It was known that every left (?,?) -derivation is a Jordan left (?,?) – derivation on ?-prime rings but the converse need not be true. In this paper we give conditions to the converse to be true

saed, Ikram A.

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Baghdad Science Journal

Baghdad Science Journal  Vol.8(3)2011  628 Jordan left  -derivations Of -prime rings  Ikram A.saed*  Received  15, March, 2010 Accepted 27 , September, 2010  Abstract: It was known that every left  -derivation is a Jordan left  – derivation on -prime rings but the converse need not be true. In this paper we give conditions to the converse to be true.  Key words: - prime rings , - square closed lie idea, left  - derivation , Jordan left  -derivations .  Introduction:  In [1] Ashraf  proved that every Jordan left  -  derivation on prime ring is a left   -  derivation on prime ring . In[2] Oukhtite and Salhi  proved that every Jordan left  derivation on -prime ring is a left derivation on -prime ring. In this paper we prove that every Jordan left  -  derivation on -prime ring is a left  -  derivation on  -prime ring.  Definition 1.1 : [2] A ring R is said to be 2-torsion-free if whenever 2x= 0 with  x R, then x = 0 . Definition 1.2 : [3] Let R be a ring . Define a lie product  [. , . ] on as follows  [ x , y ] = xy – yx , for all x,y R . Properties 1.3: [3] Let R be a ring . Then for all x,y R , we have  1. [x,yz] = y[x,z] + [x,y] z . 2. [xy,z] = x[y,z] + [x,z] y . Definition 1.4 : [4] A ring R is called a prime if for any  a,b R,  aRb = {o} implies that either a = o or  b = o Definition 1.5 : [2] A ring R with involution  is said to be - prime if aRb = aR  (b) = {o} implies that a = o or b = o  Definition 1.6 : [5] A ring R with involution  , we define   (R) = {r R /  (r) = r} .  Definition 1.7 : [3] A Lie ideal of a ring R is an additive subgroup  U of ring R  satisfying [U,R] U .  Definition 1.8 : [4] A Lie ideal U of a ring R is said to be   - lie ideal , if  (U) = U Definition 1.9 : [2] If U is a  - Lie ideal of a ring R such that  *University of Technology  Baghdad Science Journal  Vol.8(3)2011  628  U for all u U , then U is called a  – square closed Lie ideal . Definition 1.10 : [1] Let R be a ring . An additive mapping d:R→R is called a left  - derivation  where R → R is a mapping of R , if  d(xy) =  (x) d (y) +  (y) d (x) , for all x,y R and  we say that d is a Jordan left  - derivation  If d(x2) =  (x) d(x) +  (x) d(x), for all x  R.             = 2 (x) d(x), for all x  R. It is clear that every left ( - derivation of R is a Jordan left ( - derivation , but the converse is not true as the following example, shows: Example 1.11:- Let R be a commutative ring and let  a R  Such that  x R. but  ≠ 0, for some x and y  R, such that x ≠ y. Define a map d:R→ R as follows  d(x) = (x) a, for all x R  Where :R→ R is an endomorphism mapping. Then d is a Jordan left ( - derivation but not a left ( - derivation. It is clear that d is an additive mapping. Now, we have to show that d is satisfies d(x2) =  d(x) +  d(x) =  2  d(x), for all x  R d(x2) = (x2)a         = a  = 0, for all  x  R 2  d(x) =  a                   =2 a  = 0, for all x  R  d(x2) =  , for all  x  R  d is a Jordan left - derivation of R. We must prove that d is not a left - derivation of R. d(xy) = a          = a  for all  x  R. but   +   =  a +  a = a  + a =  2 a  for all x  R. Since a  ≠ 0, for some x and y  R.  d(xy) ≠   +  , for some x and y  R.  d is not a left - derivation of R.  Lemma 1.12: [2] If U  Z (R) is a  - Lie ideal of a  2- torsion – free - Prime ring R and a,b  R such that  a U b =  (a) U b = {0}, then a=0 or b=0.  Baghdad Science Journal  Vol.8(3)2011  626 Lemma 1.13: Let R be a 2- torsion-free    - prime ring and U be a  - square closed Lie ideal of R. suppose that  is an endomorphism of R. If d:R→ R is an additive mapping satisfying d(u2) = 2 (u) d(u), for all u, v  U then (i) d(uv+vu) = 2 (u) d(v) + 2 (v) d(u), for all u, v  U. (ii) d(uvu) = (u2) d(v) + 3 (u) (v) d(u) - (v) (u) d(u), for all u, v  U (iii)d(uvw + wvu) = { (u) (w) + (w) (u)} d(v) + 3 (u) (v) d(w) + 3 (w) (v) d(u) - (v) (u) d(w) - (v) (w) d(u),     for all u, v, w  U (iv) [ (u), (v)] (u) d(u) = (u) [ (u), (v)] d(u), for all u, v  U. (v) [ (u), (v)] d([u,v]) = 0, for all u, v  U. (vi) d(vu2) = (u2) d(v) + (3 (v) (u) - (u) (v) ) d(u) - (u) d([u,v]), for all u, v  U. Proof: (i) Since uv + vu = (u + v)2 – u2 – v2, we find that uv + vu  U . for all  u,v U  Hence by linearizing  d(u2) = 2 (u) d(u) on u, we get d(uv + vu) = 2 (u) d(v) + 2 (v) d(u), for all u, v  U. 1ـــــــــــــ (ii) Replacing v by uv + vu in 1, we get  d(u(uv + vu) + (uv + vu) u) =    + 2 (v) (u) d(u)            2ـــــــــــــــــــ On ht other hand,  d(u(uv + vu) + (uv + vu) u) = d(u2v + vu2) + 2d(uvu)                                        =  2 (u2) d(v) +4 (v) (u) d(u) + 2d(uvu). Combining the above equation with 2, we get  d(uvu) = (u2) d(v) + 3 (u) (v) d(u) - (v) (u) d(u),   for all u, v  U. (iii)By linearizing (ii) on u, we get d((u + w) v(u + w)) = (u2) d(v) + (w2) d(v) + { (u) (w) + (w) (u)}  d(v) + 3 (u) (v) d(w) + 3 (u) (v) d(u) + 3 (w) (v) d(w) + 3 (w) (v)  d(u) - (v) (u) d(u) - (v) (u) d(w) - (v) (w) d(u) - (v) (w) d(w).                          3ــــــــــــــــــــon the other hand,  d[(u + w) v(u + w)] = d(uvu) + d(wvw) + d(uvw + wvu)  = (u2) d(v) + 3 (u) (v) d(u) - (v) (u) d(u) + (w2) d(v) + 3 (w) (v) d(w) - (v) (w) d(w) +d(uvw + wvu).                             4ـــــــــــــــــــــــ         Combining 3 and 4, we get d(uvw + wvu) = { (u) (w) + (w) (u) } d(v) +  3 (u) (v) d(w) + 3 (w) (v) d(u) - (v) (u) d(w) - (v) (w) d(u), for all u, v  U.                                                          ـــــــــــــــــــــــ 5 (iv) Since uv + vu and uv – vu both belong to U  we find that 2uv  U for all u, v  U. Hence, by our hypothesis we find that  d((2uv)2) = 2 (2uv) d((2uv)) 4 d((uv)2) = 8 (uv) d(uv). Since char R≠ 2, we have  d((uv)2) = 2 (u) (v) d(uv). Replace w by 2uv in 5, and use the fact that char R≠ 2, to get Baghdad Science Journal  Vol.8(3)2011  628 d(uv(uv) + (uv)vu) = { (u2) (v) + (u) (v) (u)} d(v) +3 (u) (v) d(uv) + 3 (u) (v2) d(u) - (v) (u) d(uv) - (v) (u) (v) d(u). 6ـــــــــــــ On the other hand, d((uv)2 + uv2u) = 2 (u) (v) d(uv) + 2 (u2) (v) d(v) + 3 (u) (v2) d(u) - (v2) (u) d(u).          7ـــــــــــــ     Combining 6 and 7, we get [ (u), (v)] d(uv) = (u) [ (u), (v)] d(v)+ (v) [ (u), (v)] d(u)                             8ـــــــــــــ        Replacing u + v for v in 8, we have 2[ (u), (v)] (u) d(u) + [ (u), (v)] d(uv) = 2 (u) [ (u), (v)] d(u) + (u) [ (u), (v)] d(v) +  (v) [ (u), (v)] d(u). Now application of 8 yields (iv) (v) linearize (iv) on u, to get [ (u), (v)] (u) d(u) + [ (u), (v)] (v) d(v) +  [ (u), (v)] (u) d(v) + [ (u), (v)] (v) d(u) = (u) [ (u), (v)] d(u) + (u) [ (u), (v)] d(v) + (v) [ (u), (v)] d(u) + (v) [ (u), (v)] d(v), for all u, v  U. Now application of 8 and (iv) yields that  [ (u), (v)] (u) d(v) + [ (u), (v)] (v) d(u) =  [ (u), (v)] d(uv) and hence [ (u), (v)] {d(uv) - (u) d(v) - (v) d(u) } = 0 for all  u, v  U. 9ـــــــــــ Combining 1 and 9 we find that, [ (u), (v)] {d(vu) - (u) d(v) - (v) d(u)} = 0 for all  u, v  U. 10ــــــــــــ Further, combining of 9 and 10 yields the required result. (vi) replace v by 2vu in 1, and use the fact that char R ≠ 2, to get d(uvu + vu2) = 2 ( (u) d(uv) + (v) (u) d(u)) for all u, v  U. 11ــــــــــــ Again replacing v by 2uv in 1, we get  d(u2v + uvu) = 2( (u) d(uv) + (u) (v) d(u)) for all u, v  U. 12ــــــــــ Now, combining 11 and 12, we get d(u2v - vu2) = 2( (u) d([u,v]) + [ (u), (v)] d(u)), for all u, v  U. 13ــــــ Replacing u by u2 in 1, we have d(u2v + vu2) = 2( (u2) d(v) + 2 (v) (u) d(u)), for all u, v  U.  14ـــــــــــHence, subtracting 13 from 14 and using the fact that char R ≠ 2, we find that  d(vu2) = (u2) d(v) + {3 (v) (u) - (u) (v) } d(u) - (u) d([u, v]), for all u, v  U.    2 Jordan left - derivations on    - square closed Lie ideals: Theorem 2.1: Let R be a 2-torsion-free    - prime ring and let U be a   - square closed Lie ideal of R. Suppose that  is an automorphism of R. If d: R→ R is an additive mapping satisfying  d(u2) = 2 (u) d(u), for all u,  U, then  d(uv) = (u) d(v) + (v) d(u), for all u, v  U. Proof Suppose [U,U] =0 and let u, v  U. From d((u +v)2) = 2 (u + v) d(u + v), it follows that  2d(uv) = 2 (u) d(u) + 2 (v) d(v) – d(u2) – d(v2) + 2 (u) d(v) + 2 (v) d(u),  Baghdad Science Journal  Vol.8(3)2011  638 In such a way that  2d(uv) = 2( (u) d(v) + 2 (v) d(u)), for all u, v  U. As char R ≠ 2, then  d(uv) = (u) d(v) + (v) d(u). Hence we shall assume that [U,U] ≠ 0 According to Lemma 1.13 (iv) we have { (u2) (v) - 2 (u) (v) (u) + (v) (u2)} d(u) =01ـــــــــــــــــ For all u,v   U Replacing [u,w] for u in 1, where  w  U, we get [ (u), (w)]2 (v) d([u,w]) - 2[ (u), (w)] (v) [ (u), (w)] d([u,w])  + (v) [ (u), (w)]2 d([u,w]) = 0, for all u, v, w  U. Now, application of Lemma 1.13 (v) yields that  -1 ([ (u), (w)]2) U -1 (d([u,w])) = {0} which implies that [u,w]2 U -1 (d([u,w])) = {0}, for all u, w  U. let x,y   (R) ∩ U, we have [x,y]2 U -1 (d([x,y])) = {0} =     ([x,y]2) U -1 (d([x,y])) and by  Lemma 1.12 either [x,y]2 = 0 or  -1 (d([x,y])) = 0 If -1 (d([x,y])) = 0, then d([x,y] = 0, applying Lemma 1.13 (i) together with char R≠ 2, we find that d(xy) = (x) d(y) + (y) d(x). Now suppose that [x,y]2 = 0 From Lemma 1.13 (iv) it follows that { (u2) (v) - 2 (u) (v) (u) + (v) (u2)} d(v) = 0, for all u, v  U. Linearizing this relation in u, we obtain (u) (w) (v) + (w) (u) (v) - 2 (u) (v) (w) - 2 (w) (v) (u) +  (v) (u) (w) + (v) (w) (u)) d(v) = 0, for all u, v, w  U. Replacing v by [x,y] and using Lemma 1.13 (v), we conclude that (-2 (u) ([x,y]) (w) - 2 (w) ([x,y]) (w) - 2 (w) ([x,y]) (u) + ([x,y]) (w) (u)) d([x,y]) = 0  2ـــــــــــــــــــWrite u[x,y] instead of u in 2, since [x,y]2 = 0,  Lemma 1.13 (v) leads us to  ([x,y]) (u) ([x,y]) (w) d([x,y]) = 0, for all u, w  U. ([x,y]) (u) ([x,y]) U d([x,y]) = {0}, for all u  U. -1 [ ([x,y]) (u) ([x,y])] U  -1 (d([x,y]) = {0}, for all u  U. [x,y] u [x,y] U -1 (d([x,y]) = {0}, for all u  U. As [x,y]  U ∩   (R), the fact that  (U) = U yields [x,y] u [x,y] U -1 (d([x,y]) = {0} =   ([x,y] u [x,y]) U -1 (d([x,y]) and using Lemma 1.12, either  -1 d([x,y]) = 0 or [x,y] u [x,y] = 0, for all u  U. If -1 (d([x,y]) = 0 then d([x,y] = 0 by Lemma 1.13 (i) together with  char R≠2,  we find that d(xy) = (x) d(y) + (y) d(x) If [x,y] u [x,y] = 0, for all u  U, then  [x,y] U [x,y] = {0} =  ( [x,y] )U [x,y]. Once again using Lemma 1.12, we get [x,y] = 0 and  Lemma 1.13 (i) forces d(xy) = (x) d(y) + (y) d(x). Consequently, in both the cases we find that  d(xy) = (x) d(y) + (y) d(x), for all x, y   U ∩   (R) 3ــــــــــــــــــــــ Now, let u,v  U, if we set Baghdad Science Journal  Vol.8(3)2011  638 u1 = u + (u)   ,    u2 = u -   (u) v1 = v +     (v)   ,    v2 = v -   (v) then we have 2u = u1 + u2 and 2v = v1 + v2. Since u1, u2, v1, v2  U ∩   (R), application of 3 yields d(2u2v) = d((u1 + u2)( v1 + v2))               = d(u1 v1 + u1 v2 +u2 v1+u2 v2)               = (u1) d(v1) + (v1) d(u1) + (u1) d(v2) + (v2) d(u1) + (u2) d(v1) + (v1) d(u2)+ (u2) d(v2)+ (v2) d(u2)               = 2 (u) d(2v) + 2 (v) d(2u) As char R≠2, it then follows d(uv) = (u) d(v) + (v) d(u), for all u, v  U.  References:  1. Ashraf, M. 2005. On left ( , ) – derivations of prime rings, Archivum mathematicum (Brno). Toms 41:157-166. 2. Oukhtite, L. and Salhi. S. 2008. Centralizing automorphisms and Jordan left derivations of   - Prime rings, ISSN. 1(1): 19-26. 3. Ashraf, M.and Rehman.N. 2000. On Lie ideals and Jordan left derivation of prime rings, Archivum mathematicum (Brno).36:201-206. 4. Oukhtite, L., Salhi. S. and Tooufiq. L. 2007. Jordan Generalized Derivations on    - prime Rings,International Journal of Algebra. 1(5):231-234.  5. Herestein, I.N. 1976. Rings with involution. The University of Chicago press,ed. No. 2, Chicago, pp. 34.    اولية -على الحلقات  ىمشتقات يسر –   جوردان  *اكرام احمد سعيد .الجامعة التكنولوجية*  :الخالصةاولية   -مشتقة يسرى على الحلقات   مشتقة يسرى هي جوردان  – من المعروف ان كل .ون االتجاه المعاكس صحيح في هذا البحث قدمنا الشروط الكافية ليك. لكن العكس غير صحيح   

Jordan left (?,?) -derivations Of ?-prime rings

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Jordan left (?,?) -derivations Of ?-prime rings

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