A result on generalized derivations on right ideals of prime rings
Let $R$ be a prime ring of characteristic not 2 and let $I$ be a nonzero right ideal of $R$. Let $U$ be the right Utumi quotient ring of $R$ and let $C$ be the center of $U$. If $G$ is a generalized derivation of $R$ such that $[[G(x), x], G(x)] = 0$ for all $x \in I$, then $R$ is commutative or t...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508479144329216 |
|---|---|
| author | Argaç, N. Demir, Ç. Аргац, Н. Демір, Ц. |
| author_facet | Argaç, N. Demir, Ç. Аргац, Н. Демір, Ц. |
| author_sort | Argaç, N. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:29:46Z |
| description | Let $R$ be a prime ring of characteristic not 2 and let $I$ be a nonzero right ideal of $R$.
Let $U$ be the right Utumi quotient ring of $R$ and let $C$ be the center of $U$. If $G$ is a generalized derivation of $R$ such that $[[G(x), x], G(x)] = 0$ for all $x \in I$, then $R$ is commutative or there exist $a, b \in U$ such that $G(x) = ax + xb$ for all $x \in R$ and one of the following assertions is true:
$$(1)\quad (a - \lambda)I = (0) = (b + \lambda)I \;\;\text{for some}\; \lambda \in C,$$
$$(2)\quad (a - \lambda)I = (0) \;\;\text{for some}\; \lambda \in C \;\;\text{and}\; b \in C.$$ |
| first_indexed | 2026-03-24T02:25:51Z |
| format | Article |
| fulltext |
UDC 512.5
Ç. Demir, N. Argaç (Ege Univ., Izmir, Turkey)
A RESULT ON GENERALIZED DERIVATIONS
ON RIGHT IDEALS OF PRIME RINGS
(ОДИН РЕЗУЛЬТАТ) ПРО УЗАГАЛЬНЕНЕ ДИФЕРЕНЦIЮВАННЯ
НА ПРАВИХ IДЕАЛАХ ПРОСТИХ КIЛЕЦЬ
Let R be a prime ring of characteristic not 2 and let I be a nonzero right ideal of R. Let U be the right Utumi quotient
ring of R and let C be the center of U. If G is a generalized derivation of R such that [[G(x), x], G(x)] = 0 for all x ∈ I,
then R is commutative or there exist a, b ∈ U such that G(x) = ax+ xb for all x ∈ R and one of the following assertions
is true:
(1) (a− λ)I = (0) = (b+ λ)I for some λ ∈ C,
(2) (a− λ)I = (0) for some λ ∈ C and b ∈ C.
Нехай R — просте кiльце, характеристика якого не дорiвнює 2, а I — ненульовий правий iдеал R. Нехай U —
праве фактор-кiльце Утумi кiльця R, а C — центр U. Якщо G є узагальненим диференцiюванням R таким, що
[[G(x), x], G(x)] = 0 для всiх x ∈ I, то R є комутативним або iснують a, b ∈ U такi, що G(x) = ax + xb для всiх
x ∈ R i виконується одне з наступних тверджень:
(1) (a− λ)I = (0) = (b+ λ)I для деякого λ ∈ C,
(2) (a− λ)I = (0) для деяких λ ∈ C та b ∈ C.
1. Introduction. Throughout this paper R will always denote a prime ring with center Z(R),
extended centroid C, right Utumi quotient ring U (sometimes, as in [2], U is called the maximal
right ring of quotients), and two-sided Martindale quotient ring Q (see [2] for the definitions). For
any x, y ∈ R, the commutator of x and y is denoted by [x, y] and defined to be xy − yx.
An additive mapping d from R into itself is called a derivation of R if d(xy) = d(x)y + xd(y)
holds for all x, y ∈ R. An additive mapping g : R → R is called a generalized derivation of R if
there exists a derivation d of R such that g(xy) = g(x)y + xd(y) for all x, y ∈ R [10]. Obviously
any derivation is a generalized derivation. Moreover, other basic examples of generalized derivations
are the mappings of the form x 7→ ax + xb, for a, b ∈ R. A generalized derivation in this form is
called (generalized) inner. Many authors have studied generalized derivations in the context of prime
and semiprime rings (see [1, 10, 13, 14]).
In [13], T. K. Lee extended the definition of a generalized derivation as follows. By a generalized
derivation he means an additive mapping g : I → U such that g(xy) = g(x)y+xd(y) for all x, y ∈ I,
where I is a dense right ideal of the prime ring R and d is a derivation from I into U. He also proved
that every generalized derivation can be uniquely extended to a generalized derivation of U, and
moreover, there exist a ∈ U and a derivation d of U such that g(x) = ax+ d(x) for all x ∈ U [13]
(Theorem 3).
In [7], De Filippis proved that if R is a prime ring of characteristic not 2 and G is a generalized
derivation of R such that [[G(x), x], G(x)] = 0 for all x ∈ R, then either R is commutative or there
exists λ ∈ C such that G(x) = λx for all x ∈ R. In the same paper, he uses his result to prove a
theorem concerning noncommutative Banach algebras. More precisely, he proves the following:
c© Ç. DEMIR, N. ARGAÇ, 2012
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2 165
166 Ç. DEMIR, N. ARGAÇ
Let R be a noncommutative Banach algebra with a continuous generalized derivation G = La+d,
where La denotes the left multiplication by a ∈ R and d is a derivation of R. If [[G(x), x], G(x)] ∈
∈ rad(R) (the Jacobson radical of R) for all x ∈ R, then [a,R] ⊆ rad(R) and d(R) ⊆ rad(R).
In [6], V. De Filippis and M. S. Tammam El-Sayiad considered this time a similar problem on
a non-central Lie ideal L of a prime ring R of characteristic not 2. It was proved that if G is a
generalized derivation of R such that [[G(u), u], G(u)] ∈ Z(R) for all u ∈ L, a non-central Lie ideal
of R, then either there exists λ ∈ C such that G(x) = λx for all x ∈ R or G(x) = ax+ xa+ λx for
all x ∈ R and for some a ∈ U, λ ∈ C and R satisfies the standard identity s4.
The aim of the present paper is to extend Filippis’ main result in [7] to the right ideals in prime
rings. Precisely, we will prove the following theorem.
Main theorem. Let R be a prime ring of characteristic different from 2 with the extended
centroid C and I be a nonzero right ideal of R. If G is a generalized derivation of R such that
[[G(x), x], G(x)] = 0
for all x ∈ I, then R is commutative or there exist a, b ∈ U such that G(x) = ax+ xb for all x ∈ R
and one of the following holds:
(i) (a− λ)I = (0) = (b+ λ)I for some λ ∈ C,
(ii) (a− λ)I = (0) for some λ ∈ C and b ∈ C.
Before we proceed, we give some illustrative examples.
Example 1. Let R = Mn(F ) be the ring of all (n × n)-matrices over a field F, and I be the
right ideal of R generated by the matrix unit e11, that is I = e11R. We note that the extended centroid
C of R coincides with its center Z(R) = F which consists of all scalar matrices (here we identify
F with the set of all scalar matrices up to isomorphism).
1. Let a, b ∈ R be such that ai1 = 0 = bi1 for all 2 ≤ i ≤ n and a11 = λ = −b11. Then
(a − λ)I = (0) = (b + λ)I (here of course we identify λ with the scalar matrix λ · 1). Define the
generalized derivation of R by G(r) = ar + rb for all r ∈ R. Then
[[G(x), x], G(x)] = [[ax+ xb, x], ax+ xb] =
= [[x(b+ λ), x], x(b+ λ)] = [−x2(b+ λ), x(b+ λ)] = 0
for all x ∈ I.
2. Let c, d ∈ R with d ∈ Z(R) and ci1 = 0 for all 2 ≤ i ≤ n, c11 = λ. Define now G(r) =
= cr + rd = (c+ d)r for all r ∈ R. Then since (c− λ)I = (0) and d ∈ Z(R), it is readily verified
that
[G(x), x] = [cx+ xd, x] = [λx, x] = 0
for all x ∈ I, and hence [[G(x), x], G(x)] = 0 follows.
2. Preliminaries. In what follows, R will be a prime ring. The related object we need to mention
is the right Utumi quotient ring U of R. The definitions, the axiomatic formulations and the properties
of this quotient ring U can be found in [2].
In any case, when R is a prime ring, all we need to know about U is that
(1) R ⊆ U ;
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A RESULT ON GENERALIZED DERIVATIONS ON RIGHT IDEALS OF PRIME RINGS 167
(2) U is a prime ring;
(3) The center of U, denoted by C, is a field which is called the extended centroid of R.
We will make a frequent use of the theory of generalized polynomial identities and differential
identities (see [2, 11, 12, 15]). In particular we need to recall the following:
Remark 1 [4]. If R is a prime ring and I is a non-zero right ideal of R, then I, IR and IU
satisfy the same generalized polynomial identities with coefficients in U.
Remark 2 [11]. Let R be a prime ring, d a nonzero derivation of R and I a nonzero two-sided
ideal of R. Let f(x1, . . . , xn, d(x1), . . . , d(xn)) be a differential identity in I, that is
f(r1, . . . , rn, d(r1), . . . , d(rn)) = 0
for all r1, . . . , rn ∈ I. Then one of the following holds:
(i) d is an inner derivation of Q, in the sense that there exists q ∈ Q such that d(x) = [q, x] for
all x ∈ R, and I satisfies the generalized polynomial identity
f(r1, . . . , rn, [q, r1], . . . , [q, rn]),
(ii) I satisfies the generalized polynomial identity
f(x1, . . . , xn, y1, . . . , yn).
We also need to mention the following fact about generalized polynomials. It enables us to decide
whether a given generalized identity of a prime ring is a trivial identity or not.
Remark 3. Denote by T = U ∗C C{X} the free product over C of the C-algebra U and
the free C-algebra C{X}, with X a countable set consisting of non-commuting indeterminates
{x1, . . . , xn, . . .}. The elements of T are called generalized polynomials with coefficients in U. Let
a1, . . . , ak ∈ U be linearly C-independent, and
a1g1(x1, . . . , xn) + . . .+ akgk(x1, . . . , xn) = 0 ∈ T
for some g1, . . . , gk ∈ T. If gi(x1, . . . , xn) =
∑n
j=1
xjhj(x1, . . . , xn) and hj ∈ T, then g1, . . . , gk
are the zero element of T. The conclusion holds if
g1(x1, . . . , xn)a1 + . . .+ gk(x1, . . . , xn)ak = 0 ∈ T
and gi(x1, . . . , xn) =
∑n
j=1
hj(x1, . . . , xn)xj for hj ∈ T (see [4]).
2. Results. We start with an easy lemma that will be used in the sequel.
Lemma 1. Let R be a prime ring, I a nonzero right ideal of R. If a ∈ R is such that [ax, x] = 0
for all x ∈ I, then (a− λ)I = (0) for some λ ∈ C.
Proof. Linearizing [ax, x] = 0, one gets
[a, x]y + [a, y]x = 0 (1)
for all x, y ∈ I. Letting y = yr in (1) with r ∈ R and using (1) again, it follows
[a, y][x, r] = y[a, r]x. (2)
Letting now x = xs in (2) with s ∈ R, we get [a, I]I[R,R] = (0). Hence [a, I]I = (0) or R is
commutative. Of course [a, I]I = (0) if R is commutative. Then (a− λ)I = (0) for some λ ∈ C by
[3] (Lemma).
The following lemma is crucial and will be used in the proof of the inner case.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
168 Ç. DEMIR, N. ARGAÇ
Lemma 2. Let R be a prime ring of characteristic different from 2, I a nonzero right ideal of
R and a, b ∈ R.
(i) If [[ax, x], ax] = 0 for all x ∈ I, then (a− λ)I = (0) for some λ ∈ C.
(ii) If [[xb, x], xb] = 0 for all x ∈ I, then bI = (0) or b ∈ Z(R).
Proof. (i) By the hypothesis
[[ax, x], ax] = 0 (3)
for all x ∈ I. By Theorem 2 in [4] we see that (1) holds for all x ∈ IU. Replacing R and I with
U and IU respectively, we may assume that IC = I and R is centrally closed over its center C. In
case C is infinite, set R = R⊗C C and I = I ⊗C C where C is the algebraic closure of C. Then R
is centrally closed over its center C by [8], and (3) holds for all x ∈ I by a standard argument. Thus,
replacing R, I and C with R, I and C respectively, we may assume further that C is either finite or
algebraically closed. We proceed to show that (a− λ)I = (0) for some λ ∈ C.
Let u ∈ I, then
[[aux, ux], aux] = 0
for all x ∈ R. Assume on the contrary that au and u are C-independent for some u ∈ I. We claim
that
[[auX, uX], auX] (4)
is a non-trivial generalized polynomial identity (GPI for short) for R. For otherwise,
au(XuXauX −XauXuX +XuXauX)− u(XauXauX)
is the zero element of T = U ∗C C{X}. Then by Remark 3
uXauXauX = 0 ∈ T = U ∗C C{X}
implying au = 0, contrary to our assumption on au and u. Therefore (4) is a nontrivial GPI for R.
Thus R is a primitive ring with a nonzero socle soc(R) = H with C as the associated division ring
by Martindale’s theorem [15]. Now I and IH both satisfy (3), and so replacing I with IH, we may
assume that I ⊆ H.
Let e = e2 ∈ I be any idempotent. Then
[[aere, ere], aere] = 0 (5)
for all r ∈ R. Left multiplying (5) by e yields that
[[(eae)(ere), (ere)], (eae)(ere)] = 0
for all r ∈ R. Since eRe is a prime ring, char(eRe) = char(R) 6= 2 and eae ∈ eRe, we conclude
that either eRe is commutative or eae ∈ Z(eRe) = Ce by [7] (Proposition 1). In any case we have
eae ∈ Ce. On the other hand,
[[aer(1− e), er(1− e)], aer(1− e)] = 0
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
A RESULT ON GENERALIZED DERIVATIONS ON RIGHT IDEALS OF PRIME RINGS 169
for all r ∈ R. Expanding the commutator we arrive at
er(1− e)aer(1− e)aer(1− e) = 0
for all r ∈ R. Therefore ((1 − e)aer)4 = 0 for all r ∈ R, and so (1 − e)aeR is a nil right ideal of
bounded index. Hence (1− e)ae = 0 by Levitzki’s theorem [9] (Lemma 1.1). Now ae = eae ∈ Ce
for every idempotent e ∈ I. Since I is completely reducible right H-module, every element of I is
contained in fH for some f = f2 ∈ I. Then, for any x ∈ I, there exists an idempotent f ∈ I such
that x = fx, and so, it follows that
ax = afx = fafx ∈ Cfx = Cx.
Hence we see that [ax, x] = 0 for all x ∈ I, and then by Lemma 1 we have (a−λ)I = (0) for some
λ ∈ C.
(ii) Even if the proof of this part is very similar to the one in (i), we give its proof here for the
sake of completeness.
We now have
[[xb, x], xb] = 0 (6)
for all x ∈ I by the hypothesis. Again by Theorem 2 in [4] we see that (6) holds for all x ∈ IU.
Replacing R and I with U and IU respectively, we may assume that IC = I and R is centrally
closed over its center C. As in (i) replacing R, I and C with R, I and C respectively, when C is
infinite, we may assume further that C is either finite or algebraically closed.
Let u ∈ I, then
[[uxb, ux], uxb] = 0 (7)
for all x ∈ R. Assume on the contrary that b /∈ C and bI 6= (0). Then there exists u ∈ I such that
bu 6= 0. We claim that
[[uXb, uX], uXb]
is a non-trivial GPI for R. If not,
(uXbuXuX − uXuXbuX + uXbuXuX)b− (uXbuXbuX)
is the zero element of T = U ∗C C{X}. Then by Remark 3 again,
uXbuXbuX = 0 ∈ T = U ∗C C{X},
and hence bu = 0, contrary to our assumption. Therefore (7) is a non-trivial GPI for R. In the
present case, R is a primitive ring with a nonzero socle Soc(R) = H [15]. Moreover, since (6) is
also satisfied by IH, we may assume further that I ⊆ H by replacing I with IH. Similar to above,
let e = e2 ∈ I be an idempotent. Then
[[ereb, ere], ereb] = 0 (8)
for all r ∈ R. Right multiplying (8) by e yields that
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
170 Ç. DEMIR, N. ARGAÇ
[[(ere)(ebe), (ere)], (ere)(ebe)] = 0
for all r ∈ R. Since eRe is a prime ring, char(eRe) = char(R) 6= 2 and ebe ∈ eRe, we conclude
that either eRe is commutative or ebe ∈ Z(eRe) = Ce by [7] (Proposition 1). In any case we have
ebe ∈ Ce. On the other hand,
[[er(1− e)b, er(1− e)], er(1− e)b] = 0
for all r ∈ R. Expanding the commutator we arrive at
er(1− e)ber(1− e)ber(1− e) = 0
for all r ∈ R. Therefore (1− e)beR is a nil right ideal of bounded index. Hence (1− e)be = 0 again
by Levitzki’s theorem [9] (Lemma 1.1). Thus, be = ebe ∈ Ce for every idempotent e ∈ I. Since I is
completely reducible right H-module, every element of I is contained in fH for some f = f2 ∈ I.
Then, for any x ∈ I, there exists an idempotent f ∈ I such that x = fx. Therefore, it follows that
bx = bfx = fbfx ∈ Cfx = Cx
for all x ∈ I. Hence we see that [bx, x] = 0 for all x ∈ I, and so (b− µ)I = (0) for some µ ∈ C by
Lemma 1. Now (6) reduces to
0 = [[xb, x], xb] = x3µ(b− µ)
for all x ∈ I. In particular, eµ(b − µ) = 0 and (e + er(1 − e))µ(b − µ) = 0 for all e = e2 ∈ I and
r ∈ R. This implies eRµ(b− µ) = 0, that is to say µ = 0 or b = µ ∈ C. We must have µ = 0 since
b /∈ C. But then bI = (0), again a contradiction.
Lemma 3. Let R be a prime ring of characteristic different from 2, I a nonzero right ideal of
R and a, b ∈ R. If
[[ax+ xb, x], ax+ xb] = 0 (9)
for all x ∈ I, then one of the following holds:
(i) (a− λ)I = (0) = (b+ λ)I for some λ ∈ C,
(ii) (a− λ)I = (0) for some λ ∈ C and b ∈ Z(R).
Proof. Let u ∈ I. Then
[[aux+ uxb, ux], aux+ uxb] = 0 (10)
for all x ∈ R, and hence for all x ∈ U. Replacing R and I with U and IU, we may assume that C is
just the center of R. We want to show that either R is a GPI-ring or the lemma holds. Therefore we
assume that R is not a GPI-ring. Assume further that au and u are C-independent for some u ∈ I.
Then R satisfies
[[auX + uXb, uX], auX + uXb].
Expansion of (10) yields that
auf(x) + ug(x) = 0
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A RESULT ON GENERALIZED DERIVATIONS ON RIGHT IDEALS OF PRIME RINGS 171
for all x ∈ R, where
f(x) = 2xuxaux+ 2xuxuxb− xauxux− xuxbux
and
g(x) = 2xbuxaux+ 2xbuxuxb− xauxaux− xauxuxb− xuxbaux−
−xuxbuxb− xbauxux− xbuxbux.
Since R satisfies no non-trivial GPI, we must have
auf(X) = 0 ∈ T = U ∗C C{X}
by Remark 3. Hence
2auXuXauX + 2auXuXuXb− auXauXuX − auXuXbuX (11)
is the zero element of T = U ∗C C{X}. If now 1 and b are C-dependent, that is b ∈ C, then (9)
reduces to
[[(a+ b)x, x], (a+ b)x] = 0
for all x ∈ I. It follows from Lemma 2(i) that (a+b−α)I = (0) for some α ∈ C. Set λ = α−b ∈ C,
and so (a− λ)I = (0) for some λ ∈ C and b ∈ Z(R) (since b ∈ R). This gives (ii).
Therefore we may assume that 1 and b are C-independent. We rewrite (11) in the form
(2a(uX)2auX − auXa(uX)2 − a(uX)2buX) + (2a(uX)3)b = 0 ∈ T.
We conclude as above that 2a(uX)3b = 0 which is impossible unless charR = 2 or b = 0 or au = 0,
a contradiction. Until now we have shown that if au and u are C-independent for some u ∈ I, then
either the lemma holds or R is a GPI-ring. So we may assume that au and u are C-dependent for all
u ∈ I. Then [au, u] = 0 for all u ∈ I, and this implies (a−λ)I = (0) for some λ ∈ C by Lemma 1.
Now (9) reduces to
[[x(b+ λ), x], x(b+ λ)] = 0
for all x ∈ I. Hence by Lemma 2(ii), we have b ∈ C = Z(R) or (b+ λ)I = (0), giving (i) and (ii)
simultaneously.
We are now in a position to consider the case when R is a GPI-ring. Then R is a primitive
ring with a nonzero socle H with C as the associated division ring by Martindale’s theorem [15].
Moreover, since I and IH both satisfy (9), after replacing I with IH we may assume that I ⊆ H.
Let e = e2 ∈ I be any idempotent element. Then
[[aere+ ereb, ere], aere+ ereb] = 0 (12)
for all r ∈ R. Now left and right multiplying (12) by 1− e yields that
2(1− e)aererereb(1− e) = 0,
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
172 Ç. DEMIR, N. ARGAÇ
and so
(1− e)aererereb(1− e) = 0
for all r ∈ R since char(R) 6= 2. It follows by the primeness of R that (1−e)ae = 0 or eb(1−e) = 0
by the Theorem in [16]. If (1− e)ae = 0, then right multiplication of (12) by e yields[
[(eae)(ere) + (ere)(ebe), ere], (eae)(ere) + (ere)(ebe)
]
= 0 (13)
for all r ∈ R. Similarly, if eb(1− e) = 0, then the left multiplication of (12) by e gives us the same
identity in (13). Thus in any case we have[
[a′x+ xb′, x], a′x+ xb′
]
= 0 (14)
for all x ∈ eRe, where a′ = eae and b′ = ebe. Since eRe is a prime ring, char(eRe) = char(R) 6= 2
and a′, b′ ∈ eRe, (14) implies that either eRe is commutative or a′, b′ ∈ Z(eRe) = Ce by [7]
(Proposition 1). In any case we have a′, b′ ∈ Ce.
Now we claim that for a given e = e2 ∈ I, if eb(1− e) = 0, then we must have (1− e)ae = 0,
too. So assume on the contrary that eb(1 − e) = 0 but (1 − e)ae 6= 0 for some e = e2 ∈ I. Pick
any α ∈ C, r ∈ R and set q = αer(1− e). Then q2 = 0 and the mapping ϕ(x) = (1 + q)x(1− q),
x ∈ R, defines a C-automorphism of R such that ϕ(I) ⊆ I. Thus[
[ϕ(a)x+ xϕ(b), x], ϕ(a)x+ xϕ(b)
]
= 0 (15)
for all x ∈ I. As above (15) implies that (1− e)ϕ(a)e = 0 or eϕ(b)(1− e) = 0. If (1− e)ϕ(a)e = 0,
then one gets that
0 = (1− e)ϕ(a)e = (1− e)ae
which is a contradiction. So we must have eϕ(b)(1− e) = 0. By calculation we arrive at
α2er(1− e)ber(1− e) + αeber(1− e)− αer(1− e)b(1− e) = 0. (16)
In particular, taking α = 1 in (16) it follows that
er(1− e)ber(1− e) + eber(1− e)− er(1− e)b(1− e) = 0.
In a similar fashion, taking this time α = −1 in (16) one gets
er(1− e)ber(1− e)− eber(1− e) + er(1− e)b(1− e) = 0.
Comparing these last two equations and using the fact that char(R) 6= 2, we obtain
er(1− e)ber(1− e) = 0
for all r ∈ R. Hence (1− e)be = 0, and so
eb = ebe = be.
Let s ∈ R and f = e + es(1 − e) ∈ I. We note that (1 − f)af 6= 0, and so we must have
fb(1− f) = 0. But this implies bf = fb as above. Hence
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
A RESULT ON GENERALIZED DERIVATIONS ON RIGHT IDEALS OF PRIME RINGS 173[
b, e+ es(1− e)
]
= 0 (17)
for all s ∈ R. Now (17) implies b ∈ C by [5] (Lemma 1). So (9) reduces to[
[(a+ b)x, x], (a+ b)x
]
= 0
for all x ∈ I. Then for any r ∈ R, we have
0 = [[(a+ b)er(1− e), er(1− e)], (a+ b)er(1− e)],
that is
er(1− e)aer(1− e)aer(1− e) = 0.
Therefore (1− e)ae = 0 which is a contradiction. This proves our claim. So we have (1− e)ae = 0,
that is ae = eae ∈ Ce for all e = e2 ∈ I. Then since I is completely reducible right H-module,
every element of I is contained in fH for some idempotent f ∈ I. Let x ∈ I, then fx = x for some
f = f2 ∈ I. Hence
ax = afx = fafx ∈ Cfx = Cx.
This means [ax, x] = 0 for all x ∈ I, and therefore (a − λ)I = (0) for some λ ∈ C by Lemma 1.
From (9) we see that [
[x(b+ λ), x], x(b+ λ)
]
= 0
for all x ∈ I. Henceforth we have (b + λ)I = (0) or b ∈ Z(R) by Lemma 2(ii). This proves the
lemma.
We are now ready to prove our main theorem.
Main theorem. Let R be a prime ring of characteristic different from 2 with the extended
centroid C and I be a nonzero right ideal of R. If G is a generalized derivation of R such that[
[G(x), x], G(x)
]
= 0 (18)
for all x ∈ I, then R is commutative or there exist a, b ∈ U such that G(x) = ax+ xb for all x ∈ R
and one of the following holds:
(i) (a− λ)I = (0) = (b+ λ)I for some λ ∈ C,
(ii) (a− λ)I = (0) for some λ ∈ C and b ∈ C.
Proof. As we have already noted that every generalized derivation G on a dense right ideal of R
can be uniquely extended to U and assumes form G(r) = pr+d(r) for some p ∈ U and a derivation
d of U. Then [
[px+ d(x), x], px+ d(x)
]
= 0 (19)
for all x ∈ I, and hence for all x ∈ IU since I and IU satisfy the same differential identities [12].
If d = 0, then we get that [
[px, x], px
]
= 0
for all x ∈ IU. This last equation implies that (p − λ)IU = (0) for some λ ∈ C by Lemma 2(i).
Therefore g(r) = ar for all r ∈ R and (a− λ)I = (0) where a = p. So we may assume that d 6= 0.
In light of Kharchenko’s theorem (Remark 2), we divide the proof into two cases:
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
174 Ç. DEMIR, N. ARGAÇ
Case 1. Let d be the X-inner derivation induced by the element q ∈ U − C. Then by (19) we
see that [
[(p+ q)x− xq, x], (p+ q)x− xq
]
= 0 (20)
for all x ∈ I. As we noted above (20) is also satisfied by IU. Therefore replacing R and I with U
and IU respective, we may assume that p, q ∈ R. Set a = p + q and b = −q for simplicity. Now it
follows from Lemma 3 that either (a− λ)I = (0) = (b+ λ)I for some λ ∈ C or (a− λ)I = (0) for
some λ ∈ C and b ∈ C.
Case 2. Let now d be an outer derivation of U. To continue the proof we first linearize (12). By
replacing x with x+ y in (18) and using (18) again, we end up with
[[G(x), x], G(y)] + [[G(x), y], G(x)] + [[G(y), x], G(x)]+
+[[G(x), y], G(y)] + [[G(y), x], G(y)] + [[G(y), y], G(x)] = 0 (21)
for all x, y ∈ I. Replacing x with −x in (21) and adding up the resulting equation to (21) yields that
[[G(x), x], G(y)] + [[G(x), y], G(x)] + [[G(y), x], G(x)] = 0 (22)
for all x, y ∈ I since charR 6= 2. Take xr instead of x in (22) with r ∈ R to get
[[G(x)r + xd(r), xr], G(y)] + [[G(x)r + xd(r), y], G(x)r + xd(r)]+
+[[G(y), xr], G(x)r + xd(r)] = 0 (23)
for all x, y ∈ I and r ∈ R. By Kharchenko’s theorem, since d is an outer derivation, R satisfies the
identity:
[[G(x)r + xs, xr], G(y)] + [[G(x)r + xs, y], G(x)r + xs] + [[G(y), xr], G(x)r + xs] = 0
for all x, y ∈ I and r, s ∈ R. In particular, R satisfies the blended component
[[xs, y], xs] = 0
for all x, y ∈ I and s ∈ R (and hence for all s ∈ U ). So for s = 1 in this last equation we have
[[x, y], x] = 0 for all x, y ∈ I. Then for any x, y, z ∈ I we have
0 = [[x, yz], x] = 2[x, y][z, x],
and so
[x, y][x, z] = 0
since charR 6= 2. Let now z = zr in this last equation to get
[x, y]z[x, r] = 0
for all x, y, z ∈ I and r ∈ R. Therefore for any x ∈ I, we see that [x, I]I = (0) or x ∈ Z(R). Thus
we conclude that [I, I]I = (0) or R is commutative. If the first possibility holds, then it follows from
[[x, y], x] = 0, x, y ∈ I, that x[x, y] = 0. This clearly implies the commutativity of R, and so the
theorem is proved.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
A RESULT ON GENERALIZED DERIVATIONS ON RIGHT IDEALS OF PRIME RINGS 175
We finish with an example which shows that the characteristic assumption in the theorem cannot
be removed.
Example 2. Let F be a field with charF = 2, R = M2(F ) and a be any element of R. Then
for the mapping G(x) = [a, x], x ∈ R, one can easily see that for every x ∈ R, [[G(x), x], G(x)] =
= [G(x)2, x] = 0 since G(x)2 ∈ Z(R) for all x ∈ R.
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Received 16.02.11,
after revision — 25.01.12
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
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| id | umjimathkievua-article-2563 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:25:51Z |
| publishDate | 2012 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/99/3c988257f53b499528ebd16f95fb4999.pdf |
| spelling | umjimathkievua-article-25632020-03-18T19:29:46Z A result on generalized derivations on right ideals of prime rings (Один результат) про узагальнене диференцiювання на правих iдеалах простих кiлець Argaç, N. Demir, Ç. Аргац, Н. Демір, Ц. Let $R$ be a prime ring of characteristic not 2 and let $I$ be a nonzero right ideal of $R$. Let $U$ be the right Utumi quotient ring of $R$ and let $C$ be the center of $U$. If $G$ is a generalized derivation of $R$ such that $[[G(x), x], G(x)] = 0$ for all $x \in I$, then $R$ is commutative or there exist $a, b \in U$ such that $G(x) = ax + xb$ for all $x \in R$ and one of the following assertions is true: $$(1)\quad (a - \lambda)I = (0) = (b + \lambda)I \;\;\text{for some}\; \lambda \in C,$$ $$(2)\quad (a - \lambda)I = (0) \;\;\text{for some}\; \lambda \in C \;\;\text{and}\; b \in C.$$ Нехай $R$ — просте кiльце, характеристика якого не дорiвнює 2, а $I$ — ненульовий правий iдеал $R$. Нехай $U$ — праве фактор-кiльце Утумi кiльця $R$, а $C$ — центр $U$. Якщо $G$ є узагальненим диференцiюванням $R$ таким, що $[[G(x), x], G(x)] = 0$ для всiх $x \in I$, то $R$ є комутативним або iснують $a, b \in U$ такi, що $G(x) = ax + xb$ для всiх $x \in R$ i виконується одне з наступних тверджень: $$(1)\quad (a - \lambda)I = (0) = (b + A)I \;\;\text{для деякого}\; \lambda \in C,$$ $$(2)\quad (a - \lambda)I = (0) \;\;\text{для деякого}\; \lambda \in C \;\;\text{і}\; b \in C.$$ Institute of Mathematics, NAS of Ukraine 2012-02-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2563 Ukrains’kyi Matematychnyi Zhurnal; Vol. 64 No. 2 (2012); 165-175 Український математичний журнал; Том 64 № 2 (2012); 165-175 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2563/1885 https://umj.imath.kiev.ua/index.php/umj/article/view/2563/1886 Copyright (c) 2012 Argaç N.; Demir Ç. |
| spellingShingle | Argaç, N. Demir, Ç. Аргац, Н. Демір, Ц. A result on generalized derivations on right ideals of prime rings |
| title | A result on generalized derivations on right ideals of prime rings |
| title_alt | (Один результат) про узагальнене диференцiювання на правих iдеалах простих кiлець |
| title_full | A result on generalized derivations on right ideals of prime rings |
| title_fullStr | A result on generalized derivations on right ideals of prime rings |
| title_full_unstemmed | A result on generalized derivations on right ideals of prime rings |
| title_short | A result on generalized derivations on right ideals of prime rings |
| title_sort | result on generalized derivations on right ideals of prime rings |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2563 |
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