International Journal of Pure and Applied Mathematics Volume 109 No. 6 2016, 41-47 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu doi: 10.12732/ijpam.v109i6.6 PAijpam.eu GENERALIZED SEMIDERIVATIONS IN σ-prime RINGS D. Bharathi 1, V. Ganesh 2 1,2 Department of Mathematics S.V. University Tirupati, A.P., 517502, INDIA Abstract: In this paper, we prove that some results for generalized semiderivations in σ- prime rings. If R is a σ-prime ring with char 2, 0 I is a σ-ideal of R and f is a nonzero generalized semi derivation associated with a semi derivation d of R which commutes with σ, then either f[x,y] = [x,y] or f(xoy) = xoy, x,y I then R is a commutative ring. Next we prove the commutativity of R admitting generalized semiderivations f 1 and f 2 satisfy any one of the properties: (i) f 1(x)y +f 1(y)x = xf 2(y)+yf 2(x), (ii) [f 1(x),y] = [x,f 2(y)], (iii) f(x 2 ) = x 2, (iv) d(x)f(y)±xy = 0, (v) f[x,y] = [f(x),y]+[d(y),x] x,y I. Then R is commutative. AMS Subject Classification: 16W25 Key Words: commutator, σ-prime ring, σ-ideal, generalized semiderivation 1. Introduction In [1], M.N. Daif and H.E. Bell established that a prime ring R must be commutative if it admits a derivation d such that d([x,y]) = [x,y] for all x,y in a nonzero ideal of R. Recently, M.A. Quadri [2] proved that the Daif and Bellś result obtained by considering a generalized derivation instead of the derivation in a prime ring, is still true. Received: October 1, 2016 Published: November 16, 2016 c 2016 Academic Publications, Ltd. url: www.acadpubl.eu
42 D. Bharathi, V. Ganesh Prime ring. A ring R is called prime if arb = 0 implies either a = 0orb = 0, a,b R. σ-prime ring. A ring R equipped with involution σ is said to be σ-prime ring if for any a,b R,aRb = arσ(b) = 0 a = 0 or b = 0. σ-ideal. An ideal L of R is said to be σ-ideal if σ(l) = L. Semiderivation. An additive mapping d : R R is called semiderivation if there exists a function g : R R such that: d(xy) = d(x)g(y) +xd(y) = d(x)y +g(x)d(y), d(g(x)) = g(d(x)), x,y R. (1.1) Generalized semiderivation. An additive mapping f : R R is called a generalized semiderivation if there is a semiderivation d : R R associated with the function g : R R such that f(xy) = f(x)y +g(x)d(y) = d(x)g(y)+xf(y), Basic commutative identities. f(g(x)) = g(f(x)), x,y R. (1.2) (i) [x,yz] = y[x,z]+[x,y]z, x,y,z R, (1.3) (ii) [xy,z] = [x,z]y +x[y,z], x,y,z R. (1.4) 2. Results Lemma 2.1. Let R be a σ-prime ring and let I be a nonzero σ-ideal of R. If a,b in R satisfy aib = aiσ(b) = 0, then a = 0 or b = 0. Proof. Suppose a 0, there exists some x I such that ax 0. Otherwise a Rx = 0 and a Rσ(x) = 0, x I and therefore a = 0. Since airb = 0 and airσ(b) = 0, we then obtain a x Rb = a x Rσ(b) = 0. In view of the σ-primeness of R this yields b = 0. Lemma 2.2. Let R be σ-prime ring with char 2, let f be an additive mapping on R, let d be a semiderivation which commutes with σ and let I be a nonzero σ-ideal of R. If [f(y),z]id(y) = 0, y,z I, then either d = 0 or [f(y),z] = 0 y,z I.
GENERALIZED SEMIDERIVATIONS IN... 43 Lemma 2.3. LetI beanonzeroσ ideal ofrand0 dbeasemiderivation with surjective function g on R which commutes with σ. If [x,r]id(x) = 0, x I, then R is commutative. Theorem 2.4. Let R be a σ-prime ring with char 2 and let I be a with semiderivation d and surjective function g, commutes with σ such that f[x,y] = [x,y], x,y I, then R is commutative. Proof. By hypothesis f[x,y] = [x,y], x,y I.By (1.2) f(x)y +g(x)d(y) f(y)x g(y)d(x) [x,y] = 0 x,y I. (2.1) Replacing y by yz in (2.1), yields f(x)yz +g(x)d(y)z +g(x)g(y)d(z) f(y)zx+g(y)d(z)x g(yz)d(x) [x, yz] = 0. Using (2.1) and g is on-to map, we have f(y)[x,z]+y[d(x),z]+[x,y]d(z)+y[x,d(z)] y[x,z] = 0, x,y,z I. (2.2) Replacing z by zx in (2.2), we have [x,yz]d(x) = 0, x,y,z I. (2.3) Writing z by zt in (2.3), we get [x,y]ztd(x) = 0, by equation (1.3): Then [x,y]zid(x) = 0, x,y,z I. (2.4) Let x I Sa σ (R), since d commutes with σ, the relation (2.4) yields [x,y]zid(x) = 0 = [x,y]ziσ(d(x)), x,y,z I. Using Lemma 2.1, either d(x) = 0 or [x,y]z = 0. If [x,y]z = 0, y,z I, then [x,y]i = 0. Again using Lemma 2.1 [x,y] = 0. Therefore x I Sa σ (R) we have d(x) = 0 or [x,y] = 0, y I. Letm I, thefactthatm σ(m) I Sa σ (R)assuresthatd(m σ(m)) = 0 or [m σ(m),y] = 0, y I. If d(m σ(m)) = 0, then d(m) Sa σ (R) and in view of (2.4) this yield d(m) = 0 or [m,y] = 0, for any y I. Now, supposethat [m σ(m),y] = 0, y I. Since m+σ(m) Sa σ (R), then d(m+σ(m)) = 0 or [m+σ(m),y] = 0, y I. If [m + σ(m),y] = 0, then 2[m,y] = 0. Since R is a char 2, we have [m,y] = 0, y I. If d(m + σ(m)) = 0, then d(m) Sa σ (R). Again by equation (2.4) d(m) = 0 or [m,y] = 0, y I. In conclusion, m I, we have either d(m) = 0 or [m,n] = 0, n I. This means that I is the
44 D. Bharathi, V. Ganesh union of two its additive subgroups A = {m I/[m,n] = 0, n I} and B = {m I/d(m) = 0}. Since a group cannot be the union of two proper subgroups, then I = B or I = A. The fact that d 0 forces I = A and thus [m,n] = 0 m,n I. A similar reasoning as in ([6], proof of Theorem 1.1) assures that R is commutative. Theorem 2.5. Let R be a σ-prime ring with char 2 and let I be a with semiderivation d and surjective function g, commutes with σ such that f(xoy) = xoy, x,y I, then R is commutative. Proof. By hypothesis f(xoy) = xoy, x,y I, which can be written as f(xy+yx) = xoy.by (1.2) f(x)y +g(x)d(y)+f(y)x+g(y)d(x) xoy = 0 x,y I (2.5) Replace y by yx in (2.5), where x I, we get f(x)yx+g(x)d(y)x+g(x)g(y)d(x)+f(y)x 2 +g(y)d(x)x+g(yx)d(x) (xoyx) = 0 x,y I Since g is on to and using equation (2.5), we get (xoy)zd(x) = 0, x,y I Replace y by yz in the above equation, we conclude that [x,y]zd(x) = 0, x,y,z I, by equation xo(yz) = (xoy)z y[x,z] = y(xoz)+ [x, y]z and therefore [x,y]id(x) = 0, xy I. Now using Lemma 2.2 for the particular case f = 1, we get [x,y] = 0 x,y I, because d 0. By ([6], proof of theorem 1.1) this yields that R is commutative. Theorem 2.6. Let R be a σ-prime ring with char 2 and let I be a nonzero σ-ideal of R. If R admits a generalized semiderivations f 1 and f 2 with associated semiderivations d 1 and d 2 surjective functions g 1 and g 2 which commute with σ such that if R satisfies one of the properties (i) (f 1 (x)y +f 1 (y)x)±(xf 2 (y)+yf 2 (x)) = 0 and (ii) [f 1 (x),y]±[x,f 2 (y)] = 0, x,y I, then R is commutative. Proof. (i)byhypothesis(f 1 (x)y+f 1 (y)x)±(xf 2 (y)+yf 2 (x)) = 0, x,y I. f 1 (x)y +f 1 (y)x = xf 2 (y)+yf 2 (x), x,y I (2.6) replace x by xy in (2.6), we get f 1 (x)y 2 +g 1 (x)d 1 (y)+f 1 (y)xy = xyf 2 (y)+yf 2 (x)y +yg 2 (x)d 2 (y) (by eq 1.2) (2.7)
GENERALIZED SEMIDERIVATIONS IN... 45 Multiplying (2.6) with y from the right, we get f 1 (x)y 2 +f 1 (y)xy = xf 2 (y)y +yf 2 (x)y, x,y I (2.8) Combining equations (2.7) and (2.8), we obtain g 1 (x)d 1 (y) = yg 2 (x)d 2 (y)+x[y,f 2 (y)], x,y I (2.9) Replace x by rx in (2.9), where r R, we have g 1 (rx)d 1 (y) = yg 2 (rx)d 2 (y)+rx[y,f 2 (y)], x,y I and r R (2.10) Left multiplying (2.9) by r, it gives rg 1 (x)d 1 (y) = ryg 2 (x)d 2 (y)+rx[y,f 2 (y)], x,y I (2.11) Combining (2.10) and (2.11) and since g 1 and g 2 are on to, yields that [y,r]xd 2 (y) = 0, x,y I and r R Therefore [y,r]id 2 (y) = 0, x I and r R (2.12) Since I is a σ- ideal and d 2 σ = σd 2, for any y I Sa σ (R) By Lemma 2.1 we have either [y,r] = 0 or d 2 (y) = 0. Using the fact that y +σ(y) Sa σ (R) I, y I, then [y +σ(y),r] = 0 or d 2 (y +σ(y)) = 0, y I and r R. Now, two cases arise, Case 1: If [y +σ(y),r] = 0 and y σ(y) Sa σ (R) I, yields [y σ(y),r] = 0 or d 2 (y σ(y)) = 0, r R. If [y σ(y),r] = 0 then 0 = [y σ(y),r]+[y +σ(y),r] = 2[y,r] Since R is of Char 2, which implies [y,r] = 0 If d 2 (y σ(y)) = 0, r R, then d 2 (y) = d 2 (σ(y)) = σ(d 2 (y)) An application of Lemma 2.1 equation (2.12) implies [y,r] = 0 or d 2 (y) = 0. Case 2: If d 2 (y +σ(y)) = 0, then d 2 (y) = d 2 (σ(y)) = σ(d 2 (y)), and view of (2.12) [y,r]id 2 (y) = 0 = [y,r]iσ(d 2 (y))applyinglemma2.1, wehave[y,r] = 0ord 2 (y) = 0. If d 2 (y) = 0, then for any r in R, replace y by yr, we have d 2 (yr) = 0 implies d 2 (y)r +g 2 (y)d 2 (r) = 0 which gives (by equation 1.1) g 2 (y)d 2 (r) = 0, y in I. Hence Id 2 (r) = IRd 2 (r) = σ(i)rd 2 (r) = 0. Since I 0 and R is a σ-prime, we obtain d 2 (R) = 0, (i.e., d 2 = 0) yields a contradiction. Next, suppose that [y,r] = 0, then for any t in r, we have 0 = [ty,r] = [t,r]y = [t,r]i = [t,r]ri = [t,r]rσ(i) = 0. Since I 0 and R is a σ-prime, we obtain [t,r] = 0, r,t R. Hence R is commutative. (ii) Similarly we can prove that R is commutative, if R satisfies [f 1 (x),y]±[x,f 2 (y)] = 0. Corollary 2.7. Let R be a σ-prime ring with char 2 and let I be a
46 D. Bharathi, V. Ganesh with semiderivation d and surjective function g, commute with σ such that [f(x),y]+[f(y),x] = 0, x,y I or if f(x)oy+f(y)ox = 0, x,y I, then R is commutative. Theorem 2.8. Let R be a σ-prime ring with char 2 and let I be a with semiderivation d and surjective function g, commute with σ such that if R satisfies one of the properties. (i) f(x 2 )±x 2 = 0 (ii) d(x)f(y)±xy = 0, x,y I, then R is commutative. Proof. (i) By the hypothesis f(x 2 ) = x 2, x I. Replacing x by x+y in the above equation, we get f((x+y)(x+y)) = (x+y)(x+y) f(x 2 +xy +yx+y 2 ) = x 2 +xy +yx+y 2 Using the hypothesis, we get f(xoy) = xoy, x,y I. Now using the Theorem 2.5 we get the required result. If f(x 2 ) + x 2 = 0, x I, then as (i) we get f(xoy) + (xoy) = 0, x,y I. Following the same technique as used in Theorem 2.5, we get the required result. (ii) Similarly we can prove that R is commutative, if R satisfies d(x)f(y)±xy = 0. Theorem 2.9. Let R be a σ-prime ring with char 2 and let I be a with semiderivation d and surjective function g, commutes with σ such that if R satisfies f([x,y]) = [f(x),y]+[d(y),x], x,y I, then R is commutative. Proof. By hypothesis f([x,y]) = [f(x),y]+[d(y),x], x,y I (2.13) Replace y by yx in (2.13), we obtain f([x,yx]) = [f(x),yx]+[d(yx),x] f([x,y]x) = [f(x),y]x + y[f(x),x] + [d(y)x + g(y)d(x),x], x,y I (by eqs. 1.3 and 1.1) f[x,y]x+g[x,y]d(x) = [f(x),y]x+y[f(x),x]+[d(y),x]x+g(y)[d(x),x] +[g(y),x]d(x) (2.14) Using (2.13) in (2.14) g[x,y]d(x) = y[f(x),x]+g(y)[d(x),x] +[g(y),x]d(x), x,y I. (g[x,y] [g(y),x])d(x) = y[f(x),x]+g(y)[d(x),x], x,y I. Since g is on to we have 2[x,y]d(x) = y[f(x),x]+y[d(x),x] (2.15) Replace y by ry in(2.15), where r R, we obtain
GENERALIZED SEMIDERIVATIONS IN... 47 2[x,r]yd(x)+2r[x,y]d(x) = ry[f(x),x]+ry[d(x),x] (by equation 1.3) Using (2.15) in the above equation, we get 2[x,r]yd(x) = 0 Since R is a char 2, which gives [x,r]yd(x) = 0, x,y I and r R. Therefore, [x,r]id(x) = 0, x I and r R. By using Lemma 2.3, we conclude that R is commutative. References [1] Daif, M. N., Bell, H. E., Remarks on derivations on semi prime rings, International J. Math. and Math. Sci., 15(1), 205-206, 1992. [2] M. A. Quadri, M. Shadab Khan and N. Rehman, Generalized derivations and commutativity of prime rings, Indian J. Pure Appl. Math. 34, no. 9, 1393-1396, 2003. [3] L. Oukhtite and S. Salhi, On generalized derivations of s-prime rings, African Diaspora J. Math. 5, no. 1, 19-23, 2006. [4] Bergen. J., Derivations in prime rings, Cand. Math. Bull., 26, 267-270, 1983. [5] Chang, J.C., On semiderivations of prime rings, Chinese J. Math., 12, 255-262, 1984. [6] M. Ashraf, N. Rehman, On commutativity of rings with derivations, Results Math.42, 3-8, 2002.
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