813 Internat. J. Math. & Math. Sci. VOL. 20 NO. 4 (1997) 813-816 RESEARCH NOTES NOTES ON (,fl)-DERIVATIONS NEET AYDIN Adnan Menderes University Faculty of Arts and Sciences Department of Mathematics 0910 Aydin, TURKEY (Received December 18, 1995 and in revised form April 2, 1996) ABSTRACT. Let R be a prime ring of characteristic not 2, U a nonzero ideal of R and 0 # d a (a, #)derivation of R where c and/ are automorphisms of R. i) [d(U), a] 0 then a E Z ii) For a, b E R, the following conditions are equivalent (I) t(a)d(x)=d(z)(b), for all z U (II) Either a(a) #(b) CR(d(U)) or Ca(a) Ca(b) R’ and a[a,x] [a,x]b (or a[b,x] [b,x]b) for all :r U Let R be a 2-torsion free semiprime ring and U be a nonzero ideal of R iii) Let d be a (a,/)derivation of R and g be a (7, 6)-derivation of R. Suppose that dg is a (aT, 6)-derivation and g commutes both 7 and 6 then g(x)Uo-ld(y) 0, for all x, y U. iv) Let Ann(U) 0 and d be an (a, g)-derivation of R and g be a (’7, 6)-derivation of R such that g commutes both q, and 6 If for all x, y e U, -l(d(x))Ug(y) 0 g(x)Ua (d(y)) then dg is a (aT, 6)-derivation on R - KEY WORDS AND PHRASES: Derivation, semiprime ring, prime ring, commutative 1991 AMS SUBJECT CLASSIFICATION CODES: 16A15, 16A70 INTRODUCTION Let R be a ring and X be a subset of R. Let Ann,.(X)= {a R lxa 0 all x X} and Anne(X) {a R lax 0 all x X} be the right and left annihilators, respectively, of the subset of R If R is a semiprime ring then the lett and right and two-sided annihilators of an ideal X coincide It will be denoted by Ann(X). Let U be an ideal of R Note that if a is an automorphism of R and Ann(U) 0 then Ann(a(U)) 0. Let R be a ring and c, be two automorphisms of R An additive mapping d R R is called an (c, )-derivation if d(xy) a(x)d(y) + d(x)(y) holds for all pairs x, 1. Throughout this note R will represent an associative ring Let R’= {z Rid(x)= 0} The centralizer of a subset A of R is Ca(A) {y Rlau Ua, Va e A} Ca(R) Z, the center of R There are two motivations for this research Herstein [1 has proved Let R be a prime ring of characteristic not 2, and 0 :/: d be a derivation of R Then any element a R satisfying ad(x) d(z)a for all x E R, should be central In [2], Daif has proved the following theorem Let R be a prime ring and a, b E R Then the following conditions are equivalent (i) ad(x) d(z)b, V e R (ii) Either a b Ca(d(R)) or Ca(a) CR(b) R’ and a[a,x] [a,x]b (or a[b,x] [b,x]b) for all x E R In the first part of this note we generalized these two theorems for an ideal U and (a, )-derivation of R N AYDIN 814 In the second part, Bresar and Vukman [3] give some results concerning two derivations in semiprime rings We will generalize some of these results by taking an ideal of R instead of R and extend to more general mappings As a result of this, we will give a generalization of a well-known result of Posner which states that if R is a prime ring of characteristic not 2 and d, g are nonzero derivation of R then dg cannot be a derivation - RESULTS LEMMA 1. Let R be a prime ring of characteristics not 2, (0) U an ideal of R, 0 d R R a (a,/)-derivation such that ad da, d/ =/d and a E R. Ifa E Ca(d(U)) then a Z PROOF. Since a Ca(d(U)), ad(x) d(x)a for all x U Replacing x by xy, y E U, we 2. obtain aa(z)d(u) Taking yr, r + aa(z)/(U) a(x)a(U)a + a(z)/(u)a. _ d(x)[a, /(y)] Using hypothesis we have [c(x), a]d(y). R, instead of y, we obtain d(x)Z(y)[a, fl(r)] [a(x), a]a(y)d(r) for all x,y e U,r R. If we replace r by fl-l(d(z)),z U, we get d(x)fl(y)[a,d(z)] [a(x),a]a(y)/-(d2(z)). Since a_Ca(d(U)) we have [a(x),a]a(y)-l(d2(z)) 0 for all x,y,z e U Since a(U) is an ideal of R and / is prime we get a_Z or d2(U) 0. Ifd2(U) 0 then O=d2(xy) a2(x)d2(y)+2d(a(x))d(fl(y)) and so d(a(x))d((y)) 0. By [4, Lemma 3] we have a contradiction Thus a Z. THEOREM 1. Let R be a prime ring of characteristic not 2, 0 d R R a (c,/)-derivation, (0) U and ideal of R and a, b E R. Then the following conditions are equivalent (I) a(a)d(x) d(x)/(b), for all x e U. (II) Either/(b)=c(a)Ca(d(U)) or Ca(a)=Ca(b)= R’ and a[a,x] [a,x]b (or a[b, c] [b,x]b) for all x U. we get a(a)e Z. (I) gives If a e CR(d(U)) then by Lemma PROOF. (I) (II) that /(b) a(a). Similarly, if d(x)((b) a(a)) 0, for all x E U. By [4, Lemma 3] it implies (b) Ca(d(U)) then (b) a(a). We assume henceforth that neither a(a) nor/(b) in Ca(d(U)). Let in (I) x be rx, where r R, and using (I), we have a(a)a(r)d(x) + a(a)d(r)l(x) a(r)d(x)/(b) + d(r)(x)/(b) and so - = __ c([a, r])d(x) d(r)(xb) c(a)d(r)/(x). (2.1) Taking y instead of r where y e U, in (2.1) and using (I) we obtain a([a, y])d(x) d(y)([x, b]), for all x, y U. (2.2) Now ifd(x) =0 then (2.2) gives us d(y)([x,b]) 0 for all y U By [4, Lemma 3], we get xCR(b). Conversely, if x CA(b), then (2.2) gives us a([y, a])d(x) 0. Since by [4, Lemma 3] a Z, we have d(x) =0 Therefore CA(b) R’. Similarly, we can show that CA(a) R’. In particular, d(a) =d(b) =0 and ab ha. Replace r by yb, y U, in (2.1) we have a([a,y])a(b)d(x) d(y)(b)(xb) a(a)d(y)(bx) a(a)d(y)(bx) a(a)d(y)fl(xb) a(a)d(y)/(bx) c(a)d(y)/([x, b]) and using (2.2) we get a([a, y])a(b)d(x) a(a)a([a, y])d(x) and so a([a, y]b a[a, y])d(x) 0 for all x, y By [4, Lemma 3 we obtain a[a, y] [a, y]b for all y E U. e U. NOTES ON (a,/)-DERIVATIONS 815 Furthermore, replacing z by az in (2.2) and using (2 2) and hypothesis we also have a[b, z] [b, z]b (II) (I) If a(a) =/3(b) Ca(d(U)) it is obviously a(a)d(z) d(z)/(b) for all z U Therefore it suffices to show that if Ca(a)= Ca(b)= R’ and a[a,z] [a,z]b for all z U then a(a)d(x) d(x)(b) for all z U. Since d(a) d(b) O, ab ba, [a, ax xb] a[a,x] [a,x]b 0 It gives ax xb R’ and so 0 d(ax xb) a(a)d(x) d(x)(b). This proves the theorem For the second part we begin with LEMMA 2 [3, Lemma 1]. Let R be a 2-torsion free semipfime ring and a, b the elements of R Then the following conditions are equivalent" for all x R (i) axb 0 for all x R (ii) bxa 0 (iii) axb+bxa=0 for all xR If one of these conditions is fulfilled then ab ba 0 too. LEMMA :3. Let R be a semiprime ring and U a nonzero ideal of R such that Ann(U)= 0 Let d be an (a,/)-derivation of R and g be a (-),,6)-derivation of R. If d(U)Ug(U)= 0 then d(R)Ug(R) =0. PROOF. For all x, y, z U, d(x)yg(z) 0 Replace x by xs, s R we have 0 d(xs)yg(z) a(x)d(s)yg(z) + d(x)5(s)yg(z) Since/3(s)y U, the last equation implies that c(x)d(s)yg(z) O, for all x, y, z U and 8 R Taking tz instead of z, where R, we have 0 c(x)d(s)y3,(t)g(z) + a(z)d(s)yg(t)6(z) Since y3"(t) U, it gives c(x)d(s)yg(t)6(z) 0 for all x, y, z U and s, R Therefore d(8)yg(t)6(z) Ann(a(U))= 0. Thus we get d(s)yg(t)6(z)=0 for all y,z U and R Hence d(s)yg(t) Ann(6(U)) 0. As a result of this, it implies that d(R)Ug(R) 0 s, LEMMA 4. Let R be a semiprime ring and U be a nonzero ideal of R such that Ann(U) O. Let a, b R be such that aUb 0 then aRb O. PROOF. For all x U 0 axb. Replace x by tbxrat, where t, r r we have atbxratbx 0 R. Thus atb Ann(U) 0 we get Since R is semiprime ring, this implies that atbU 0 for all aRb 0 TItEOREM 2. Let R be a 2-torsion free semiprime ring and U be a nonzero ideal of R with O. Let d be a (o,/3)-derivation of R and g be a (3’, 6)-derivation of R. Suppose that dg is a (c3,,/36)-derivation and g commutes both 3, and 6. Then g(x)Ua-ld(y) 0, for all x, y U. PROOF. Since g commutes both 3’ and 6, from the first par to the proof of [5, Lemma 1] there is no loss of generality in assuming/3 1 and 6 1 For all x, y U, dg(xy! d(3,(x)g(y) + g(x)y) a7(x)dg(y) +d(3,(x))g(y)+a(g(x))d(y)+dg(x)y. On the other hand, since dg is an (a3,,1)derivation we have dg(xy) a/(x)dg(y)+ dg(x)y. Comparing the two expressions so obtained for dg(xy), we see that (2 3) d(3,(x))g(y) + a(g(x))d(y) 0 for all x,y U. Ann(U) R in (2.3) we obtain O=d(3,(x))g(yz)+a(g(x))d(yz)=d(3,(x))3,(y)g(z)+ +a(g(x) )d(y)z {d (3,(x))g(y)+a(g(x))d(y)} z +d(3,(x) )3,(y)g(z) + +a(g(x) )a(y)d(z) )g(y)z d(3,(x) a(g(x))a(y)d(z). This relation reduces to (2.4) d(3,(x))3,(y)g(z) + a(g(x))a(y)d(z) 0 for all x,y U,z R. Replacing y by yz where z Replace U by yg(t), U and take z U we have d(3,(x))3,(y)3,(g(t))g(z)+a(g(z))a(y)a(g(t))d(z)=O. a(g(z))a(y)d (-r(t))g(z) Considering this relation (2.4) and (2.3) we obtain d(7(x))’r(y)Tfg(t))g(z) Comparing the last two relations we get a(g(x))a(y)a(g(t))d(z) for all x,y,zU. 2a(g(x))a(y)a(g(t))d(z) 0. Since R is 2-torsion free, it gives N AYDIN 816 (=)e(,)o-d(=) 0 for al =, , =, u. by tu, u E U it follows O=g(z)V()g(u)a-(d(z)+g(z)Vg()ua-(d(z)) Since V() U this relation reduces to g(z)Ug()ua-(d(z)) 0 for all z,,u,z U By Lemma 4 we have for all z, u, z U, g(z)Rg()ua-(d(z)) 0. In particular g(z)ua-X(d(z))Rg(z)ua-(d(z))=0 for all z, u,z U. Since R is semipfime we obtn g(z)Ua-(d(z)) 0 for 1 z,z U COROLLARY. Let R be a prime ring of characteristic not 2, d be an (a, )-devation of Rmd g be a (% g)-defivation of R such that g commutes both and 6 If the composition dg is a (a?, derivation then d 0 or 9 0. EOM 3. Let R be a 2-torsion flee semipfime ring d U be a nonzero ide of R such that Ann(U) 0. Let d be a (a, )-derivation of R and g be a (% 6)-defivation of R such that g coutes both 7 and 6. If for all z,V U, fl-(d(z))Ug()= 0 g(z)uo-l(d()) then d9 is a (aT, B6)derivation on R PROOF. From Lena 3 d Lena 4, we get -l(d(x))yg(z)= 0 g(x)o-(d(z)) for z,y,z R On the other hd, since fl-(d(x))Vg(z)=0 for 1 x,V,z R d since automorpsm of R we obtn d(7(x))(V)(g(z)) 0 for 1 x, z R. Since R is a sepfime ring, by Lemma 2 we get d(7(x))(g(z)) 0 for 1 x,z R. Similly from g(x)Ua-d(y) O, we get a(g(x))d(6(V)) 0 Therefore dg is m (a, 6)-defivation on R Replacing , , FENCES RS, I.N., A note on derivations, Canad. Math. Bull. 22 (1979), 509-511. [2] D, M.N, A prime ring satisng a condition for derivations, Bull. Cal. Math. S. 78 (1986), 303-304. [3] BS M. d , J., Ohogon derivation d eension of a theorem of Posner, Rovi MatematicM 5 (1989), 237-246 [4] A, N. d Y K., Some generition in prime ring th (a, r)-defivation, Doga-Tr. J. ofMath. 16 (1992), 169-176. [5] BS M, On the compositions of (a,)-defivations of tings, d applications to yon Neumn gebras, Acta Sct. Math. 56 (1992), 369-375. [6] POS E.C., Derivations in prime tings, Pr. Amer. Math. 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