Ratio Mathematica Volume 38, 2020, pp. 367-375

On commutativity of prime and semiprime
rings with generalized derivations

Md Hamidur Rahaman*

Abstract

Let R be a prime ring, extended centroid C and m,n,k ≥ 1 are fixed
integers. If R admits a generalized derivation F associated with a
derivation d such that (F(x) ◦ y)m + (x ◦ d(y))n = 0 or (F(x) ◦m
y)k +x◦nd(y)=0 for all x,y ∈ I, where I is a nonzero ideal of R, then
either R is commutative or there exist b ∈ U, Utumi ring of quotient
of R such that F(x) = bx for all x ∈ R. Moreover, we also examine
the case R is a semiprime ring.
Keywords: Prime rings, Semiprime rings, Extended centroid, Utumi
quotient rings, Generalized derivations.
2010 AMS subject classifications: 16W25, 16N60, 16U80. 1

1 Introduction
R is always an associative ring with centre Z(R), extended centroid C and

Utumi quotient ring U. For further information, definitions and properties on
these concepts refer to Beidar and Martindale [1996]. For any x,y ∈ R, the sym-
bols [x,y] and x◦y stand for the Lie commutator xy−yx and Jordan commutator
xy + yx respectively. Given x,y ∈ R, we set x◦0 y = x, x◦1 y = x◦y = xy + yx
and inductively x◦m y = (x◦m−1 y)◦y for m > 2. Recall that a ring R is prime if
for any a,b ∈ R, aRb = {0} implies that either a = 0 or b = 0 and is semiprime
if for any a ∈ R, aRa = {0} implies that a = 0. Every prime ring is a semiprime
ring but semiprime ring need not be prime ring. The socle of a ring R denoted by
Soc(R) is the sum of the minimal left (right) ideals of R, if R has minimal left

*Department of Mathematics, Aligarh Muslim University, Aligarh, India; rahamanhamid-
math@gmail.com.

1Received on February 15th, 2020. Accepted on June 1st, 2020. Published on June 30th, 2020.
doi: 10.23755/rm.v38i0.502. ISSN: 1592-7415. eISSN: 2282-8214. ©Md Hamidur Rahaman
This paper is published under the CC-BY licence agreement.

367



Md Hamidur Rahaman

(right) ideals, otherwise Soc(R) = (0). The goal of this paper is to establish that
there is a relationship between the structure of the ring R and the behaviour of
suitable additive mappings defined on R that satisfy certain special identities. In
particular we study the case when the map is a generalized derivation of R. We re-
call that an additive map d : R → R is a derivation of R if d(xy) = d(x)y +xd(y)
for all x,y ∈ R. In particular, d is an inner derivation induced by an element
a ∈ R, if Ia(x) = [a,x] for all x ∈ R and d is outer derivation if it is not inner
derivation. An additive map F : R → R is said to be a generalized derivation if
there is a derivation d of R such that for all x,y ∈ R, F(xy) = F(x)y + xd(y).
All derivations are generalized derivations but generalized derivation need to be
derivation. During the past few decades, there has been an ongoing interest con-
cerning the relationship between the commutativity of a ring and the existence of
certain specific types of derivations (see Ashraf and Rehman [2002] , where fur-
ther references can be found). In Argaç and Inceboz [2009], Argaç and Inceboz
proved that: If R is a prime ring, I a nonzero ideal of R, k a fixed positive integer
and R admits a nonzero derivation d with the property (d(x◦y))k = x◦y for all
x,y ∈ I, then R is commutative. In [Quadri et al., 2003, Theorem 2.3], Quadri
et al. discussed the commutativity of prime rings with generalized derivations.
More precisely, Quadri et al. proved that if R is a prime ring, I a nonzero ideal of
R and F a generalized derivation associated with a nonzero derivation d such that
F(x◦y) = x◦y for all x,y ∈ I, then R is commutative. In Ashraf and Rehman
[2002], Ashraf and Rehman proved that if R is a 2-torsion free prime ring, I a
nonzero ideal of R and d a nonzero derivation of R such that d(x) ◦d(y) = x◦y
for all x,y ∈ I, then R is commutative.

The present paper is motivated by the previous results and we here generalized
the result obtained in Ashraf and Rehman [2002] . Moreover, we continue this
line of investigation by examining what happens if a ring R satisfies the following
identities:

(i)(F(x) ◦y)m + (x◦d(y))n = 0, (ii)(F(x) ◦m y)k + x◦n d(y) = 0

for all x,y ∈ I, a nonzero ideal of R. We obtain some analogous results for
semiprime ring in the case I = R. More precisely, we shall prove the following
theorems:

Theorem 1.1. Let R be a prime ring, I a nonzero ideal of R and m,n are fixed
positive integers. If R admits a generalized derivation F associated with a nonzero
derivation d such that (F(x) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ I, then either
R is commutative or there exist b ∈ U, Utumi ring of quotient of R such that
F(x) = bx for all x ∈ R.

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On commutativity of prime and semiprime rings with generalized derivations

Theorem 1.2. Let R be a prime ring, I a nonzero ideal of R and m,n,k are fixed
positive integers. If R admits a generalized derivation F associated with a nonzero
derivation d such that (F(x) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ I, then either
R is commutative or there exist b ∈ U, Utumi ring of quotient of R such that
F(x) = bx for all x ∈ R.

Theorem 1.3. Let R be a semiprime ring, U the left Utumi quotient ring of R and
m,n are fixed positive integers. If R admits a generalized derivation F associated
with a nonzero derivation d such that (F(x)◦y)m + (x◦d(y))n = 0 for all x,y ∈
R, then R is commutative.

Theorem 1.4. Let R be a semiprime ring, U the left Utumi quotient ring of R
and m,n,k are fixed positive integers. If R admits a generalized derivation F
associated with a nonzero derivation d such that (F(x) ◦m y)k + x ◦n d(y) =
0 for all x,y ∈ R, then R is commutative.

2 Preliminary results
Before starting our results, we state some well known facts which are very

crucial for developing the proof of our main result. In particular, we will make
frequent use of the following facts.

Fact 2.1. [Lee, 1992, Theorem 2] If I is a two-sided ideal of R, then R,I and U
satisfy the same differential identities.

Fact 2.2. [Lee, 1999, Theorem 4] Let R be a semiprime ring. Then every gen-
eralized derivation F on a dense right ideal of R is uniquely extended to U and
assumes the form F(x) = ax + d(x) for some a ∈ U and a derivation d on U.
Moreover, a and d are uniquely determined by the generalized derivation F .

Fact 2.3. [Chuang, 1988, Theorem 2] If I is a two-sided ideal of R, then R, I and
U satisfy the same generalized polynomial identities with coefficients in U.

Fact 2.4. Let R be a prime ring and d a nonzero derivation on R and I be a
nonzero ideal of R. By Kharchenko’s Theorem [Kharchenko, 1978, Theorem 2], if
I satisfies the differential polynomial identity P(x1,x2, ....,xn,d(x1),d(x2), ....,d(xn)) =
0, then either d is an inner derivation or d is outer derivation and I satisfies the
generalized polynomial identity P(x1,x2, ....,xn,y1,y2, ....,yn) = 0.

Fact 2.5. [Chuang, 1994, Page no. 38] If R is semiprime then so is its left Utumi
quotient ring. The extended centroid C of a semiprime ring coincides with the
center of its left Utumi quotient ring.

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Md Hamidur Rahaman

Fact 2.6. [Lee, 1992, Lemma 2] Any derivation of a semiprime ring R can be
uniquely extended to a derivation of its left Utumi quotient ring U, and so any
derivation of R can be defined on the whole U.

Fact 2.7. [Chuang, 1994, Page no. 42] Let B be the set of all the idempotents in
C, the extended centroid of R. Assume R is a B −algebra orthogonal complete.
For any maximal ideal P of B, PR forms a minimal prime ideal of R, which is
invariant under any nonzero derivation of R.

3 Main Results
Theorem 1.1 Let R be a prime ring, I a nonzero ideal of R and m,n are fixed
positive integers. If R admits a generalized derivation F associated with a deriva-
tion d such that (F(x) ◦ y)m + (x ◦ d(y))n = 0 for all x,y ∈ I, then either
R is commutative or there exist b ∈ U, Utumi ring of quotient of R such that
F(x) = bx for all x ∈ R.

Proof. By hypothesis

(F(x) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ I. (1)

By the Fact 2.1 I, R and U satisfy the same generalized polynomial identity (GPI),
we have

(F(x) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ U. (2)

Since R is a prime ring and F a generalized derivation of R, by Fact 2.2,
F(x) = ax + d(x) for some a ∈ U and a derivation d on U. Then U satisfies

((ax + d(x)) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ U. (3)

That is

(ax◦y + d(x) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ U. (4)

In the light of Kharchenko’s theorem [Kharchenko, 1978, Theorem 2], we divide
the proof into two cases:-
Case I Let d be an inner derivation of U, that is, d(x) = [q,x] for all x ∈ U and
for some q ∈ U. Then U satisfies

F(x,y) = (ax◦y + [q,x] ◦y)m + (x◦ [q,y])n = 0 for all x,y ∈ U. (5)

In case C is infinite, we have F(x,y) = 0 for all x,y ∈ U
⊗

C C̄, where C̄
is the algebraic closure of C. Since both U and U

⊗
C C̄ are prime and centrally

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On commutativity of prime and semiprime rings with generalized derivations

closed Erickson et al. [1975], we may replace R by U or U
⊗

C C̄ according to
C is finite or infinite. Thus we may assume that R is centrally closed over C
which is either finite or algebraically closed and F(x,y) = 0 for all x,y ∈ R.
By Martindale’s theorem [Martindale, 1969, Theorem 3], R is then a primitive
ring having nonzero socle, soc(R) with D as associative division ring. Hence by
Jacobson’s theorem [Jacobson, 1956, p.75], R is isomorphic to dense ring of linear
transformations of vector space V over C. Then the density of R on V implies
that R ∼= Mk(D), where k = dimDV . Assume that dimDV > 2, otherwise we
are done. Suppose that there exists v ∈ V such that v and qv are linearly D-
independent. Since dimDV > 2, then there exists w ∈ V such that {v,qv,w} is
linearly independent over D. By density of R there exist x,y ∈ R such that

xv = 0,xqv = w,yw = v,xw = 0,yv = 0,yqv = v. (6)

Multiplying equation (5) by v from right and using conditions in equation (6), we
get (−1)mv = 0, a contradiction. Now we want to show that qv = vβ for some
β ∈ D. Let v,w be linearly independent. Then by the precedant argument, there
exist βv, βw, βv+w ∈ D, such that qv = vβv, qw = wβw, q(v+w) = (v+w)βv+w.
Moreover, vβv +wβw = (v+w)βv+w and hence v(βv−βv+w)+w(βw−βv+w) = 0.
Since v,w are linearly independent, we have βv = βw = βv+w that is β does not
depend on the choice of v.

Let now for r ∈ R, v ∈ V , by precedant calculation, qv = vα, r(qv) = r(vα)
and also q(rv) = (rv)α. Thus 0 = [q,r]v for any v ∈ V , that is, [q,r]V = 0
. Since V is a left faithful irreducible R -module, [q,r] = 0 for all r ∈ R i.e.,
q ∈ Z(R) and d = 0.
Case 2 Let d be an outer derivation of R. Then by Fact 2.4, I satisfies the gener-
alized polynomial identity

(ax◦y + t◦y)m + (x◦ t)n = 0 for all x,y,t ∈ I. (7)

In particular, for y = 0, I satisfies (xt + tx)n = 0. By Chuang [Chuang, 1988,
Theorem 2], this polynomial identity is also satisfied by Q and hence R as well.
By Lemma 1 Lanski [1993], there exists a field F such that R ⊆ Mk(F), the ring
of k × k matrices over a field F , where k ≥ 1. Moreover, R and Mk(F) satisfy
the same polynomial identity, that is (xt + tx)n = 0 for all t,x ∈ Mk(F). Let eij
be the usual matrix unit with 1 in the (i,j) entry and zero elsewhere. By choosing
t = e12,x = e21, we see that (xt + tx)n = (e11 + e22)n 6= 0, a contradiction.

Theorem 1.2 Let R be a prime ring, I a nonzero ideal of R and m,n,k are fixed
positive integers. If R admits a generalized derivation F associated with a nonzero
derivation d such that (F(x) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ I, then either

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Md Hamidur Rahaman

R is commutative or there exist b ∈ U, Utumi ring of quotient of R such that
F(x) = bx for all x ∈ R.
Proof. By hypothesis

(F(x) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ I. (8)

By the Fact 2.1 I, R and U satisfy the same generalized polynomial identity (GPI),
we have

(F(x) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ U. (9)

By Fact 2.2, F(x) = ax + d(x) for some a ∈ U and derivation d on U. Then U
satisfies

((ax + d(x)) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ U. (10)

That is

(ax◦m y + d(x) ◦m y)k + x◦n d(y) = 0 for all x,y ∈ U. (11)

In the light of Kharchenko’s theorem [Kharchenko, 1978, Theorem 2], we divide
the proof into two cases:-
Case I Let d be an inner derivation of U that is d(x) = [q,x] for all x ∈ U and for
some q ∈ U. Then U satisfies

(ax◦m y + [q,x] ◦m y)k + x◦n [q,y] = 0 for all x,y ∈ U. (12)

As in the proof of Theorem 1.1, we have

(ax◦m y + [q,x] ◦m y)k + x◦n [q,y] = 0 for all x,y ∈ R. (13)

where R is a primitive ring with D as the associated division ring. If V is finite
dimensional over D, then the density of R implies that R ∼= Mk(D), where k =
dimDV . Assume that dimDV > 2, otherwise we are done. Suppose that there
exists v ∈ V such that v and qv are linearly D-independent. Since dimDV > 2,
then there exists w ∈ V such that {v,qv,w} is linearly independent over D. By
density of R there exist x,y ∈ R such that

xv = 0,xqv = w,yw = v,xw = 0,yv = 0,yqv = v.

Multiplying equation (13) by v from right, we get (−1)mkv = 0 which is a contra-
diction to the linearly independent of the set {v,qv}. Therefore, {v,qv} is linearly

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On commutativity of prime and semiprime rings with generalized derivations

dependent and so q ∈ Z(R), i.e, d = 0.
Case 2 Let d be an outer derivation. Then

(ax◦m y + t◦m y)k + x◦n s = 0 for all x,y,s,t ∈ I. (14)

In particular, choosing y = 0, we have s ◦n x = 0. By Chuang [Chuang, 1988,
Theorem 2], this polynomial identity is also satisfied by Q and hence R as well.
By Lemma 1 Lanski [1993], there exists a field F such that R ⊆ Mk(F), the ring
of k × k matrices over a field F , where k ≥ 1. Moreover, R and Mk(F) satisfy
the same polynomial identity, that is, s ◦n x = 0 for all s,x ∈ Mk(F). Denote
eij the usual matrix unit with 1 in (i,j)-entry and zero elsewhere. By choosing
s = e12,x = e11, we see that s◦n x = e12 6= 0, a contradiction.

The following examples demonstrate that R to be prime can not be omitted in
the hypothesis of Theorem 1.1 and Theorem 1.2.

Example 3.1. For any ring S, let R =
{(

x y
0 0

)
| x,y ∈ S

}
and I ={(

0 y
0 0

)
| y ∈ S

}
. Then R is a ring under usual addition and multipli-

cation of matrices and I is a nonzero ideal of R. Define maps F,d : R → R

by F(
(
x y
0 0

)
) =

(
x 2y
0 0

)
and d(

(
x y
0 0

)
) =

(
0 y
0 0

)
. Then F is a

generalized derivation on R associated with the nonzero derivation d satisfying
(F(x)◦y)m + (x◦d(y))n = 0 for all x,y ∈ I. However R is not commutative as
well as F can not be written as F(x) = bx for all x ∈ R as d is nonzero. Hence
Theorem 1.1 is not true for arbitrary rings.

Example 3.2. Let R =
{(

x y
o z

)
| x,y,z ∈ Z2

}
and I =

{(
0 y
0 0

)
|

y ∈ Z2
}

. Then R is a ring under usual addition and multiplication of matrices

and I is a nonzero ideal of R. Define maps F,d : R → R by F(
(
x y
0 z

)
) =(

x 0
0 0

)
and d(

(
x y
0 z

)
) =

(
0 y
0 0

)
. Then F is a generalized derivation on

R associated with the nonzero derivation d satisfying (F(x)◦m y)k + x◦n d(y) =
0 for all x,y ∈ I. However R is not commutative as well as F can not be written
as F(x) = bx for all x ∈ R as d is nonzero . Hence Theorem 1.2 is not true for
arbitrary rings.

Theorem 1.3 Let R be a semiprime ring, U the left Utumi quotient ring of R and
m,n are fixed positive integers. If R admits a generalized derivation F associated

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Md Hamidur Rahaman

with a nonzero derivation d such that (F(x)◦y)m + (x◦d(y))n = 0 for all x,y ∈
R, then R is commutative.

Proof. By Fact 2.6, any derivation of a semiprime ring R can be uniquely extended
to a derivation of its left Utumi quotient ring U and so any derivation of R can be
defined on the whole of U. Moreover, by Fact 2.3 I, R and U satisfy the same
GPIs and by Fact 2.1 I, R, U satisfy same differential identities. Also by Fact 2.2,
we have F(x) = ax + d(x) for some a ∈ U and derivation d of U. Then

((ax + d(x)) ◦y)m + (x◦d(y))n = 0 for all x,y ∈ U. (15)

By Fact 2.5, we have Z(U) = C. Let M(C) be the set of all maximal ideals of
C and P ∈ M(C). By Fact 2.7, we have PU is a prime ideal of U invariant under
all derivations of U. Moreover,

⋂
{PU | P ∈ M(C) } = 0. Set U = U/PU.

Then derivation d canonically induce derivation d on U defined by d(x) = d(x)
for all x ∈ Ū. Therefore,

((āx̄ + d(x)) ◦ ȳ)m + (x◦d(y))n = 0

for all x,y ∈ U. It is obvious that U is prime. Therefore, by Theorem 1.1,
we have for each P ∈ M(C) either [U,U] ⊆ PU or d(U) ⊆ PU. In any case
d(U)[U,U] ⊆ PU for all P ∈ M(C). Thus d(U)[U,U] ⊆

⋂
{PU | P ∈ M(C)

} = 0, we obtain d(U)[U,U] = 0. Therefore, [U,U] = 0 since
⋂
{PU | P ∈

M(C) } = 0 and d 6= 0. Since R is subring of U, so in particular [R,R] = 0.
Hence R is commutative. This completes the proof of the theorem.

Using the similar arguments as used in the proof of the above theorem, we can
prove the following theorem.

Theorem 1.4 Let R be a semiprime ring, U the left Utumi quotient ring of R
and k,m,n are fixed positive integers. If R admits a generalized derivation F
associated with a nonzero derivation d such that (F(x) ◦m y)k + x ◦n d(y) =
0 for all x,y ∈ R, then R is commutative.

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