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Kyungpook Mathematical Journal 2020; 60(2): 415-421

Published online June 30, 2020

Copyright © Kyungpook Mathematical Journal.

Correction to "On prime near-rings with generalized (σ,Ƭ)- derivations, Kyungpook Math. J., 45(2005), 249-254"

Hassan J. Al Hwaeer*, Gbrel Albkwre, Neşet Deniz Turgay

Department of Mathematics and Computer applications, Al-Nahrain University,Iraq
e-mail : hjh@sc.nahrainuniv.edu.iq
Department of Mathematics, West Virginia University, Morgantown, WV 26506, USA
e-mail : gmalbkwre@mix.wvu.edu
Department of Mathematics, Faculty of Arts and Sciences, Eastern Mediterranean University, Famagusta, North Cyprus via Mersin 10, Turkey
e-mail : neset.turgay@emu.edu.tr

Received: June 12, 2019; Revised: September 24, 2019; Accepted: November 18, 2019

In the proof of Theorem 3 on p.253 in [4], both right and left distributivity are assumed simultaneously which makes the proof invalid. We give a corrected proof for this theorem by introducing an extension of Lemma 2.2 in [2].

Keywords: prime near-rings, derivation, generalized (σ,Ƭ,)-derivation

Over the last few decades lots of work has been done on commutativity of prime rings with derivations. It is natural to look for comparable results on near rings (see for example [6]). Historically speaking, the study of derivations of prime near-rings was initiated by H. E. Bell and G. Mason in 1987 [3]. An analogue of Posner’s result on prime near-rings was obtained by Beidar in [1]. Results concerning prime near-rings with derivations have since been generalized in several ways. Throughout the paper N is a zero-symmetric left near-ring with a multiplicative center Z. For x, yN, the symbol [x, y] will denote the commutator xyyx, while the symbol (x, y) will denote the additive group commutator x + yxy. Our main aim in this paper is to give a correction for Theorem 3 on p.253 in [4].

The proof of Theorem 3 of [4], stated as Theorem 2.12 in this paper, is incorrect because of the assumption of both left and right distributivity. In Section 3, we first introduce Lemma 3.1, which is an extension of [2, Lemma 2.2], then give a correction for Theorem 2.12.

We first recall some of the background which will be used throughout this work.

Definition 2.1

A left near-ring is a set N together with two binary operations + and. such that (N, +) is a group (not necessarily abelian) and (N, .) is a semigroup satisfying the distributive law x(y + z) = xy + xz for all x, y, zN.

Definition 2.2

A near-ring N is said to be zero-symmetric if 0.x = 0 ∀aN.

Definition 2.3

A left near-ring N is called a prime near-ring if a, bN and aNb = 0 imply that a = 0 or b = 0.

Definition 2.4

A derivation on a near-ring N is defined to be an additive endomorphism satisfying the “product rule” d(xy) = xd(y) + d(x)y for all x, yN.

Definition 2.5

Let σ, τ be two endomorphisms of a near-ring N. An additive mapping d : NN is called a (σ, τ )-derivation if d(xy) = σ(x)d(y) + d(x)τ (y) for all x, yN.

Definition 2.6

Let N be a near-ring, and d a derivation of N. An additive mapping f : NN is said to be left generalized (σ, τ )-derivation of N associated with d if

f(xy)=d(x)τ(y)+σ(x)f(y)x,yN.

f is said to be a right generalized (σ, τ )-derivation of N associated with d if

f(xy)=f(x)τ(y)+σ(x)d(y)x,yN.

f is said to be a generalized (σ, τ )-derivation associated with d if it is both left and right generalized (σ, τ )-derivation associated with d.

We now recall some properties about near-rings with derivation and generalized (σ, τ )-derivation.

Lemma 2.7.([4, Lemma 1])

  • (i) Let f be a right generalized (σ, τ )-derivation of near-ring N associated with d. Then f(xy) = σ(x)d(y) + f(x)τ (y) ∀x, yN.

  • (ii) Let f be a left generalized (σ, τ )-derivation of near-ring N associated with d. Then f(xy) = σ(x)f(y) + d(x)τ (y) ∀x, yN.

Lemma 2.8.([4, Lemma 3])

Let N be a prime near-ring, f a generalized (σ, τ )-derivation of N associated with d, and aN.

  • (i) If af(N) = 0, then a = 0.

  • (ii) If f(N)τ (a) = 0, then a = 0.

Lemma 2.9.([5, Lemma 3])

Let N be a prime near-ring, d a non-zero (σ, τ )-derivation of N, and aN.

  • (i) If d(N)σ(a) = 0, then a = 0.

  • (ii) If ad(N) = 0, then a = 0.

Lemma 2.10.([4, Lemma 2])

  • (i) Let f be a right generalized (σ, τ )-derivation of near-ring N associated with d. Then

    (f(x)τ(y)+σ(x)d(y))τ(z)=f(x)τ(y)τ(z)+σ(x)d(y)τ(z),,

    for all x, yN.

  • (ii) Let f be a generalized (σ, τ )-derivation of near-ring N associated with d. Then

    (d(x)τ(y)+σ(x)f(y)τ(z))=d(x)τ(y)τ(z)+σ(x)f(y)τ(z),

    for all x, yN.

Theorem 2.11.([4, Theorem 2])

Let N be a prime near-ring with a non-zero generalized (σ, τ )-derivation f associated with d. If f(N) ⊂ Z then (N, +) is abelian. Moreover, if N is 2-torsion free then N is a commutative ring.

Theorem 2.12.([4, Theorem 3])

Let N be a prime near-ring with a non-zero generalized (σ, τ )-derivation f associated with non-zero (σ, τ )-derivation d such that τf = fτ, fσ = σf, σd = dσ, and τd = dτ. If [f(N), f(N)] = 0 then (N, +) is abelian. Moreover, if N is 2-torsion free, then N is commutative.

We first give the following auxiliary lemma.

Lemma 3.1

Let N be a prime near-ring and f be a generalized (σ, τ )-derivation associated with non-zero (σ, τ )-derivation d such that σf = fσ, τf = fτ and σd = dσ, τd = dτ. If d(f(N)) = {0}, then f(d(N)) = {0}.

Proof

Assume that d(f(x)) = 0. It follows for all x, yN,

0=d(f(xy))=d(f(x)τ(y))+d[σ(x)d(y))=d(f(x)τ(y))+d(σ(x)d(y));0=σ(f(x))d(τ(y))+σ2(x)d2(y)+d(σ(x))τ(d(y)).

Applying d to (3.1), we get

0=d(σ(f(x))d(τ(y)))+d(σ2(x)d2(y))+d(d(σ(x))τ(d(y))).0=σ2(f(x))d2(τ(y))+σ3(x)d3(y)+d(σ2(x))τ(d2(y))+d(σ2(x))τ(d2(y))+d2(σ(x))τ2(d(y)).

Substituting d(y) for y in (3.1) gives

0=σ(f(x))d2(τ(y))+σ2(x)d3(y)+d(σ(x))τ(d2(y)).

Replacing x by σ(x) in (3.3), we get

0=σ2(f(x))d2(τ(y))+σ3(x)d3(y)+d(σ2(x))τ(d2(y)).

Hence (3.2) yields

0=d(σ2(x))τ(d2(y))+d2(σ(x))τ2(d(y)).

Replacing x by d(x) in (3.1) we obtain

0=σ(f(d(x)))d(τ(y))+σ2(d(x))d2(y)+d2(σ(x))τ(d(y)).

Taking τ (y) for y in (3.6) gives

0=σ(f(d(x)))d(τ2(y))+σ2(d(x))d2(τ(y))+d2(σ(x))τ2(d(y)).

By using (3.5) we have

0=σ(f(d(x)))d(τ2(y)).

Replacing x by σ −1(x) in (3.8), we get

f(d(x))d(τ2(y))=0.

x, yN. Thus, f(d(x)) = 0 ∀xN by using Lemma 2.9 (ii). This finishes the proof.

The preceding lemma will now be used to establish the correction.

Proof of Theorem 2.12

We start with the same argument given in the proof of [4, Theorem 3]. If z and z+z commute element wise with f(N), then for any x, yN,

(z+z)(f(x)+f(y))=(f(x)+f(y))(z+z).

So we get zf(x) + zf(y) − zf(x) − zf(y) = 0 and this gives

zf(x,y)=0x,yN.

Substituting f(t), tN, for z in (9) we have

f(t)f((x,y))=0.

Since τ is an automorphsim of N we have τ (f(t))τ (f(x, y)) = 0. By using the assumption τf = , we get

f(τ(t))τ(f(x,y))=0x,y,tN.

By Lemma 2.8, we get f(x, y) = 0 ∀x, yN. Now for wN, we have

0=f(wx,wy)=f(w(x,y))=d(w)τ(x,y)+σ(w)f(x,y).

and this gives

d(w)τ(x,y)=0.

Replacing w by wr in (10) and using Lemma 2.10 we get

d(w)τ(r)τ(x,y)+σ(w)d(r)τ(x,y)=0,

and this gives

d(w)Nτ(x,y)=0x,y,wN.

Since N is a prime near-ring, d ≠ 0, we get (x, y) = 0, so x + yxy = 0, and this gives x + y = y + x. Hence (N, +) is abelian.

Now, assume that N is 2-torsion free. By the assumption [f(N), f(N)] = 0, we have

f(τ(z))f(f(x)y)=f(f(x)y)f(τ(z))x,y,zN.

Using τf = , = σf and using Lemma 2.7, we obtain

f(τ(z))d(f(x))τ(y)+f(τ(z))σ(f(x))f(y)=d(f(x))τ(y)τ(f(z))+σ(f(x))f(y)τ(f(z)).

and this gives

f(τ(z))d(f(x))τ(y)=d(f(x))τ(y)τ(f(z)).

If we take yw instead of y in (11), then

f(τ(z))d(f(x))τ(y)τ(w)=d(f(x))τ(y)τ(w)τ(f(z)).

So we have

d(f(x))τ(y)f(τ(z))τ(w)=d(f(x))τ(y)τ(w)τ(f(z))x,y,wN.

Hence d(f(x))N(f(τ (z)), τ(w)) = 0. Since N is prime, we have

d(f(x))=0or f(N)Z.

If f(N) ⊆ Z, then N is a commutative ring by Theorem 2.11. It remains to show that d(f(x)) = 0 is impossible.

Assume, by way of contradiction, that d(f(x)) = 0, then

0=d(f(xy))=d(d(x)τ(y)+σ(x)f(y))=d(d(x)τ(y))+d(σ(x)f(y)).

So, we have

d2(x)τ2(y)+σ(d(x))d(τ(y))+d(σ(x))τ(f(y))=0x,yN.

Replacing y by τ−1(y) in the last equation, we get

d2(x)τ(y)+d(σ(x))d(y)+d(σ(x))(f(y))=0x,yN.

We claim that f(d(x)y) = 0 ∀x, yN. Indeed, by using the definition of f, we have

f(d(x)y)=d2(x)τ(y)+σ(d(x))f(y).

Also

f(d(x)y)=f(d(x))τ(y)+σ(d(x))d(y).

By Lemma 3.1 we get

f(d(x)y)=σ(d(x))d(y).

Substituting (13) and (14) in (12), we get 2f(d(x)y) = 0, but N is 2-torsion free which implies f(d(x)y) = 0. This proves the claim.

Now, substituting yz instead of y in (12), we get

d2(x)τ(yz)+σ(d(x))d(yz)+d(σ(x))f(yz)=0.

and from this we obtain

d2(x)τ(yz)+σ(d(x))d(y)τ(z)+σ(d(x))σ(y)d(z)+d(σ(x))f(y)τ(z)+d(σ(x))σ(y)d(y)=0.

Rewrite the equation above and use Lemma 2.10, to find

(d2(x)τ(y)+d(σ(x))f(y))τ(z)+σ(d(x))d(y)τ(z)+2σ(d(x))σ(y)d(z)=0.

By using our claim and the equations (13),(14), and (15) we get

2σ(d(x))σ(y)d(z)=0x,y,zN.

Since N is 2-torsion free, we have d(N)Nd(N) = 0. It follows that d(N) = 0 which is a contradiction. This completes the proof.

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