IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 (,)- Strongly Derivations Pairs o n Rings I. A. Saed Unive rsity of Technology Received in : 30 November 2010 Accepte d in : 27 February 2011 Abstract Let R be an associative ring. In this p aper we p resent t he definition of (,)- Strongly derivation p air and Jordan (,)- st rongly derivation p air on a ring R, and st udy the relation between them. Also, we study p rime rings, semip rime rings, and rings that have commutator left nonzero divisior with (,)- st rongly derivation p air, to obtain a (,)- derivation. Wher e ,: RR are two mapp ings of R. Keywords Prime ring, semip rime ring, (,)-d erivation, (,)- Strongly derivation p air, Jordan (,)- Strongly derivation p air. §1 Basic Concepts Deinition 1.1: [1] A nonemp ty set R is said to be associative ring if in R there are defined two op erations, denoted by + and . resp ectively, 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 exist s 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 rin g if for any a, bR, a R b = {0}, implies that either a=0 or b=0. De finition 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 p rime ring is semiprime rin g, but the converse in general is not true. The following examp le justifies this remark. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 Exam ple 1.5: [1] R = Z 6 is a semiprime ring but is not p rime. Let a  R such that aRa = {0}, i mplies that a 2 = 0, hence a=0, therefore R is a semip rime ring. But R is not p rime, since 20 and 30 imp lies that 2R3 = {0}. De finition 1.6:[2] A ring R is said to be n-torsion free, where n≠0 is an inte ger if when ever n a=0, with aR, then a=0. De finition 1.7:[2] Let R be a ring. A Lie product[,] on R is defined as[x,y ]=xy – y x, for all x,y  R. De finition 1.8:[2] Let R be a ring. An additive map p ing d:RR is called a deriv ation 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. De finition 1.9:[3] Let R be a ring. An additive map p ing d:RR is called a (,)-deriv ation, 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. De finition 1.10:[4] Let R be a ring, additive map p ings d,g : RR is called S-deriv ation p air (d,g) if satisfies t he followin g 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( x 2 )= g( x) x + xd( x), for all x  R. Exam ple 1.11:[4] Let R be a non commutative rin g and let a,b  R, such that xa= xb=0, for all x  R. Define d, g : RR, as follows: d(x) = a x, g( x) = b x Then (d,g) is a S-deriv ation p air of R. Remark 1.12:[4] Every S-derivation p air is a Jordan S-deriv ation p air, but the converse is in general not t rue. The following examp le illust rates t his remark. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 Exam ple 1.13:[4] Let R be a 2-torsion free non commutative rin g, and let a  R, such that xax = 0, for a ll 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-d erivation p air, but not a S-derivation p air. De finition 1.14:[5] A ring R is sa id to be a co mmutator ri ght (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 d efinition of (,)- Strongly derivation p air, and we denoted by (,)-S-d erivation p air, and Jordan (,)- Strongly derivation p air and we denoted by Jordan (,)-S-derivation pair, also we will give the relation between them. Where , : R  R are two mapp ings on R. Now, in this section we introduce the p rinciple definition. De finition 2.1 Let R be a ring, additive mapp ings 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-d erivation p air if: d(x 2 )=d( x)σ(x) + ( x) g( x), for all x  R. g( x 2 )= g( x)σ(x) + ( x)d( x), for all x  R. The following examp le exp lains the principle definition: Exam ple 2.2 Let R be a non commutative rin g 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 endomorp hism 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) b σ(y) =d(x)σ(y) + (x) g(y ) IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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 p air is a Jordan (,)- S-derivation p air, but the Converse is in general not t rue. The following examp le illust rates t his: Exam ple 2.4 Let R be a 2-torsion free non commutative rin g, 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 add itive 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 endomorp hism mappings. Then (d,g) is a Jordan (,)- S-derivation pair, but not a (,)- S-deriv ation p air. Let x,y  R, so: d(x 2 )=( x 2 ) a + aσ(x 2 ) 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) =(x 2 )a + aσ(x 2 ) Hence d( x 2 ) =d(x) σ(x) + ( x) g( x) Also: g( x 2 ) = ( x 2 )a – aσ( x 2 ) = 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 t he other hand: g( xy )=(xy )a - aσ(xy ) Since ( x) aσ(y )≠0, for some x≠y R, the two exp ressions are not equ al, hence we get (d, g) is not (,)- S-derivation p air. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 Proposi tion 2.5 Let R be a semip rime ring. Sup p ose that , are automorp hisms of R. If R admits a (,)- S- derivation p air (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) Replacin g y x for y in (1) and using (1), we h ave: d(x) g(y x)=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 a ll x,y  R._____(2) By semip rimeness 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) Replacin g y x for y in (3) and using (3), we h ave: g( x)d(y x)=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 a ll x,y  R_____(4) By semip rimeness of R, (4) gives: g( x)=0, for all x  R. Proposi tion 2.6 Let R be a semip rime ring. Sup p ose that , are automorp hisms of R. If R admits a (,)- S- derivation p air (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) Replacin g 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 semip rimeness of R, (3) gives: g(y )=0, for all y  R. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 Similar ly, 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. Proposi tion 2.7 Let R be any ring and , are two mapp ings on R. T hen 1- If (d,g) is a (,)-S-d erivation p air on R, then d+g is a (,)-derivation. 2- If (d,g) is a Jordan (,)-S-d erivation p air on R, then d+g is a Jordan (,)-derivation. Proof 1- We have (d,g) is a (,)-S-d erivation p air, 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 (,)-d erivation 2- We have (d,g) is a Jordan (,)-S-d erivation p air, so d(x 2 )=d( x)( x) + ( x) g( x), for all x  R _____(3) g( x 2 )= 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 (,)-d erivation. §3 Relation Be tween (,)-S-Derivation pairs and (,)-Derivations In this section, we st udy p rime rings, semip rime rings, and rings that have a commutator left nonzero divisor with (,)-S-d erivation p air, to obtain a (,)-derivation. The orem 3.1 Let R be a 2-torsion free semip rime ring, and (d, g) be a (,)-S-d erivation p air on R, then d and g are (,)-d erivations. Where , are automorp hisms of R. Proof Sup p ose that (d,g) is (,)- S-deriv ation p air. Then: d(xy x)=d( x(y x)) = d( x)y x) + ( x) g(y x), for all x,y  R _____(1) That is: d(xy x)=d( x)(y x) + ( x) g(y )( x) + ( x)(y )d(x), for all x,y  R____(2) Also: IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.24 (3) 2011 d(xy x)=d(( xy )x)=d( xy )( x) + ( xy )g( x), for all x,y  R_____(3) That is: d(xy x)=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 a ll 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 semip rime, 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( x 2 )= 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 (,)-deriv ations on R. The orem 3.2 Let R be a p rime, and (d,g) be a (,)- S-derivation p air on R, then d and g are (,)- derivations. Where , are automorp hisms of R. Proof Since (d, g) is (,)- S-deriv ation p air, we have (see how relation (5) was obtained from relation (1) in the p roof of Theorem 3.1) (xy )(d(x)- g( x)=0, for a ll x,y  R_____(1) And, by p rimeness of R, we get: d(x)= g( x), for all x  R_____(2) And hence d and g are (,)-derivations on R. The orem 3.3 Let R be a rin g which has a commutator left nonzero divisor and (d,g) b e a (,)- S-derivation p air on R, then d and g are (,)-deriv ations.Where , are automorp hisms of R. Proof 1. That is We have: 2. d(y x 2 )=d(y )(x 2 ) + (y )g( x 2 ), for all x,y  R____(1) 3. That is: 4. d(y x 2 )=d(y )(x 2 )+ (y )g( x)( x)+(y )( x)d( x), for all x,y  R__(2) 5. On t he other hand: 6. d(y x 2 )=d(y x)( x)+ (y x) g( x), for all x,y  R_____(3) 7. 8. d(y x 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. S CI. VOL.24 (3) 2011  (y )((x)d( x)-( x) g( x))=0, for all x,y  R_____(5) Replacin g y by y r in (5), to get: (y r)((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) fro m (6), we get: [(y ),(r)]((x)d( x)-( x) g( x))=0, for all x,y ,r  R_____(8) Since R has a co mmutator left nonz ero divisor, we get: (x)d( x)=( x) g( x), for all x  R_____(9) Linear izing (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 a ll x,y  R_____(11) Replacin g y by ry in (11), to get: (x)(d- g)(ry ) + (ry )(d-g)(x)=0, for a ll 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) fro m (13), we get: (rx) (d- g)(y ) - (x)(d- g)(ry )=0, for all x,y ,r  R _____(14) Replacin g 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 multip ly ing of (14) by (s), t o get: (srx)(d- g)(y )-(sx)(d- g)(ry )=0, for all x,y ,r,s  R_____(16) Subtracting (16) fro m (15), we get: [(r),(s)] (x)(d- g)(y )=0, for all x,y ,r,s R _____(17) Since R has a co mmutator left nonz ero 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,)-d erivations. Re ferences 1. Herst ien, I.N., (1969), T OPICS IN RING THEORY, The University of Chicago Press, Chicago. 2. Ashraf, M ., Ali, S. and Haetinger, C., (2006), " On Derivations in Rings and their App lications", The Aligarh Bull of M ath., 25(2), 79-107. 3. Hamdi, A.D., (2007), "(σ,)-Derivations on p rime Rings", M Sc. Thesis, Baghd ad University . 4. Yass, S., (2010), "Strongly Derivation Pairs on Prime and Semiprime Rings", M Sc. Thesis, Baghdad University . 5. Cortes, W. and Ha etinger, C., (2005),"On Jordan Generalized Hi gh er Derivations in Rings", T urkish 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 ة الجامعة التكنولوجی 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 قوي ),(-حلقة اولیة، حلقة شبھ اولیة، مشتقة .),(-، اشتقاق جوردان المزدوج القوي ),(-، األشتقاق المزدوج ال