ON PRIME AND SEMIPRIME NEAR-RINGS WITH DERIVATIONS NURCAN ARGA

Let N be a semiprime right near-ring, A a subset of N such that 0 E A and AN C_ A, and d a derivation ofN The purpose of this paper is to prove that if d acts as a homomorphism on A or as an anti-homomorphism on A, then d(A) {O}


INTRODUCTION
Throughout this paper N will be a right near-ring A derivation on N is defined to be an additive endomorphism satisfying the "product rule" d(zU) zd(U) + d(z)U for all z, U E N According to Bell   and Mason [1], a near-ring N is said to be prime ifzNu {0} for z,U E N implies z 0 or U 0, and semiprime if zNz {0} for z N implies z 0 Let S be a nonempty subset of N and d be a derivation of N If d(:rU) d(z)d(u) or d(zl) d(u)d(z) for all z, U S, then d is said to act as a homomorphism or anti-homomorphism on S, respectively.As for terminologies used here without mention, we refer to Pilz [2].
Bell and Kappe [3] proved that if d is a derivation of a semiprime" ring R which is either an endomorphism or anti-endomorphism, then d 0 They also showed that if d is a derivation of a prime ring R which acts as a homomorphism or an anti-homomorphism on 1, where I is a nonzero right ideal, then d 0 on R

TIE RESULTS
It is our aim in this paper to prove that the above conclusions hold for near-rings as follows TIIEOREM Let N be a semiprime right near-ring, and d a derivation on N Let A be a subset of N such that 0 A and AN c_C_ A If d acts as a homomorphism on A or as an anti-homomorphism on A, then d(A) {0}.
In order to give the proof of the above theorem we need the following lemmas LEMMA 1.If N is a right near-ring and d a derivation of N, then c(ud(z) + d(u)z) cud(z) + cd(u)z for all :r, 9, c N.
A proof can be given by using a similar approach as in the proof of 1, Lemma LEMMA 2. Let N be a right near-ring, d a derivation of N, and A a multiplicative subsemigroup of N which contains 0 If d acts as an anti-homomorphism on A, then a0 0 for all a A PROOF.Since 0a 0 for all a A and d acts as an anti-homomorphism on A then we have d(a)O 0 for all a A Taking a0 instead of a, one can obtain aO + d(a)O 0 for all a A Thus we get a0 0 for all a A LEMMA 3. Let N be a right near-ring, and A a multiplicative subsemigroup of N (a) If d acts as a homomorphism on A, then d(y)xd(y) yxd(y) d(y)xy for all x, y A.
(  x, y A. From this relation we arrive at d(x)xy d(x)yd(x) 0 for all x, y A Similarly taking xy instead of x in (2 5), one can prove the relation d(y)xd(y) xyd(y) for all x, y A. PROOF OF TItEOREM.(a) First suppose that d acts as a homomorphism on A. By Lemma 3 (a), we have d(y)xd(y) d(y)xy for all x, y A. ( Right-multiplying (2.6) by d(z), where z A, and using the hypothesis that d acts as a homomorphism on A together with Lemma 1, we obtain d(y)xd(y)z 0 for all x, y, z A Taking xr instead of x, where r N, we have d(y)xrd(y)z 0 for all x,y,z A and r N. Hence d(y)xNd(y)x {0} for all x, y A; and by semiprimeness d(y)x=O for all x,yA. ( Substituting yr for y in (2 7), where r N, leads to yd(r)x + d(y)rx 0 for all x, y A, r N. (2.8) Left-mukiplying (2.8) by d(z), where z A, we have that d(z)yd(r)x + d(z)d(y)rx 0 According to (2 7) this relation reduces to d(zy)rx 0. Hence we get zd(y)rx 0 for all z, y, z A and r N By semiprimeness, we get zd(y) o zrd(y) for all y, z A and r N. (2.9) Combining (2 7) and (2.9) shows that d(yz)= 0 for all V,z E A In partmular, d(xrz)= 0 for all z E A, r 6 N, and since d acts as a homomorphism on A, a(zr)a(z) o za(r)a(z) + d(z)rd(z) for all x A, r N.
In view of(2.9), this gives d(z)Nd(x) {0} and hence d(:c) 0 for all z A (b) Now assume that d acts as an anti-homomorphism on A Note that a0 0 for all a E A by Lemma 2. According to Lemma 3 (b), a(u)za(u) zud(u) for all z, y A, (2 10) d(y)xd(y) d(y)yx for all x, y A.
(2 11) Replacing x by xd(y) in (2.10) and using Lemma we get d(y)xyd(y) + d(y)xd(y)y xd(y)yd(y) for all x, y E A. ( Substituting xy for x in (2 10), we have d(y)zyd(y) zy2d(y) for all x, y A. ( Right-multiplying (2 10) by y we arrive at d(y)xd(y)v xyd(y)y for all x,y A. ( Replacing x by y in (2.10), we have d(y)yd(y)= y2d(y), and left-multiplying this relation by x, we obtain xd(y)yd(y) xy2d(y) for all x, y e A. ( Using ( 213), (214), and (2.15) in (2.12) one obtains zyd(y)y 0 for all x,y A, hence yryd(y)y 0 and yd(y)yryd(y)y yd(y)O 0 for all y E A, r N, and by semiprimeness yd(y)y=O for all yA.
According to (2.14) we get d(y)xd(y)y 0 for all x,y A Using this relation in (2 11), we arrive at d(y)yxy 0 for all x, y A. ( Replacing x by xd(y) in (2.16), we have d(y)yxd(y)y 0 d(y)yxrd(y)yx for all x, y fi A, r N, hence d(y)yx=O for all z,yA.
Using (217) in (211), one obtains d(y)xd(y) 0 d(y)xrd(y)x for all x, y E A, r E N, hence d(y)x=O for all x,yA. ( Therefore, xd(z)d(yn)x 0 for all x, y, z 6 A, n N, xd(z)(yd(n) + d(y)n)x 0 for all z, y, z A, n N, xd(z)yd(n)x + xd(z)d(y)nx 0 for all z, y, z A, n E N. x, y, z A Since d acts as an anti-homomorphism on A, we have xd(yz) 0 for all x, y, z A, so that xyd(z) + xd(y)z 0 for all x, y, z E A By (2.18) we now get xyd(z) 0 zd(z)rud(z) for all x,y,z E A and r N; and taking x instead ofy we get xd(z) 0 for all x,z _ A Recalling (2.18), we now have d(xy) 0 for all x, y E A, so d(xxr) 0 for all x A and r N. Thus d(xr)d(x) O, and we finish the proof as in case (a).
We now state some consequences of the theorem COROLLARY I. Let N be a semiptime right near-ring, and d a derivation of N If d acts as a homomorphism on N or as an anti-homomorphism on N, then d 0 COROLLARY 2. Let N be a prime right near-ring, and d a derivation of N Let A be a nonzero subset of N such that 0EA and ANC_A.Ifd acts as a homomorphism on A or as an anti- homomorphism on A then, d 0 PROOF.Bythe theorem, we have d(a) 0 for all a E A Then d(az) ad(z)+d(a)z ed(z) 0 ayd(z) for all a A,z,y N, and by primeness we get a 0 or d(z) 0 for all a A,z N Since A is nonzero, we have d(x) 0 for all z E N ) (b) If d acts as an anti-homomorphism on A, then d(y)xd(v) d(y)vx xyd(y) for all x, y A. (22) PROOF.(a) Let d act as a homomorphism on A Then d(xy) xd(y) + d(x)y d(x)d(y) for all x,y A.
4)By Lemma 1, d(y)d(xy) d(y)xd(y) + d(y)d(x)y d(y)xd(y) + d(yx)y Using this relation in (2 4), we obtain yxd(y) d(y)xd(y) for all x,y A Similarly, taking yx instead ofy in (2 3) one can prove the relation d(y)xd(y) d(y)xy for all x, y A.(b) Since d acts as an anti-homomorphism on A, we have d(xy) xd(y) + d(x)y d(y)d(x) for all x,y A.