Math. J. Okayama Univ. 49 (2007), 53–58

ON IDEALS AND ORTHOGONAL GENERALIZEDDERIVATIONS OF SEMIPRIME RINGS
Emine ALBAŞAbstract. In this paper, some results concerning orthogonal generalized derivations are generalized for a nonzero ideal of a semiprime ring.
These results are a generalization of results of M. Bres̆ar and J. Vukmanin [3], which are related to a theorem of E. Posner for the product ofderivations on a prime ring.

1. Introduction
Throughout R will represent an associative ring. In [2], Bres̆ar defined
the following notion. An additive mapping D : R → R is said to be a generalized derivation if there exists a derivation d : R → R such that
D(xy) = D(x)y + xd(y)for all x, y ∈ R.

By the above notion it is easily seen that the concept of a generalized derivations covers both the concepts of a derivation and of a left multiplier. This
notion is found in P. Ribenboim [9], where some module structures of thesehigher generalized derivations were treated. Other properties of generalized
derivations were given by B. Hvala [4], T-K. Lee [5] and A. Nakajima [6],[7] and [8].
Two additive maps d, g : R → R are called orthogonal ifd(x)Rg(y) = 0 = g(y)Rd(x)for all x, y ∈ R.

In [3] Bres̆ar and Vukman introduced the notion of orthogonality for two
derivations d and g on a semiprime ring, and they presented several necessary
and sufficient conditions for d and g to be orthogonal. In [10] the authors
replaced R by a non zero ideal I of R, then they showed that some properties
in [[3], Theorem] are also valid in this subconstruction. Finally, in [1] theauthors introduced orthogonal generalized derivations on a semiprime ringand they presented some results concerning two generalized derivations ona semiprime ring. Their results are a generalization of results of M. Bres̆arMathematics Subject Classification. 16W25, 16N60, 16D25.
Key words and phrases. derivation, generalized derivation, orthogonal generalizedderivations, semiprime ring, ideal.
5354E. ALBAŞ

and J. Vukman in [3]. In this paper, our aim is to extend their results toorthogonal generalized derivations on a nonzero ideal I of R.
For a semiprime ring R and an ideal I of R, it is well-known that theleft and right annihilators of I in R coincide. Note that I ∩ `(I) = 0 (orI ∩ r(I) = 0) where `(I) and r(I) denote the left annihilator and the rightannihilator of I, respectively.
Throughout this paper we assume that R is a 2-torsion free semiprime
ring and I is a nonzero ideal of R unless stated otherwise.
2. Preliminaries
In the following, we give the notation of orthogonal generalized derivations.
Definition. Two generalized derivations (D, d) and (G, g) of R are called
orthogonal if
D(x)RG(y) = 0 = G(y)RD(x)for all x, y ∈ R.

The following example shows that there are many pairs of generalized
derivations which are orthogonal.


x y
Example 1. Let R =: x, y, z ∈ Z, the set of integers . Let m,0 z
p and s be fixed two nonzero elements of Z and the additive maps D, G, dand g
define thefollowing; 
x y0 mx + mz
x y0 mx − mz
D=, d=,0 z000 z00  x y0 pz − ysx y0 −ys
G=and g=. Then0 z000 z00
it is easy to see that d and g are derivations of R and that (D,d) and (G,g)
are a generalized derivation on R such that (D,d) and (G,g) are orthogonal.
Now, to obtain the main result, we need the following lemmas:
Lemma 1. ([10], Lemma 1). Let R be a 2-torsion free semiprime ring, I anonzero ideal of R and a, b the elements of R. Then the following conditions
are equivalent.
(i) axb = 0 for all x ∈ I.
(ii) bxa = 0 for all x ∈ I.
(iii) axb + bxa = 0 for all x ∈ I.
Moreover, if one of the three conditions is fulfilled and `(I) = 0, then ab =ba = 0.

ON IDEALS AND ORTHOGONAL GENERALIZED DERIVATIONS

55

Lemma 2. ([10], Lemma 3). Let R be a semiprime ring and I be a nonzero
ideal of R. Suppose that additive mappings F and H of R into itself satisfy
F (x)IH(x) = 0 for all x ∈ I. Then F (x)IH(y) = 0 for all x, y ∈ I.
3. The Results
The main goal of this section is to prove the following theorem, whichcorresponds to [[3], Theorem 1].
Theorem 1. Let (D, d) and (G, g) be generalized derivations of R and I bea nonzero ideal of R such that `(I) = 0. Then the following conditions areequivalent.
(i) (D, d) and (G, g) are orthogonal.
(ii) For all x, y ∈ I, the following relations hold.
(a) D(x)G(y) + G(x)D(y) = 0.
(b) d(x)G(y) + g(x)D(y) = 0.
(iii) D(x)G(y) = d(x)G(y) = 0 for all x, y ∈ I.
(iv) D(x)G(y) = 0 for all x, y ∈ I and dG(x) = dg(x) = 0 for all x, y ∈ I.
(v) (DG, dg) is a generalized derivation on I and D(x)G(y) = 0 for allx, y ∈ I.
For the proof of the Theorem 1 we need the following lemmas. In all that
follows x, y, z ∈ I and r, s, t ∈ R.
Lemma 3. Let (D, d) and (G, g) be generalized derivations of R and `(I) =0. If D(I)IG(I) = 0, then D(R)RG(R) = 0.
Proof. By 0 = D(x)zG(y) = G(y)zD(x) for all x, y, z ∈ I and Lemma
1, we have 0 = D(x)g(r) = g(r)D(x) and by g(r)D(x) = 0, we get 0 =g(r)d(s) = d(s)g(r). Using these relations, we have D(s)xg(r) = 0 and soby 0 = D(xz)G(y), we obtain d(z)G(y) = 0. Therefore 0 = D(rx)G(sy) =D(r)xG(s)y, which shows D(r)xG(s) = 0. Replace x by r 0 G(s)xD(r)r 0 forsome r 0 ∈ R, we have D(r)r 0 G(s) = 0, as desired.
Moreover, we have the following:
Lemma 4. Let (D, d) and (G, g) be generalized derivations of R and I an
ideal of R such that `(I) = 0. Then the following conditions are equivalent.
(i) For any x, y ∈ I, the following relations hold.
(a) D(x)G(y) + G(x)D(y) = 0.
(b) d(x)G(y) + g(x)D(y) = 0.
(ii) D(x)G(y) = d(x)G(y) = 0 for all x, y ∈ I.
(iii) D(x)G(y) = 0 for all x, y ∈ I and dG = dg = 0 for all x, y ∈ I.
(iv) (DG, dg) is a generalized derivation from I to R and D(x)G(y) = 0
for all x, y ∈ I.

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E. ALBAŞ

Proof. (i) ⇔ (ii). By (a), (b) Lemmas 1 and 2, we have 0 = D(x)zG(y) =D(x)G(y) and using this d(z)G(y) = 0. This shows (ii). And the converse
is easily obtained by the relations D(x)G(y) = G(y)D(x) = 0 and Lemma 1.
(ii) ⇒ (iii). By assumption, D(x)zG(y) = d(x)zg(y) = 0. Then by
Lemma 3, d and g are orthogonal, which shows dg = 0. Moreover, by
0 = d(x)G(y) and Lemma 1, we have 0 = d(d(r)sG(y)) = d(r)sdG(y) forr, s ∈ R. Take r = G(y), we have dG(y) = 0. Since D(x)G(y) = G(y)d(x) =d(x)g(y) = 0, using Lemma 3 we obtain dG(r) = 0, this gives (iii).
(iii) ⇒ (iv). By dG = dg = 0, we haveG(x)d(y) + d(x)g(y) = 0 = g(x)d(y) + d(x)g(y).
Then by the proof of (i) ⇒ (ii), we see that d(x)g(y) = 0 and so G(x)d(y) =0. Moreover, by 0 = D(x)G(y), we have D(x)g(z) = 0. Therefore DG(xy) =DG(x)y and thus (DG, dg = 0) is a generalized derivation from I to R.
(iv) ⇒ (ii). (DG, dg) is a generalized derivation if and only if
G(x)d(y) + D(x)g(y) = 0 = d(x)g(y) + g(x)d(y).
So we obtain dg = 0. Furthermore by 0 = D(x)G(y), we get D(x)g(y) = 0
and by the above relation, we see G(x)d(y) = 0. Therefore G(x)zd(y) = 0
and by Lemma 1, we arrive at d(y)G(x) = 0. This shows (ii).
Using Lemma 3 and Lemma 4, the proof of Theorem 1 is easily seen asfollows:
Proof of Theorem 1. (i) ⇒ (ii), (iii), (iv) and (v) are clear by [[1], Theorem
1]. Since (ii), (iii), (iv) and (v) are equivalent by Lemma 4, we assume (iii).
This implies that 0 = (D(x)z + xd(z))G(y) = D(x)zG(y). Then we have
D(I)IG(I) = 0. Thus by Lemma 3, we have Theorem 1 (iii) ⇒ (i).
Remark 1. If (DG, dg) is a generalized derivations on I and `(I) = 0 then
(DG, dg) is a generalized derivations on R.
Proof. It is easily seen that (DG, dg) is a generalized derivations on I if
and only if
G(x)d(y) + D(x)g(y) = 0,d(x)g(y) + g(x)d(y) = 0.

Then by the second relation, we have d and g are orthogonal. By the first
relation 0 = G(x)d(y) + D(x)g(y), we get 0 = G(x)zd(y) + D(x)zg(y).
Hence replacing z by g(y)z in this relation and using the orthogonality

ON IDEALS AND ORTHOGONAL GENERALIZED DERIVATIONS

57

of the derivations d and g, we obtain 0 = D(x)g(y)zg(y) which implies
that D(x)g(y) = G(x)d(y) = 0. Moreover by 0 = D(x)g(yr), we get
0 = D(x)g(r) for all r ∈ R. Using this relation we have D(s)xg(r) = 0and similarly we can see that D(s)g(r) = G(s)d(r) = 0. Thus we obtainDG(rs) = DG(r)s for all r, s ∈ R which completes the proof.
Theorem 2. Let (D, d) and (G, g) be generalized derivations of R and`(I) = 0. Then the following conditions are equivalent.
(i) (DG, dg) is a generalized derivation on I.
(ii) (GD, gd) is a generalized derivation on I.
(iii) D and g are orthogonal, and G and d are orthogonal.
The proof of the Theorem 2 is clear by Remark 1 and [[1], Theorem 2].
Corollary 1. Let (D, d) be a generalized derivations of R and `(I) = 0. If(D 2 , d2 ) is a generalized derivation on I, then d = 0.
Proof. The fact that (D 2 , d2 ) is a generalized derivation on I is implies
that d and d are orthogonal. Therefore we get d = 0 by the semiprimenessof R.
Corollary 2. Let (D, d) be a generalized derivations of R and `(I) = 0. If
D(x)D(y) = 0 for all x, y ∈ I, then D = d = 0.
Proof. By the hypothesis we have 0 = D(x)D(yz) = D(x)D(y)z +D(x)yd(z) = D(x)yd(z) for all x, y, z ∈ I. In particular, we see thatd(z)D(x) = 0 for all x, z ∈ I by Lemma 1. Replacing x by xy in the last
relation we get 0 = d(z)D(x)y + d(z)xd(y) = d(z)xd(y) for all x, y, z ∈ I.
By [[10], Lemma 2, (a) and (b)], we obtain d(s)Rd(r) = 0 for all s, r ∈ R.
In particular d(s)Rd(s) = 0 for all s ∈ R. Thus d = 0 by the semiprimeness of R. Then we have 0 = D(xz)D(y) = D(x)zD(y) for all x, y, z ∈ I.
By Lemma 3, we arrive at D(r)RD(s) = 0 for all r, s ∈ R. In particular,D(r)RD(r) = 0 for all r ∈ R which implies D = 0, as desired.
Acknowledgement. The author would like to thank the referee forhis/her valuable suggestions which simplify some arguments of this paper.
References[1] N. Argaç, A. Nakajima and E. Albaş: On Orthogonal generalized derivations ofsemiprime rings, Turk J. Math. 28 (2004), 185-194.
[2] M. Bres̆ar: On the distance of the compositions of two derivations to the generalizedderivations, Glasgow Math. J. 33 (1991), 80-93.

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E. ALBAŞ

[3] M. Bres̆ar and J. Vukman: Orthogonal derivation and extension of a theorem of
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Ege University,
Department of Mathematics,
Science Faculty
35100 Bornova, Izmir-Turkey
e-mail address: emine.albas@ege.edu.tr, eminealbas@yahoo.com(Received July 29, 2005 )