Math. J. Okayama Univ. 46 (2004), 1–7 RESULTS ON PRIME NEAR-RING WITH (σ, τ )-DERIVATION Öznur GÖLBAŞI and Neşet AYDIN Abstract. Let N be a prime left near-ring with multiplicative center Z, and D be a (σ, τ )-derivation such that σD = Dσ and τ D = Dτ. (i) If D(N ) ⊂ Z, or [D(N ), D(N )] = 0 or [D(N ), D(N )]σ,τ = 0, then (N, +) is abelian. (ii) If N is 2-torsion free, d1 is a (σ, τ )-derivation and d2 is a derivation on N such that d1 d2 (N ) = 0, then d1 = 0 or d2 = 0. 1. Introduction Recently, some results concerning commutativity in prime near-rings with derivation have been generalized in several ways. The primary purpose of this paper is to generalize some results obtained by H. E. Bell and G. Mason [1], and A. A. M. Kamal[2]. Throughout this paper, N will denote a zero-symetric left near-ring with multiplicative center Z. N is called a prime near-ring if aN b = {0} implies that a = 0 or b = 0. Let σ and τ be two near-ring automorphisms of N . An additive mapping D : N → N is called a (σ, τ )-derivation if D(xy) = τ (x)D(y) + D(x)σ(y) holds for all x, y ∈ N . For x, y ∈ N , the symbol [x, y] will denote xy − yx, while the symbol (x, y) will denote the additive-group commutator x+y −x−y. Given x, y ∈ N , we write [x, y]σ,τ = xσ(y)−τ (y)x; in particular [x, y]1,1 = [x, y], in the usual sense. As for terminologies used here without mention, we refer to G. Pilz [3]. 2. Results We begin with two quite general and useful lemmas. Lemma 1. Let D be a (σ, τ )-derivation of near ring N. Then D(xy) = D(x)σ(y) + τ (x)D(y) for all x, y ∈ N. Proof. Note that D(x(y + y)) = τ (x)D(y + y) + D(x)σ(y + y) = τ (x)D(y) + τ (x)D(y) + D(x)σ(y) + D(x)σ(y), and D(xy + xy) = τ (x)D(y) + D(x)σ(y) + τ (x)D(y) + D(x)σ(y). Mathematics Subject Classification. 16Y30, 16N60,16W25. Key words and phrases. Prime Near-Ring, Derivation, (σ, τ )-Derivation. 1 2 Ö. GÖLBAŞI AND N. AYDIN Comparing these two expressions, one can obtain τ (x)D(y) + D(x)σ(y) = D(x)σ(y) + τ (x)D(y) and so, D(xy) = D(x)σ(y) + τ (x)D(y), for all x, y ∈ N. ¤ Lemma 2. Let D be a (σ, τ )-derivation on a near-ring N and a ∈ N. Then for all x, y ∈ N, (τ (x)D(y) + D(x)σ(y))σ(a) = τ (x)D(y)σ(a) + D(x)σ(y)σ(a). Proof. For all x, y ∈ N, we get D((xy)a) = τ (xy)D(a) + D(xy)σ(a) = τ (x)τ (y)D(a) + (τ (x)D(y) + D(x)σ(y))σ(a). On the other hand, D(x(ya)) = τ (x)D(ya) + D(x)σ(ya) = τ (x)τ (y)D(a) + τ (x)D(y)σ(a) + D(x)σ(y)σ(a). For these two expressions of D(xya), we obtain that, for all x, y ∈ N, (τ (x)D(y) + D(x)σ(y))σ(a) = τ (x)D(y)σ(a) + D(x)σ(y)σ(a). ¤ Lemma 3. Let N be a prime near-ring, D a nonzero (σ, τ )-derivation of N and a ∈ N. i) If D(N )σ(a) = 0 then a = 0. ii) If aD(N ) = 0 then a = 0. Proof. i) For all x, y ∈ N, we get 0 = D(xy)σ(a) = τ (x)D(y)σ(a) + D(x)σ(y)σ(a). Using hypothesis and σ is an automorphism of N, we have D(x)N σ(a) = 0. Since N is prime near-ring and D is a nonzero (σ, τ )-derivation of N , we obtain a = 0. ii) A similar argument works if aD(N ) = 0. ¤ Lemma 4. Let D be a (σ, τ )-derivation which commute σ and τ. If N is a 2-torsion free near-ring and D2 = 0 then D = 0. RESULTS ON PRIME NEAR-RINGS 3 Proof. For arbitrary x, y ∈ N, we have 0 = D2 (xy) = D(D(xy)) = D(τ (x)D(y) + D(x)σ(y)) = τ 2 (x)D2 (y) + D(τ (x))σ(D(y)) + τ (D(x))D(σ(y)) + D2 (x)σ 2 (y). By hypothesis, 2D(τ (x))D(σ(y)) = 0 for all x, y ∈ N. Since N is 2-torsion free near-ring and σ is an automorphism on N, we get D(τ (x))D(N ) = 0. It gives D = 0 by Lemma 3 (ii). ¤ Theorem 1. Let N be a near-ring and D a nonzero (σ, τ )-derivation of N. If u ∈ N is not a left zero divisor and [D(u), u]σ,τ = 0 then (x, u) is constant (that is, D(x, u) = 0) for every x ∈ N. Proof. Since u(u + x) = u2 + ux, we have D(u(u + x)) = D(u2 + ux). Expanding this equation, we have τ (u)D(u + x) + D(u)σ(u + x) = D(u2 ) + D(ux) and so τ (u)D(u) + τ (u)D(x) + D(u)σ(u) + D(u)σ(x) = τ (u)D(u) + D(u)σ(u) + τ (u)D(x) + D(u)σ(x) which reduces to τ (u)D(x) + D(u)σ(u) − τ (u)D(x) − D(u)σ(u) = 0. Therefore τ (u)D(x, u) = 0 by using the assumption [D(u), u]σ,τ = 0. Since u is not a left zero divisor, we get D(x, u) = 0. Thus (x, u) is a constant for every x ∈ N. ¤ Theorem 2. Let N be a prime near-ring with a nonzero (σ, τ )-derivation D such that σD = Dσ and τ D = Dτ. If D(N ) ⊂ Z then (N, +) is abelian. Moreover, if N is 2-torsion free, then N is a commutative ring. Proof. Suppose that a ∈ N such that D(a) 6= 0. So, D(a) ∈ Z\{0} and D(a) + D(a) ∈ Z\{0}. For all x, y ∈ N, we have (x + y)(D(a) + D(a)) = (D(a) + D(a))(x + y), that is, xD(a) + xD(a) + yD(a) + yD(a) = D(a)x + D(a)y + D(a)x + D(a)y. 4 Ö. GÖLBAŞI AND N. AYDIN Since D(a) ∈ Z, we get D(a)x + D(a)y = D(a)y + D(a)x, and so, D(a)(x, y) = 0 for all x, y ∈ N. Since D(a) ∈ Z\{0} and N is a prime near-ring, it follows that (x, y) = 0, for all x, y ∈ N. Thus (N, +) is abelian. Using hypothesis, for any b, c ∈ N , σ(c)D(ab) = D(ab)σ(c). By Lemma 2, we have σ(c)τ (a)D(b) + σ(c)D(a)σ(b) = τ (a)D(b)σ(c) + D(a)σ(b)σ(c). Comparing these two expressions, using D(N ) ⊂ Z and (N, +) is abelian, we obtain that σ(c)τ (a)D(b) + D(a)σ(c)σ(b) = τ (a)D(b)σ(c) + D(a)σ(b)σ(c) so we have D(b)[τ (a), σ(c)] = D(a)σ([c, b]) for all b, c ∈ N. Suppose now that N is not commutative. Choosing b, c ∈ N such that [b, c] 6= 0 and a = D(x) ∈ Z, we get D2 (x)σ([c, b]) = 0 for all x ∈ N. Since the central element D2 (x) can not be a nonzero divisor of zero, we conclude D2 (x) = 0 for all x ∈ N. By Lemma 4, this cannot happen for nontrivial D. ¤ Theorem 3. Let N be a prime near-ring admitting a nonzero (σ, τ )-derivation D such that σD = Dσ and τ D = Dτ. If [D(N ), D(N )] = 0, then (N, +) is abelian. Moreover, if N is 2-torsion free, then N is a commutative ring. Proof. The argument used in the proof of Theorem 2 shows that if both z and z + z commute elementwise with D(N ),then we have (2.1) zD(x, y) = 0 for all x, y ∈ N. Substituting D(t), t ∈ N for z in (2.1), we get D(t)D(x, y) = 0. Since σ is an automorphism of N , we have σ(D(t))σ(D(x, y)) = 0. Using σD = Dσ, we get D(σ(t))σ(D(x, y)) = 0 for all x, y, t ∈ N. By Lemma 3 (i), we obtain that D(x, y) = 0 for all x, y ∈ N. For w ∈ N , we have 0 = D(wx, wy) = D(w(x, y)) and so we obtain D(w)σ((x, y)) = 0. RESULTS ON PRIME NEAR-RINGS 5 Again, applying Lemma 3 (i), we get (x, y) = 0 for all x, y ∈ N. Now, assume that N is 2-torsion free. By the assumption [D(N ), D(N )] = 0, D(σ(z))D(D(x)y) = D(D(x)y)D(σ(z)) for all x, y, z ∈ N. Hence, we get D(σ(z))τ (D(x))D(y) + D(σ(z))D2 (x)σ(y) = τ (D(x))D(y)D(σ(z)) + D2 (x)σ(y)D(σ(z)) by Lemma 2. Using D(τ (x))D(σ(z)) = D(σ(z)) D(τ (x)), σD = Dσ and τ D = Dτ , we have D(τ (x))D(σ(z))D(y) + D(σ(z))D2 (x)σ(y) = D(τ (x))D(y)D(σ(z)) + D2 (x)σ(y)D(σ(z)) Since (N, +) is abelian, we conclude that D(τ (x))[D(σ(z)), D(y)] = D2 (x)σ([D(z), y]) for all x, y, z ∈ N. The left term of this equation is zero by the hypothesis, so we get D2 (x)σ(D(z))σ(y) = D2 (x)σ(y)σ(D(z)) for all x, y, z ∈ N. (2.2) Replacing y by yt, (t ∈ N ) in (2.2) and using (2.2), we have D2 (x)σ(y)σ(t)σ(D(z)) = D2 (x)σ(D(z))σ(y)σ(t) = D2 (x)σ(y)σ(D(z))σ(t) and so, (2.3) D2 (x)N σ([t, D(z)]) = 0 for all x, t, z ∈ N. Since N is a prime near-ring, we have D2 (N ) = 0 or D(N ) ⊂ Z by Brauers’s Trick. If D2 (N ) = 0, then it contradicts that D is a nonzero (σ, τ )-derivation of N by Lemma 4. So, D(N ) ⊂ Z. Thus, N is a commutative ring by Theorem 2. ¤ Theorem 4. Let N be a 2-torsion free prime near-ring, d1 a (σ, τ )-derivation of N and d2 a derivation of N. If d1 d2 (N ) = 0, then d1 = 0 or d2 = 0. Proof. For x, y ∈ N , we have 0 = d1 d2 (xy) = d1 (xd2 (y) + d2 (x)y) = τ (x)d1 d2 (y) + d1 (x)σ(d2 (y)) + τ (d2 (x))d1 (y) + d1 d2 (x)σ(y). 6 Ö. GÖLBAŞI AND N. AYDIN That is, (2.4) d1 (x)σ(d2 (y)) + τ (d2 (x))d1 (y) = 0 for all x, y ∈ N. If we take d2 (x) instead of x in (2.4), then τ (d22 (x))d1 (y) = 0 for all x, y ∈ N. Using Lemma 3 (ii) one can obtain d1 = 0 or d22 = 0. If d22 = 0, we have d2 = 0 by Lemma 4. This completes the proof of theorem. ¤ Theorem 5. Let N be a 2-torsion free prime near-ring, d1 a derivation and d2 be a (σ, τ )-derivation of N such that τ d2 = d2 τ and τ d1 = d1 τ. If d1 d2 (N ) = 0, then d1 = 0 or d2 = 0. Proof. The same argument in the proof of Theorem 4, we can write (2.5) d1 (τ (x))d2 (y) + d2 (x)d1 (σ(y) = 0 for all x, y ∈ N. Replacing x by d2 (x) in (2.5) and using τ d2 = d2 τ and τ d1 = d1 τ , we have d22 (x)d1 (σ(y) = 0 for all x, y ∈ N. Applying [1, Lemma 3 (ii)], we obtain d1 = 0 or d22 = 0. If d22 = 0, then d2 = 0 by Lemma 4. ¤ Theorem 6. Let D be a nonzero (σ, τ )-derivation of a prime near-ring N and a ∈ N. If [D(N ), a]σ,τ = 0 then D(a) = 0 or a ∈ Z. Proof. By hypothesis, D(ax)σ(a) = τ (a)D(ax) for all x ∈ N and so, (τ (a)D(x) + D(a)σ(x))σ(a) = τ (a)(τ (a)D(x) + D(a)σ(x)). Since N satisfies the partial distributive law by Lemma 2, we get τ (a)D(x)σ(a) + D(a)σ(x)σ(a) = τ (a)τ (a)D(x) + τ (a)D(a)σ(x). Using the hypothesis, we have τ (a)τ (a)D(x) + D(a)σ(x)σ(a) = τ (a)τ (a)D(x) + D(a)σ(a)σ(x), that is, (2.6) D(a)σ([x, a]) = 0 for all x ∈ N. Substituting xy, (y ∈ N ) for x and using (2.6), we have D(a)σ(x)σ([y, a]) = 0 for all x, y ∈ N. Since σ is automorphism of prime near-ring of N , we get D(a) = 0 or a ∈ Z. This completes the proof. ¤ RESULTS ON PRIME NEAR-RINGS 7 Theorem 7. Let D be a nonzero (σ, τ )-derivation of a prime near-ring N such that σD = Dσ and τ D = Dτ. If [D(N ), D(N )]σ,τ = 0, then (N, +) is abelian. Moreover, if N is 2-torsion free then N is a commutative ring. Proof. By Theorem 6, we have N = {x ∈ N | D2 (x) = 0} ∪ {x ∈ N | D(x) ∈ Z}. By Brauer’s Trick, we get D2 (N ) = 0 or D(N ) ⊂ Z. Since D is a nonzero (σ, τ )-derivation of N, we get D(N ) ⊂ Z. By Theorem 2, we prove the theorem. ¤ References [1] Bell, H: E. and Mason G., On Derivations in near-rings, Near-rings and Near-fields, North-Holland Mathematical Studies 137, (1987). [2] Kamal, Ahmed A. M., σ-derivations on prime near-rings, Tamkang J. Math. 32(2001), no.2, 89–93. [3] Pilz, G., Near-rings 2nd Ed., North Holland, Amsterdam, (1983). Öznur Gölbaşi Cumhuriyet University Faculty of Arts and Science Department of Mathematics Sivas - TURKEY e-mail address: ogolbasi@cumhuriyet.edu.tr URL: http://www.cumhuriyet.edu.tr Neşet Aydin Çanakkale 18 Mart University Faculty of Arts and Science Department of Mathematics Çanakkale - TURKEY e-mail address: neseta@comu.edu.tr URL: http://www.comu.edu.tr (Received June 4, 2003 ) Math. J. Okayama Univ. 52 (2010), 199–200 CORRECTION: RESULTS ON PRIME NEAR-RINGS WITH (σ, τ )-DERIVATION Math. J. Okayama Univ. 46 (2004), 1–7. Öznur GÖLBAŞI and Neşet AYDIN In the proof of Theorem 7 on pp.7 in [1], Brauer’s Trick method is used wrongly, in which case the corrected should read as follows: Theorem 7. Let N be a 2−torsion free prime left near-ring, D be a nonzero (σ, τ )-derivation of N such that σD = Dσ, τ D = Dτ. If [D(N ), D(N )]σ,τ = 0 then N is commutative ring. Proof. It is correctly shown in [1, Theorem 6] that D 2 (x) = 0 or D(x) ∈ Z, for all x ∈ N. Choosing x such that D(x) ∈ Z. If D(x) = 0 then D 2 (x) = 0, so we get D(x) ∈ Z\{0}. It follows D(y +z)σ(D(x+x)) = τ (D(x+x))D(y + z), for all y, z ∈ N, by the hypothesis. That is (D(y) + D(z))σ(D(x)) + (D(y) +D(z))σ(D(x)) = τ (D(x+x))D(y) +τ (D(x+x))D(z). Using D(x) ∈ Z and the hypothesis, we can arrive at σ(D(x))D(y) + σ(D(x))D(z) + σ(D(x))D(y) + σ(D(x))D(z) = D(y)σ(D(x + x)) + D(z)σ(D(x + x)). Computing this equation, we have σ(D(x))D(z, y) = 0, for all y, z ∈ N. Since D(x) ∈ Z\{0} and N is prime near-ring, we conclude that D(z, y) = 0, for all y, z ∈ N. For any w ∈ N, we can write 0 = D(wz, wy) = D(w(z, y)), and so we obtain D(w)σ(z, y) = 0, for all y, z, w ∈ N. By [1, Lemma 3 (i)], (z, y) = 0, for all y, z ∈ N. Thus (N, +) is abelian. Now, we have [D(D(x)y), D(z)]σ,τ = 0, for all y, z ∈ N. We calculate this equation using [1, Lemma 2], D(x) ∈ Z and (N, +) is abelian, we have τ (D(x))[D(y), D(z)]σ,τ = τ (D(z))D2 (x)σ(y) − D 2 (x)σ(y)σ(D(z)). Since the left term of this equation is zero by the hypothesis and σ is an automorphism of N, we conclude that τ (D(z))D2 (x)y = D2 (x)yσ(D(z)), for all y, z ∈ N. Replacing y by yt, t ∈ N in this equation and using this, we obtain that D2 (x)y[σ(D(z)), t] = 0, for all y, z, t ∈ N. By the primeness of N, we infer D2 (x) = 0 or D(N ) ⊂ Z, for all x ∈ N. In the first case, D2 = 0, and so D = 0 by [1, Lemma 4], contrary to our original hypothesis. Hence D2 (x) = 0 does not in fact occur. Thus we get D(N ) ⊂ Z, then N is commutative ring by [1, Theorem 2]. This completes the proof.  The above proof, stemming from the authors’ oversight in editing and proofreading, is immaterial for the other results, proofs and discussions of the article. Mathematics Subject Classification. 16A72, 16A70, 16Y30. Key words and phrases. Prime Near-Ring, Derivation,(σ, τ )−Derivation. 199 200 ÖZNUR GÖLBAŞI AND NEŞET AYDIN References [1] Gölbaşı, Ö., Aydın, N., Results on Prime Near-Rings with (σ, τ )−Derivation, Math. J. of Okayama Univ., 46, 1-7, (2004). Cumhuriyet University, Faculty of Arts and Science, Department of Mathematics, Sivas - TURKEY e-mail address: ogolbasi@cumhuriyet.edu.tr URL: http://www.cumhuriyet.edu.tr Çanakkale 18 Mart University, Faculty of Arts and Science, Department of Mathematics, Çanakkale - TURKEY e-mail address: neseta@comu.edu.tr URL: http://www.comu.edu.tr (Received May 8, 2008 ) (Revised July 9, 2008 )