On Ideals via Generalized Reverse Derivation On Factor Rings

On Ideals via Generalized Reverse Derivation On
Factor Rings
Zakia Z. Al-Amery
Department of Mathematics, Aden University, Aden, Yemen:
Abstract. In current article, for a prime ideal P of any ring R, we study the commutativity of the
factor ring R/P, whenever R equipped with generalized reverse derivations F and G associated with
reverse derivations d and g, respectively. That satisfies certain differential identities involving in P that
connected to an ideal of R. Additionally, we show that, for some cases, the range of the generalized reverse
derivation F or G repose in the prime ideal P. Moreover, we explore several consequences and special
cases. Throughout, we provide examples to demonstrate that various restrictions in the assumptions of our
outcomes are essential.
Keywords: prime ideal; integral domain; generalized reverse derivation; factor ring.
1 Introduction
In current article, R is an associative ring with center Z(R). R is prime ring if and only
if the set {0} is prime ideal of R. On other words, R is a prime ring if xRy = {0} then
x = 0 or y = 0. P is a prime ideal of R if P ̸= R and for any two ideals I and J of R such
that IJ ⊆ P, then one at least of I or J is involving in a prime ideal P.
An additive mapping d : R −→ R is called a derivation if that satisfied the function
d(xy) = d(x)y + xd(y) for all x, y ∈ R. An additive mapping F : R −→ R is called a gen-
eralized derivation associated with the derivation d if the function F(xy) = F(x)y +xd(y)
for all x, y ∈ R. One of most famous examples of derivation is dm = [m, x] for any x ∈ R,
which is called the inner derivation prompted by m. For a non-trivial example of a gen-
eralized derivation on a non-commutative ring, the interesting reader can refer to [1] and
[2].
The concept of a reverse derivation was initially defined by Herstein in [3] when he had
been proved that the prime ring R is a commutative integral domain whenever the im-
posed derivation is a Jordan derivation. It had been defined to be an additive mapping
d : R −→ R that satisfies the function d(xy) = d(y)x + yd(x) for any x, y ∈ R. Several
studies in reverse derivation field are appeared that studied the commutativity of a ring
R; prime, semiprime or arbitrary ring. Samman et al. in [4] explored the reverse deriva-
tions in semiprime rings. In [5], Aboubakr et al. studied the relationship between gener-
alized reverse derivations and generalized derivations in semiprime rings. A generalized
reverse derivation associated with reverse derivation d is defined as an additive mapping
F : R −→ R satisfied the function F(xy) = F(y)x + yd(x) for all x, y ∈ R. In case R is
commutative ring, then generalized derivations and generalized reverse derivations is co-
incide. However, the converse does not always hold, as illustrated [[6], Example 1]. In [7],
Ibraheem in its paper proved the prime ring commutativity under influences of generalized
reverse derivation that associated with derivation d such that [F(x), x] contained in a cen-
ter of a ring, for all element x in a right ideal I of a ring R, provided that the intersection
of a right ideal I and the center Z(R) does not equal to zero. In [8] Bulak et al. determined
a detailed study on generalized reverse derivations. In the first part of their article, they
Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.12, No.3, September 2025
DOI : 10.5121/mathsj.2025.12301

examined that the prime ring commutativity under the influence of functional identities
that were involving two generalized reverse derivations. In the second part, they explored
the relationships between r-generalized reverse derivations and l-generalized derivations,
as well as the interplay between l-generalized reverse derivations and r-generalized deriva-
tions in the context connected to a non-central square-closed Lie ideal when a ring R was
a semiprime ring.
Building on these previous studies, researchers have made significant progress in under-
standing the conditions under which commutativity holds in various algebraic structures.
Several of these results have been derived by applying appropriate mappings such as deriva-
tions, generalized derivations, and generalized reverse derivations that satisfy specific func-
tional identities within suitable subsets of the ring R. For more on these developments,
please see references [9], [10], [11] and [12].
More recently, attention has turned to the behavior of factor rings R/P, where P is a prime
ideal of R. Researches have investigated how these factor rings behave under influences
of derivations, generalized derivations, generalized (α, β)-derivations that satisfy certain
differential identities involving in a prime ideal. Further details are found in literature, for
instance the interested reader can be referred to [[13]– [23]]. In [24] and [25]. The concept
of generalized reverse derivations was placed instead of generalized derivations, and the
commutativity of factor rings R/P was studied under identities involving prime ideals
related to the generalized reverse derivation F. This shift in perspective has led to new
insights into the algebraic structure of factor rings.
The primary aim of current article is to further study in this direction. More precisely,
considering that an arbitrary ring R that equipped with generalized reverse derivations
F and G that associated with reverse derivations d and g, respectively. We prove that if
(F, d) and (G, g) satisfies several functional identities involving within prime ideal P, then
the factor ring R/P is an integral domain. In some cases, it comes out that the range of
the generalized reverse derivation F or G or the range of addition or difference of two as-
sociated reverse derivations are in a prime ideal P, i.e., F(R) ⊆ P, G(R) ⊆ P, d(R) ⊆ P,
g(R) ⊆ P or (d ± g)(R) ⊆ P. Moreover, some consequences as well as special cases are
concluded. Examples that illustrate the necessity of the primeness assumptions stated in
our theorems are provided.
2 Preliminaries
Fact 1 [26] Let I be a nonzero ideal of any ring R, and let P be a prime ideal of R such
that P ⊊ I. If aIb ⊆ P for all a, b ∈ R, then either a ∈ P or b ∈ P.
Fact 2 Let I be a non-zero ideal of a ring R, and let P be a prime ideal of R provided that
P ⊊ I. If d is a reverse derivation of R such that d(I) is contained in P, then d(R) ⊆ P.
Fact 3 Let I be a non-zero ideal of a ring R, and let P be a prime ideal of R provided that
P ⊊ I. If (F, d) is a generalized derivation of R with d(I) ⊆ P, that satisfies F(I) ⊆ P,
then F(R) is also contained in P.
Lemma 1. [27, Lemma 2.3] If P and I be two ideals of a given ring R where P be prime
provided that P ⊊ I, then R/P is an integral domain if [x, y] ∈ P satisfies for every two
elements x, y ∈ I.
Alsowait et al. in [Lemma 2. , [24]] proved that the reverse derivation d mapping a ring
R to a prime ideal P or the factor ring of a ring R by a prime ideal P is integral domain,
2

if [x, F(x)] ∈ P for all x ∈ R. Where F is a generalized reverse derivation that associated
with d. Following in similar tactic with minor treatment, we prove same identity for all
x ∈ I, where I is an ideal of a ring R.
Lemma 2. Let P and I be two ideals in an arbitrary ring R such that P is a prime
ideal provided that P ⊊ I. If R equipped with a generalized reverse derivation F associated
with a reverse derivation d such that [x, F(x)] ∈ P for all x ∈ I. Then, it followed that
d(R) ⊆ P or R/P is an integral domain.
Following corollary is a consequence of Lemma 2 that outcomes when we restricted a
generalized reverse derivation F to be a reverse derivation d.
Corollary 1. Let P and I be two ideals in an arbitrary ring R such that P is a prime
ideal provided that P ⊊ I. If R equipped with a reverse derivation d, such that [x, d(x)] ∈ P
for all x ∈ R. Then, it followed that d(R) ⊆ P or R/P is an integral domain.
3 Main Result
Bouchannafa et al. in [[28], Theorem 2.5] proved that either a ring R/P is an integral
domain or an associated derivation d maps a ring R to a prime ideal P, whenever the ring
R that equipped with a generalized derivation F such that F(x ◦ y) − F(x) ◦ y ∈ Z(R/P)
for all x, y ∈ R, where P is a prime ideal of a ring R. Alsowait et al. [24]. Theorem 1] had
get similar outcome when they studied the identity F(x)◦y−F(x◦y) ∈ P for all x, y ∈ R,
whenever the ring R that equipped with a generalized reverse derivation F that associated
with reverse derivation d. Also, they studied the identity F(x) ◦ y + F(x ◦ y) ∈ P for all
x, y ∈ R. The similar outcome with minor different was gotten, that was an associated
derivation d maps the arbitrary ring R to a prime ideal P, or the factor of a ring R by a
prime ideal P is an integral domain of characteristic two.
In the context of next theorems, our objective is to achieve the parallel outcomes for
expanded identities, by utilizing two generalized reverse derivations F and G, which is
associated with a reverse derivation d and g, respectively. That denoted by (F, d) and
(G, g).
Theorem 1. Consider I and P are two ideals in any ring R, where P is prime provided
that P ⊊ I. If R equipped with generalized reverse derivations (F, d) and (G, g), that
satisfies the condition F(x) ◦ y ± G(x ◦ y) ∈ P for all x, y ∈ I. Then, it followed that:
(i) R/P is an integral domain of characteristic two, or
(iii) R/P is an integral domain and (F ± G)(R) ⊆ P, or
(iii) G(R) ⊆ P and F(R) ⊆ P.
Proof. (i) The assumption is
F(x) ◦ y + G(x ◦ y) ∈ P, ∀ x, y ∈ I. (1)
Placing yx instead of x in Equation (1) to yield
(F(x) ◦ y)y + (x ◦ y)d(y) + x[d(y), y] + G(x ◦ y)y + (x ◦ y)g(y) ∈ P, ∀ x, y ∈ I. (2)
By right multiplying of Equation (2) by y and comparing it with Equation (1), we obtain
(x ◦ y)(d(y) + g(y)) + x[d(y), y] ∈ P, ∀ x, y ∈ I. (3)
Placing tx instead of x in Equation (3) to yield
t(x ◦ y)(d(y) + g(y)) − [t, y]x(d(y) + g(y)) + tx[d(y), y] ∈ P, ∀ x, y, t ∈ I.
3

By left multiplying of Equation (3) by t and comparing it with previous equation to yield
[t, y]I(d(y) + g(y)) ⊆ P, ∀ y, t ∈ I.
Utilizing Fact (1), the last equation implies that [t, y] ∈ P for all t, y ∈ I, or d(y)+g(y) ∈ P
for all y ∈ I. In First scenario, we conclude that R/P is an integral domain by applying
Lemma 1. in light of commutativity of R/P, Equation (1) can be rewritten as
2F(x)y + 2G(xy) ∈ P, for all x, y ∈ I. (4)
Placing yt instead of y in previous equation and applying it, we obtain 2xIg(t) ⊆ P for
all x, t ∈ I. By utilizing Fact 1, we deduced that char(R/P) = 2 or d(t) ∈ P for all
t ∈ I. Temporary, let us assume that char(R/P) ̸= 2, hence Equation (4) simplifies to
F(x)y + G(x)y ∈ P for all x, y ∈ I. Hence (F + G)(x)I ⊆ P for all x ∈ I. According the
primeness of P and initial assumption that P ̸= I together with Fact 3, we conclude that
(F + G)(R) ⊆ P.
In Second scenario, we have (d + g)(y) ∈ P for all y ∈ I, applying it in Equation (3),
we can easily conclude that xI[d(y), y] ⊆ P for all x, y ∈ I. Utilizing Fact 1 with initial
assumption that P ̸= P, we deduced [d(y), y] ∈ P for all y ∈ I. Hence R/P is an integral
domain or d(R) ⊆ P, by applying Corollary 1. The last result immediately leads us to
g(R) ⊆ P.
Now, we can be rewriting Equation 1 as
F(x)y + yF(x) + G(xy) + G(yx) ∈ P, for all x, y ∈ I.
By using the result g(R) ⊆ P, the last expression simplifies to
F(x)y + yF(x) + G(y)x + G(x)y ∈ P, for all x, y ∈ I. (5)
Using substituting x = yx in Equation 5, we find
F(x)yy + yF(x)y + G(y)yx + G(x)yy for all x, y ∈ I.
By right multiplying of Equation 5 by y and comparing with last equation, we are having
G(y)[x, y] ∈ P for all x, y ∈ I. Taking x = xt, we arrive to G(y)I[t, y] ∈ P for all t, y ∈ I.
Again; by utilizing Fact 1, we can determinate that either G(I) ⊆ P or [t, y] ∈ P for all
y, t ∈ I. Applying Fact 3 and Lemma 1, we conclude that G(R) ⊆ P or R/P is an integral
domain. By using first outcome in Equation 5, we can be written
F(x)y + yF(x) ∈ P, for all x, y ∈ I. (6)
Placing tx rather than x in Equation 6, we obtain F(x)ty + yF(x)t ∈ P for all x, y, t ∈ I.
By right multiplying of Equation 6 by t and comparing with last equation, we are arriving
to F(x)[t, y] ∈ P for all x, y, t ∈ I. Taking x = mx, we arrive to F(x)I[t, y] ∈ P for all
x, y, t ∈ I. we can determinate that either F(I) ⊆ P or [t, y] ∈ P for all y, t ∈ I. Applying
Fact 3 and Lemma 1, respectively. We conclude that F(R) ⊆ P or R/P is an integral
domain.
(ii) For the assumption F(x) ◦ y − G(x ◦ y) ∈ P, ∀ x, y ∈ I. It can prove in similar
technique that followed in (i).
If we restricted G to be F, then we can give the following corollaries as consequences.
4

Corollary 2. Consider I and P are two ideals in any ring R, where P is prime provided
that P ⊊ I. If R equipped with a generalized reverse derivation (F, d), that satisfies the
condition F(x) ◦ y − F(x ◦ y) ∈ P for all x, y ∈ I. Then, it followed that:
(i) R/P is an integral domain of characteristic two, or
(ii) R/P is an integral domain and d(R) ⊆ P, or
(iii) F(R) ⊆ P.
condition F(x) ◦ y + F(x ◦ y) ∈ P for all x, y ∈ I. Then, it followed that:
(i) char(R/P) = 2, or
(ii) R/P is an integral domain and d(R) ⊆ P, or
(iii) F(R) ⊆ P.
Other consequence of Theorem 1 is when we restricted the two generalized reverse
derivations F and G to be the associated reverse derivations d and g, respectively. That
we can present in following corollary.
that P ⊊ I. If R equipped with reverse derivations d and g, that satisfies the condition
d(x) ◦ y ± g(x ◦ y) ∈ P for all x, y ∈ I. Then, it followed that d(R) ⊆ P and g(R) ⊆ P, or
R/P is an integral domain of characteristic two.
Example 1. Consider the ring of real numbers R and let R = {ae12 + be13 + ce14 + be24 −
ae34 | a, b, c ∈ R}. I = {ce14} and P = {0}. Defining (F, d), (G, g) : R −→ R, by:
F(ae12 + be13 + ce14 + be24 − ae34) = −ce14 ;
d(ae12 + be13 + ce14 + be24 − ae34) = −ce14 + be24 − ae34,
and
G(ae12 + be13 + ce14 + be24 − ae34) = −2ce14 ;
g(ae12 + be13 + ce14 + be24 − ae34) = −2ce14 + 2be24 − 2ae34.
It is evident that R is a ring, I and P are ideals of R that satisfies P ⊊ I, F and G are
generalized reverse derivations associated with the reverse derivations d and g, respectively.
That satisfies the exploring identity in Theorems 1. However, R/P is noncommutative and
its characteristic does not equal two, F(R) ⊈ P, G(R) ⊈ P, d(R) ⊈ P, g(R) ⊈ P and
(F ± G)(R) ⊈ P. Moreover, P is not a prime ideal of R since ce14 (ae12 − ae34) ∈ P, but
neither ce14 ∈ P nor (ae12 − ae34) ∈ P, hence; P is not prime ideal of R. Therefore, the
assumption that P is prime in Theorems 1 cannot be omitted.
Example 2. Consider R = W3 × H , where W3 = {


0 0 0
a 0 0
b c 0

 | a, b, c ∈ C} , C is a ring of
complexes and H is a ring of quaternions whit integers coefficients. I = {(


0 0 0
0 0 0
b 0 0

 , 2H)}
and P = (0, 0). Defining (F, d), (G, g) : R −→ R, by:
F(


0 0 0
a 0 0
b c 0

 , 2H) = (


0 0 0
a 0 0
0 0 0

 , 0) ; d(


0 0 0
a 0 0
b c 0

 , 2H) = (


0 0 0
0 0 0
2c 0 0

 , 0),
5

and
G(


0 0 0
a 0 0
b c 0

 , 2H) = (


0 0 0
2a 0 0
0 0 0

 , 0) ; g(


0 0 0
a 0 0
b c 0

 , 2H) = (


0 0 0
0 0 0
c 0 0

 , 0).
It is evident that R is a ring, I and P are ideals of R that satisfies P ⊊ I, F and G are
generalized reverse derivations associated with the reverse derivations d and g, respectively.
That satisfies the exploring identity in Theorems 1. However, R/P is noncommutative and
its characteristic does not equal two, F(R) ⊈ P, G(R) ⊈ P, d(R) ⊈ P, g(R) ⊈ P and
(F ± G)(R) ⊈ P. Moreover, P is not a prime ideal of R since (


0 0 0
a 0 0
b 0 0

 , 0)(


0 0 0
0 0 0
0 0 0

 , q) ∈
P, but neither (


0 0 0
a 0 0
b 0 0

 , 0) ∈ P nor (


0 0 0
0 0 0
0 0 0

 , q) ∈ P. Hence; P is not prime ideal of R.
Therefore, the assumption that P is prime in Theorems 1 cannot be omitted.
satisfies the conditions
(1) [F(x), y] + G(x ◦ y) ∈ P for all x, y ∈ I, then
(i) (d − g)(R) ⊆ P, or
(ii) R/P is an integral domain and G(R) ⊆ P, or
(iii) R/P is an integral domain of characteristic two.
(2) [F(x), y] − G(x ◦ y) ∈ P for all x, y ∈ I, then
(ii) R/P is an integral domain and G(R) ⊆ P, or
(iii) (d + g)(R) ⊆ P.
[F(x), y] + G(x ◦ y) ∈ P, ∀ x, y ∈ I. (7)
Placing yx instead of x in Equation (7) to yield
[F(x), y]y + x[d(y), y] + [x, y]d(y) + G(x ◦ y)y + (x ◦ y)g(y) ∈ P, ∀ x, y ∈ I. (8)
x[d(y), y] + [x, y]d(y) + (x ◦ y)g(y) ∈ P, ∀ x, y ∈ I. (9)
Placing tx instead of x in Equation (9) to yield
tx[d(y), y] + t[x, y]d(y) + [t, y]xd(y) + t(x ◦ y)g(y) − [t, y]xg(y) ∈ P, ∀ x, y, t ∈ I.
Left multiplication of Equation (9) by t and comparing with last equation, we conclude
[t, y]I(d(y) − g(y)) ⊆ P, ∀ x, y, t ∈ I.
Utilizing Fact 1 in the last equation, we obtain that [t, y] ∈ P for all t, y ∈ I, or d(y)−g(y) ∈
P for all y ∈ I. First scenario, we conclude that R/P is an integral domain by applying
Lemma 1. Using this outcome in given hypothesis that becomes
2G(xy) ∈ P, ∀ x, y ∈ I. (10)
6

Placing xy instead of x in Equation (10) and using it, we can arrive to 2xIg(y) ⊆ P for
all x, y ∈ I. Utilizing Fact 1, we obtain 2x ∈ P for all x ∈ I, or g(y) ∈ P for all y ∈ I.
First case with given that P ̸= I, forces that the characteristic of the factor ring R/P is
two. Temporarily, let char(R/P) ̸= 2, then g(y) ∈ P valid for all y ∈ I. Utilizing it in
Equation (10), that can be rewritten as G(y)x ∈ P for all x, y ∈ I. Again, Fact 1 together
with the given assumption that P ̸= I force that G(y) ∈ P for all y ∈ I. Therefore;
G(R) ⊆ P, immediately by applying Fact 3.
Second scenario, (d − g)(y) ∈ P for all y ∈ I, implies that (d − g)(I) ⊆ P. Hence,
by utilizing Fact 2, we conclude (d − g)(R) ⊆ P.
(ii) For the assumption [F(x), y] − G(x ◦ y) ∈ P, ∀ x, y ∈ I. It can prove in similar
As consequences of Theorem 2, we present the following corollaries. First one if G = F,
second outcomes when we restricted F and G to be associated d and g, respectively.
condition [F(x), y] − F(x ◦ y) ∈ P for all x, y ∈ I. Then, it followed that:
(ii) F(R) ⊆ P, or
(iii) R/P is an integral domain and F(R) ⊆ P.
(1) [d(x), y] + g(x ◦ y) ∈ P for all x, y ∈ I, then
(i) (d − g)(R) ⊆ P, or
(ii) R/P is an integral domain and g(R) ⊆ P, or
(iii) R/P is an integral domain of characteristic two.
(2) [d(x), y] − g(x ◦ y) ∈ P for all x, y ∈ I, then
(ii) R/P is an integral domain and g(R) ⊆ P, or
(iii) (d + g)(R) ⊆ P.
Example 3. In Examples 1 and 2, we can note that the identity in Theorem 2 satisfied,
although; R/P is not an integral domain and G(R) ⊈ P, char(R/P) ̸= 2 and (d∓g)(R) ⊈
P. This emphasize the necessity of the primeness condition in Theorem 2.
satisfies the condition [F(x), y] ± G[x, y] ∈ P for all x, y ∈ I. Then, it followed that:
(i) F(R) ⊆ P and G(R) ⊆ P, or
(ii) R/P is an integral domain.
[F(x), y] + G[x, y] ∈ P, ∀ x, y ∈ I. (11)
Placing yx instead of x in Equation (11), we obtain
[F(x), y]y + x[d(y), y] + [x, y]d(y) + G[x, y]y + [x, y]g(y) ∈ P, ∀ x, y ∈ I. (12)
7

x[d(y), y] + [x, y](d(y) + g(y)) ∈ P, ∀ x, y ∈ P. (13)
Placing tx instead of x in Equation (13), we obtain
tx[d(y), y] + t[x, y](d(y) + g(y)) + [t, y]x(d(y) + g(y)) ∈ P, ∀ x, y, t ∈ I.
By left multiplying of Equation (13) by t and comparing it with previous equation, we are
concluding
[t, y]I(d(y) + g(y)) ⊆ P, ∀ t, y ∈ I.
By utilizing Fact 1, the last equation implies that [t, y] ∈ P for all t, y ∈ I, or d(y)+g(y) ∈
P for all y ∈ I. From first scenario, we conclude that R/P is an integral domain by applying
Lemma 1. Second one; (d + g)(y) ∈ P for all y ∈ I, implies that (d + g)(I) ⊆ P. Hence, by
utilizing Fact 2, we conclude (d + g)(R) ⊆ P. Applying this outcome in Equation 13, we
can deduce that xI[d(y), y] ∈ P for all y ∈ I. By using Fact 1 together initial assumption
that I ̸= P, we determine that [d(y), y] ∈ P for all y ∈ I. Corollary 1 is arrived us to R/P
is an integral domain or d(R) ⊆ P. Last outcome leads to g(R) ⊆ P. Utilizing this result
in Equation 11, we can be rewriting
F(x)y − yF(x) + G(y)x − G(x)y ∈ P, ∀ x, y ∈ I. (14)
Placing yx rather than x in Equation 14, we have F(x)yy−yF(x)y+G(y)yx−G(x)yy ∈ P
for all x, y ∈ I. Right multiplying of Equation 14 by y and comparing with last relation,
we can arrive to G(y)I[t, y] ∈ P for all y, t ∈ I. By applying Fact 1, we conclude that
G(I) ∈ P or [t, y] ∈ P for all y, t ∈ I. Utilizing Fact 3 and Lemma 1, respectively. We
conclude that G(R) ⊆ P or R/P is an integral domain. Finally, using first outcome in
Equation 14, we obtain
F(x)y − yF(x) ∈ P, ∀ x, y ∈ I. (15)
Using substituting x = tx in Equation 15 and applying it, we obtain F(x)[t, y] ∈ P for all
x, y, t ∈ I. Letting x = mx, we can deduced that F(x)I[t, y] ∈ P for all x, y, t ∈ I. Again,
by applying Fact 1, we conclude that F(I) ∈ P or [t, y] ∈ P for all y, t ∈ I. utilizing Fact 3
and Lemma 1, respectively. We conclude that F(R) ⊆ P or R/P is an integral domain.
(ii) For the assumption [F(x), y] − G[x, y] ∈ P, ∀ x, y ∈ I. It can prove in similar
As previously; if we restricted G to be F in Theorem 3, then we can give the following
corollaries as consequences.
condition [F(x), y] − F[x, y] ∈ P for all x, y ∈ I, then either F(R) ⊆ P or R/P is an
integral domain.
condition [F(x), y] + F[x, y] ∈ P for all x, y ∈ I, then
(ii) F(R) ⊆ P, or
8

(iii) R/P is an integral domain.
Also, other consequence of Theorem 3 is when we restricted the two generalized reverse
derivations F and G to be the associated reverse derivations d and g, respectively. That
we can present in following corollary.
[d(x), y] ± g[x, y] ∈ P for all x, y ∈ I. Then, it followed that:
(i) d(R) ⊆ P and g(R) ⊆ P, or (ii) R/P is an integral domain.
Example 4. In Examples 1 and 2, we can note that the identity in Theorem 3 satisfied,
although; R/P is not an integral domain, F(R) ⊈ P and G(R) ⊈ P. This emphasize the
necessity of the primeness condition in Theorem 3.
4 Conclusion
In current work, we went ahead studied of generalized reverse derivation related to prime
ideal, when we used the arbitrary assumption for a study ring R and the domain of taken
elements is an ideal of R. Where R equipped with two generalized reverse derivations F
and G associated with reverse derivation d and g, respectively. We proved that if (F, d)
and (G, g) satisfies several functional identities involving within prime ideal P, then the
factor ring R/P is an integral domain. In some cases, it came out that the range of
the generalized reverse derivation F or G or the range of addition or difference of two
associated reverse derivations are in a prime ideal P, i.e., F(R) ⊆ P, G(R) ⊆ P, d(R) ⊆ P,
g(R) ⊆ P or (d ± g)(R) ⊆ P. Moreover, some consequences as well as special cases were
concluded. Examples that illustrated the necessity of the primeness assumptions stated in
our theorems are provided.
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Authors
Z. Z. Al-Amery received Master of mathematics from Aden University. He did Bachelor
of mathematics and sciences from Aden University. Currently, he is pursuing his PhD
in mathematics from Aden University. His research interests include abstract algebra,
derivations in ring and algebras.
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