR is called a (,)-derivation, where ,: RR are two mappings of R, if d(xy)= d(x)σ(y) + (x) d(y), for all x,y  R, and we say that d is a Jordan (σ,)-derivation if d(x 2 )=d(x) σ(x)+ (x) d(x), for all x  R. Definition 1.10:[4] Let R be a ring, additive mappings d,g : RR is called S-derivation pair (d,g) if satisfies the following equations: d(xy) = d(x)y + xg(y), for all x,y  R. g(xy)= g(x)y + xd(y), for all x,y  R. And is called Jordan S-derivation pair if: d(x 2)=d(x)x + xg(x), for all x  R. g(x2)=g(x)x + xd(x), for all x  R. Example 1.11:[4] Let R be a non commutative ring and let a,b  R, such that xa=xb=0, for all x  R. Define d,g : RR, as follows: d(x) = ax,g(x) = bx Then (d,g) is a S-derivation pair of R. Remark 1.12:[4] Every S-derivation pair is a Jordan S-derivation pair, but the converse is in general not true. The following example illustrates this remark. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Example 1.13:[4] Let R be a 2-torsion free non commutative ring, and let a  R, such that xax = 0, for all x  R, but xay ≠ 0, for some (x ≠ y )  R. An additive pair d,g : R R is defined as d(x) = xa +ax, g(x) = [x,a] Then (d,g) is Jordan S-derivation pair, but not a S-derivation pair. Definition 1.14:[5] A ring R is said to be a commutator right (resp. left) nonzero divisior, if there exists elements a and b of R, such that c[a,b] = 0 (resp. [a,b]c = 0) implies c=0, for every c  R. §2 (,)-S-Derivation pairs In this section, we will introduce the definition of (,)-Strongly derivation pair, and we denoted by (,)-S-derivation pair, and Jordan (,)-Strongly derivation pair and we denoted by Jordan (,)-S-derivation pair, also we will give the relation between them. Where , : R  R are two mappings on R. Now, in this section we introduce the principle definition. Definition 2.1 Let R be a ring, additive mappings d,g : RR is called (,)-S-derivation pair (d,g) where , : RR are two mappings of R, if satisfy the following equations: d(xy)=d(x)σ(y) + (x)g(y), for all x,y  R. g(xy)=g(x)σ(y) + (x)d(y), for all x,y  R. And is called Jordan (,)-S-derivation pair if: d(x 2)=d(x)σ(x) + (x)g(x), for all x  R. g(x2)=g(x)σ(x) + (x)d(x), for all x  R. The following example explains the principle definition: Example 2.2 Let R be a non commutative ring and let a,b  R, such that (x) a= (x)b = 0, for all x  R. Define d,g: R  R as follows: d(x) = a (x), g(x) = b (x), for all x  R where , : R  R are two endomorphism mappings. Then (d,g) is a (,)-S-derivation pair of R. Let x,y  R, so: d(xy) = aσ(xy) = aσ(x)σ(y) = aσ(x)σ(y) + (x) bσ(y) =d(x)σ(y) + (x)g(y) IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Also: g(xy) = bσ(xy) = bσ(x)σ(y) = bσ(x)σ(y) + (x) aσ(y) = g(x)σ(y) + (x)d(y) Hence (d,g) is a (,)-S-derivation pair. Remark 2.3 Every (,)-S-derivation pair is a Jordan (,)-S-derivation pair, but the Converse is in general not true. The following example illustrates this: Example 2.4 Let R be a 2-torsion free non commutative ring, and let a  R, such that (x) a (x) = 0, for all x  R, but(x) a (y) ≠ 0, for some (x ≠ y )  R. Define an additive pair d,g: R  R, as follows: d(x) = (x) a + a (x), g(x) = (x)a- a(x), for all x  R. where ,: R  R are two endomorphism mappings. Then (d,g) is a Jordan (,)-S-derivation pair, but not a (,)-S-derivation pair. Let x,y  R, so: d(x 2)=(x2) a + aσ(x2) d(x)σ(x) + (x)g(x) = ((x)a + aσ(x)) σ(x) + (x)((x)a – aσ(x)) =(x)aσ(x) + aσ(x)σ(x) + (x)(x)a - (x)aσ(x) =(x2)a + aσ(x2) Hence d(x2) =d(x)σ(x) + (x) g(x) Also: g(x 2) = (x2)a – aσ(x2) = g(x)σ(x) + (x)d(x) Thus, (d,g) is Jordan (,)-S-derivation pair. Now, we show that (d,g) is not (,)-S-derivation pair. d(xy) = (xy)a + aσ(xy) d(x)σ(y) + (x)g(y) = x)a + aσ(x)) σ(y) + (x)((y)a – a σ(y)) =(x)aσ(y) + aσ(x)σ(y) + (x)(y)a - (x)aσ(y) =(xy)a + aσ(xy) Hence d(xy)=d(x) σ(y) + (x)g(y) But: g(xy)= g(x)σ(y) + (x)d(y) =((x)a – aσ(x))σ(y)+ (x)((y)a + aσ(y)) =(x)aσ(y) – aσ(x)σ(y) + (x)(y)a + (x)aσ(y) =(xy)a – aσ(xy) +2x)aσ(y) On the other hand: g(xy)=(xy)a - aσ(xy) Since (x)aσ(y)≠0, for some x≠y R, the two expressions are not equal, hence we get (d,g) is not (,)-S-derivation pair. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Proposition 2.5 Let R be a semiprime ring. Suppose that , are automorphisms of R. If R admits a (,)-S- derivation pair (d,g), such that d(x) g(y)=0 (resp. g(x) d(y) =0), for all x,y  R, then d=0 (resp. g=0). Proof We have d(x)g(y)=0, for all x,y  R_____(1) Replacing yx for y in (1) and using (1), we have: d(x)g(yx)=0, for all x,y  R. d(x)(g(y)σ(x) + (y)d(x))=0, for all x,y  R. d(x)g(y)σ(x) + d(x)(y)d(x)=0, for all x,y R. d(x)(y)d(x)=0, for all x,y  R._____(2) By semiprimeness of R, (2) gives: d(x)=0, for all xR. If we have g(x)d(y)=0, for all x,y  R_____(3) Replacing yx for y in (3) and using (3), we have: g(x)d(yx)=0, for all x,y R. g(x)(d(y)σ(x) + (y)g(x))=0, for all x,y  R. g(x)d(y)σ(x) + g(x) (y)g(x)=0, for all x,y  R. g(x)(y)g(x)=0, for all x,y  R_____(4) By semiprimeness of R, (4) gives: g(x)=0, for all x  R. Proposition 2.6 Let R be a semiprime ring. Suppose that , are automorphisms of R. If R admits a (,)-S- derivation pair (d,g), such that d(x)= ± (x) (resp. g(x)=± (x)), for all x  R, then g=0 (resp. d=0). Proof We have d(x)=σ(x), for all x  R_____(1) Replacing x by xy in (1) and using (1), we get: d(xy) = (xy), for all x,y  R. d(x)σ(y) + (x)g(y)=σ(xy), for all x,y  R. σ(x)σ(y)+(x)g(y)= σ(x)σ(y), for all x,y  R. (x)g(y)=0, for all x,y  R ____(2) Left multiplication of (2) by g(y), leads to: g(y)(x)g(y)=0, for all x,y  R _____(3) By semiprimeness of R, (3) gives: g(y)=0, for all y  R. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Similarly, we can show if d(x)=-(x), for all x  R, then g=0 In the same way, if g(x)=± (x), for all xR, then d=0. Proposition 2.7 Let R be any ring and , are two mappings on R. Then 1- If (d,g) is a (,)-S-derivation pair on R, then d+g is a (,)-derivation. 2- If (d,g) is a Jordan (,)-S-derivation pair on R, then d+g is a Jordan (,)-derivation. Proof 1- We have (d,g) is a (,)-S-derivation pair, so d(xy)= d(x) (y)x)g(y), for all x,y  R _____(1) g(xy)= g(x) (y)x)d(y), for all x,y  R _____(2) By adding (1) and (2), we get (d+g)(xy)=(d+g)(x)(y)+(x)(d+g)(y) Hence d+g is a (,)-derivation 2- We have (d,g) is a Jordan (,)-S-derivation pair, so d(x 2)=d(x)(x) + (x)g(x), for all x  R _____(3) g(x2)=g(x)(x) + (x)d(x), for all x  R _____(4) By adding (3) and (2), we get (d+g)(x 2 )=(d+g)(x)(x) + (x)(d+g)(x), for all x  R. Hence d+g is a Jordan (,)-derivation. §3 Relation Between ( ,)-S-Derivation pairs and ( ,)-Derivations In this section, we study prime rings, semiprime rings, and rings that have a commutator left nonzero divisor with (,)-S-derivation pair, to obtain a (,)-derivation. Theorem 3.1 Let R be a 2-torsion free semiprime ring, and (d,g) be a (,)-S-derivation pair on R, then d and g are (,)-derivations. Where , are automorphisms of R. Proof Suppose that (d,g) is (,)-S-derivation pair. Then: d(xyx)=d(x(yx)) = d(x)yx) + (x)g(yx), for all x,y  R _____(1) That is: d(xyx)=d(x)(yx) + (x)g(y)(x) + (x)(y)d(x), for all x,y  R____(2) Also: IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 d(xyx)=d((xy)x)=d(xy)(x) + (xy)g(x), for all x,y  R_____(3) That is: d(xyx)=d(x)(y)(x) + (x)g(y)(x) + (xy)g(x), for all x,y  R___(4) From (2) and (4), we get: (xy)(d(x)-g(x))=0, for all x,y  R_____(5) Replace (y) by (d(x)-g(x)) (y) (x) in (5), we get: (x)(d(x)-g(x))(y)(x)(d(x)-g(x))=0, for all x,y  R_____(6) Since R is semiprime, we get: (x)d(x)=(x)g(x), for all x  R_____(7) It follows that: d(x 2 )=d(x)(x)+(x)d(x), for all x  R_____(8) And: g(x2)=g(x)(x)+(x)g(x), for all x  R_____(9) Thus, by using [3, Theorem 2.3.7], we obtain that d and g are (,)-derivations on R. Theorem 3.2 Let R be a prime, and (d,g) be a (,)-S-derivation pair on R, then d and g are (,)- derivations. Where , are automorphisms of R. Proof Since (d,g) is (,)-S-derivation pair, we have (see how relation (5) was obtained from relation (1) in the proof of Theorem 3.1) (xy)(d(x)-g(x)=0, for all x,y  R_____(1) And, by primeness of R, we get: d(x)=g(x), for all x  R_____(2) And hence d and g are (,)-derivations on R. Theorem 3.3 Let R be a ring which has a commutator left nonzero divisor and (d,g) be a (,)-S-derivation pair on R, then d and g are (,)-derivations.Where , are automorphisms of R. Proof 1. That is We have: 2. d(yx 2)=d(y)(x2) + (y)g(x2), for all x,y  R____(1) 3. That is: 4. d(yx 2 )=d(y)(x 2 )+ (y)g(x)(x)+(y)(x)d(x), for all x,y  R__(2) 5. On the other hand: 6. d(yx 2)=d(yx)(x)+ (yx)g(x), for all x,y  R_____(3) 7. 8. d(yx 2 )=d(y)(x 2 )+ (y)g(x)(x) + (y)(x)g(x), for all x,y  R_____(4) 9. From (2) and (4), we obtain: IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011  (y)((x)d(x)-(x)g(x))=0, for all x,y  R_____(5) Replacing y by yr in (5), to get: (yr)((x)d(x)-(x)g(x))=0, for all x,y,r  R_____(6) Again, left multiplying of (5) by (r), to get: (r)(y)((x)d(x)-(x)g(x))=0, for all x,y,r  R_____(7) Subtracting (7) from (6), we get: [(y),(r)]((x)d(x)-(x)g(x))=0, for all x,y,r  R_____(8) Since R has a commutator left nonzero divisor, we get: (x)d(x)=(x)g(x), for all x  R_____(9) Linearizing (9), we get: (x)d(y) + (y)d(x)= (x)g(y) + (y)g(x), for all x,y  R_____(10) That is: (x)(d-g)(y) + (y)(d-g)(x)=0, for all x,y  R_____(11) Replacing y by ry in (11), to get: (x)(d-g)(ry) + (ry)(d-g)(x)=0, for all x,y,r  R_____(12) Again, left multiplying of (11) by (r), to get: (r)(x)(d-g)(y) + (r)(y)(d-g)(x)=0, for all x,y,r R _____(13) Subtracting (12) from (13), we get: (rx) (d-g)(y) - (x)(d-g)(ry)=0, for all x,y,r  R _____(14) Replacing x by sx in (14), to get: (rsx)(d-g)(y)-(sx)(d-g)(ry)=0, for all x,y,r,s  R _____(15) Also, left multiplying of (14) by (s), to get: (srx)(d-g)(y)-(sx)(d-g)(ry)=0, for all x,y,r,s  R_____(16) Subtracting (16) from (15), we get: [(r),(s)] (x)(d-g)(y)=0, for all x,y,r,s R _____(17) Since R has a commutator left nonzero divisor, we get: (x)(d-g)(y)=0, for all x,y R_____(18) That is: (x)d(y)=(x)g(y), for all x,y  R_____(19) Hence d and g are (σ,)-derivations. References 1. Herstien, I.N., (1969), TOPICS IN RING THEORY, The University of Chicago Press, Chicago. 2. Ashraf, M., Ali, S. and Haetinger, C., (2006), " On Derivations in Rings and their Applications", The Aligarh Bull of Math., 25(2), 79-107. 3. Hamdi, A.D., (2007), "(σ,)-Derivations on prime Rings", MSc. Thesis, Baghdad University . 4. Yass, S., (2010), "Strongly Derivation Pairs on Prime and Semiprime Rings", MSc. Thesis, Baghdad University. 5. Cortes, W. and Haetinger, C., (2005),"On Jordan Generalized Higher Derivations in Rings", Turkish J. of M ath., 29(1),1-10. 2011) 3( 24المجلد مجلة ابن الهیثم للعلوم الصرفة والتطبیقیة على الحلقات) ,(-األشتقاقات المزدوجة القویة سعید اكرام احمد الجامعة التكنولوجیة 2010تشرین الثاني 30: استلم البحث في 2011شباط 27: قبل البحث في الخالصة - واشتقاق جوردان المزدوج القوي) ,(-في هذا البحث قدمنا تعریف األشتقاق المزدوج القوي. حلقة تجمیعیة Rلتكن ), (ة في الحلقRكذلك، ندرس الحلقات األولیة، الحلقات شبه األولیة، والحلقات التي لها مبدل . ، ودراسة العالقة بینهم هما ,: R Rأي ان ). ,(-للحصول على األشتقاق ) ,(-قاسم غیر صفري أیسر مع األشتقاق المزدوج القوي .Rدالتین على الحلقة :الكلمات المفتاحیة .),(-، اشتقاق جوردان المزدوج القوي ),(-، األشتقاق المزدوج القوي ),(-حلقة اولیة، حلقة شبھ اولیة، مشتقة