EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 6084
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Characterization of Multiplicative Mixed Jordan-Type
Derivations on Ring with Involution

Md Arshad Madni1, Muzibur Rahman Mozumder1, Abu Zaid Ansari2,∗,
Faiza Shujat3

1 Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India
2 Department of Mathematics, Faculty of Science, Islamic University of Madinah, K.S.A
3 Department of Mathematics, Faculty of Science, Taibah University, Madinah, K.S.A.

Abstract. Let B be a unital ∅-ring with a 2-torsion free that contains non-trivial symmetric
idempotent. For any B1, B2, B3, . . . , Bn ∈ B, a product B1 ◦ B2 = B1B2 + B2B1 is called Jordan
product and B1 • B2 = B1B2 + B2B

∅
1 is recognized as a skew Jordan product. Characterize

mixed Jordan triple product as Q3(B1, B2, B3) = B1 ◦ B2 • B3 and mixed Jordan n-product
as Qn(B1, B2, . . . , Bn) = B1 ◦ B2 ◦ · · · • Bn for all integer n ≥ 3. The present paper deals
that a mapping which is called multiplicative mixed Jordan n-derivation, Ψ: B → B satisfies
Ψ(Qn(B1, B2, . . . , Bn)) =

∑n
i=1 Qn(B1, . . . , Bi−1,Ψ(Bi), Bi+1, . . . , Bn) for all B1, B2, . . . , Bn ∈ B

if and only if Ψ is an additive ∅-derivation. Finally, primary outcome is applicable in various
specific categories of unital ∅-rings and ∅-algebras including prime ∅-rings, prime ∅-algebras and
factor von Neumann algebras.

2020 Mathematics Subject Classifications: 16W10, 46L10, 47B47.

Key Words and Phrases: Additive ∅-derivation; Mixed Jordan-type derivation; ∅-rings; factor
von Neumann algebra.

1. Introduction

Throughout this work, B is taken to be an associative ring having Z(B) as its center.
An involution ‘∅’ is described as an anti-automorphism of order 1 or 2 and B along
with an involution ‘∅’ is known as a ∅-ring. If α2 = α = α∅, then α ∈ B is termed a
symmetric idempotent and it is considered non-trivial if α is neither 0 nor I. A mapping
Ψ: B → B is called a derivation if it satisfies Ψ(B1+B2) = Ψ(B1)+Ψ(B2) and Ψ(B1B2) =
Ψ(B1)B2+B1Ψ(B2) for all B1, B2 ∈ B. For a ring B with an involution ‘∅’ a mapping Ψ :
B → B is called an additive ∅-derivation, if it is a derivation and Ψ(B∅

1 ) = Ψ(B1)
∅ holds

for all B1, B2 ∈ B. The expressions B1 ◦B2 = B1B2+B2B1 and B1 •B2 = B1B2+B2B
∅
1 ,

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6084

Email addresses: arshadmadni7613@gmail.com (M. A. Madni), muzibamu81@gmail.com (M. R.
Mozumder), ansari.abuzaid@gmail.com (A. Z. Ansari), faiza.shujat@gmail.com (F. Shujat)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 2 of 16

define the Jordan product and skew Jordan product of B1 and B2 respectively. An additive
mapping Ψ : B → B is known as a Jordan derivation (respectively Jordan triple derivation)
if

Ψ(B1 ◦B2) = Ψ(B1) ◦B2 +B1 ◦Ψ(B2)

(resp. Ψ((B1 ◦B2) ◦B3) = (Ψ(B1) ◦B2) ◦B3 + (B1 ◦Ψ(B2)) ◦B3 + (B1 ◦B2) ◦Ψ(B3))

holds for all B1, B2, B3 ∈ B. Analogously, an additive mapping Ψ : B → B is said to be a
skew-Jordan derivation (respectively skew-Jordan triple derivation) if

Ψ(B1 •B2) = Ψ(B1) •B2 +B1 •Ψ(B2)

(resp. Ψ((B1 •B2) •B3) = (Ψ(B1) •B2) •B3 + (B1 •Ψ(B2)) •B3 + (B1 •B) •Ψ(B3))

holds for all B1, B2, B3 ∈ B. In case the mappings Ψ is not necessarily additive in the
above definitions, then Ψ is called a multiplicative Jordan derivation (respectively multi-
plicative Jordan triple derivation) and multiplicative skew Jordan derivation (respectively
multiplicative skew Jordan triple derivation. In recent years, many mathematicians have
been interested in mappings with different kinds of algebras and rings. These findings to
be becoming more and more important in a variety of research areas, and many authors
have taken an interest in studying them. (see [1–6]).

Define a sequence of polynomials as follows: Qn(B1, B2, . . . , Bn−1, Bn) = Qn−1(B1, B2,
. . . , Bn−1) •Bn (n ≥ 3) where Qn(B1, B2, . . . , Bn−1, Bn) = Qn−1(B1, B2, . . . Bn−1) •Bn is
recognized as mixed-Jordan n-product. We have Q3(B1, B2, B3) = B1 ◦B2 •B3 is referred
to as the mixed-Jordan triple product (see [7]), for n = 3. We define a mapping Ψ : B → B
(not necessarily additive) is said to be multiplicative mixed-Jordan n-derivation if

Ψ(Qn(B1, B2, . . . , Bn)) = Qn(Ψ(B1), B2, . . . , Bn) +Qn(B1,Ψ(B2), . . . , Bn)

+ · · ·+Qn(B1, B2, . . . ,Ψ(Bn))

for all B1, B2, . . . , Bn ∈ B.
Every multiplicative mixed-Jordan 3-derivations on ∅-rings is a multiplicative mixed-
Jordan n-derivation, in general the converse need not be true. Multiplicative mixed-
Jordan (3, 4, n)-derivation are en masse recognized as multiplicative mixed-Jordan type
derivations. Recently, many researchers paid more attention to the study of Lie (Jordan)
mappings involving two different kinds of products at the same time in the nonlinear set-
tings in [8–10]. Zhou in [11] studied that: Every nonlinear mixed Lie-triple derivation
on unital prime ∅-algebra is an additive ∅-derivation. Rehman in [7], proved that: On
∅-algebras, all nonlinear mixed Jordan-triple derivations qualify as additive ∅-derivation.
The work of Ferreira and Wei in [12] focused on characterizing nonlinear mixed ∅-Jordan-
type derivations on ∅-algebras. They established that any nonlinear mixed ∅-Jordan
n-derivation is an additive ∅-derivation. In their study presented in [13], Yu et al. demon-
strated that all skew Lie derivations on factor von Neumann algebras are equivalent to
additive ∅-derivations. In a recent work, Kong and Zhang [14] generalized this result to



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 3 of 16

prime ∅-rings and proved that every skew Lie derivation on 2-torsion prime ∅-ring is an
additive ∅-derivation. The assumption 1

2 ∈ B holds throughout the discussion.

Inspired by the preceding studies, this paper examines the relationship between multi-
plicative mixed-Jordan n-derivations and additive ∅-derivation in arbitrary ∅-rings. Our
results demonstrate that, under specific conditions, every multiplicative mixed-Jordan n-
derivation defined is an additive ∅-derivation, on ∅-ring with unity.

2. The principal conclusion

The key result of this work can be summarized as follows:

Theorem 1. Suppose that B is a 2-torsion free ∅-ring having unity u containing a non-
trivial symmetric idempotent α1. Write α2 = u− α1 and assume that B fullfils

Y Bαk = 0 =⇒ Y = 0 for k = 1, 2 (♠)

where Y ∈ B. Then a map Ψ : B → B (need not necessarily additive) satisfies

Ψ(Qn(B1, B2, . . . , Bn)) =

n∑
i=1

Qn(B1, . . . , Bi−1,Ψ(Bi), Bi+1, . . . , Bn) (2.1)

for all B1, B2, . . . , Bn ∈ B if and only if Ψ is an additive ∅-derivation.

Assume Bij = PiBPj with i, j = 1, 2, then by the Peirce decomposition, we have B =
B11 ⊕ B12 ⊕ B21 ⊕ B22. Clearly any R ∈ B can be written as R = B11 +B12 +B21 +B22,
where Bij ∈ Bij for i, j = 1, 2. To conclude the proof of the stated theorem, a number of
following lemmas are necessary:

Lemma 1. Ψ(0) = 0.

Proof. We have

Ψ(0) = Ψ(Qn(0, 0, . . . , 0))

= Qn(Ψ(0), 0, . . . , 0) +Qn(0,Ψ(0), . . . , 0) + · · ·+Qn(0, 0, . . . ,Ψ(0)) = 0.

Lemma 2. For any B12 ∈ B12 and B21 ∈ B21, we have

Ψ(B12 +B21) = Ψ(B12) + Ψ(B21).

Proof. Let H = Ψ(B12 + B21) − Ψ(B12) − Ψ(B21). We need to prove that H = 0.
Applying Qn(u, u, . . . , B12, α2, α1) = 0 and Lemma 1 to get

Ψ(Qn(u, u, . . . , B12 +B21, α2, α1))

= Ψ(Qn(u, u, . . . , B12, α2, α1)) + Ψ(Qn(u, u, . . . , B21, α2, α1))



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 4 of 16

= Qn(Ψ(u), u, . . . , B12, α2, α1) +Qn(u,Ψ(u), . . . , B12, α2, α1) + · · ·+
Qn(u, u, . . . ,Ψ(B12), α2, α1) +Qn(u, u, . . . , B12,Ψ(α2), α1) +Qn(u, u, . . . , B12, α2,

Ψ(α1)) +Qn(Ψ(u), u, . . . , B21, α2, α1) +Qn(u,Ψ(u), . . . , B21, α2, α1) + · · ·+
Qn(u, u, . . . ,Ψ(B21), α2, α1) +Qn(u, u, . . . , B21,Ψ(α2), α1) +Qn(u, u, . . . , B21, α2,

Ψ(α1))

= Qn(Ψ(u), u, . . . , B12 +B21, α2, α1) +Qn(u,Ψ(u), . . . , B12 +B21, α2, α1) + · · ·+
Qn(u, u, . . . ,Ψ(B12) + Ψ(B21), α2, α1) +Qn(u, u, . . . , B12 +B21,Ψ(α2), α1) +

Qn(u, u, . . . , B12 +B21, α2,Ψ(α1)).

Alternatively, we obtain

Ψ(Qn(u, u, . . . , B12 +B21, α2, α1))

= Qn(Ψ(u), u, . . . , B12 +B21, α2, α1) +Qn(u,Ψ(u), . . . , B12 +B21, α2, α1) + · · ·+
Qn(u, u, . . . ,Ψ(B12 +B21), α2, α1) +Qn(u, u, . . . , B12 +B21,Ψ(α2), α1)

+Qn(u, u, . . . , B12 +B21, α2,Ψ(α1)).

A comparison of the aforementioned expressions reveals that Qn(u, u, . . . ,H, α2, α1) = 0.
This futher implies that H21 = 0. Similarly, we can show that H12 = 0. Based on the fact
Qn(u, u, . . . , u, (α1 − α2), B21) = 0 and use Lemma 1 to have

Ψ(Qn(u, u, . . . , u, α1 − α2, B12 +B21))

= Ψ(Qn(u, u, . . . , u, α1 − α2, B12)) + Ψ(Qn(u, u, . . . , u, α1 − α2, B21))

= Qn(Ψ(u), u, . . . , u, α1 − α2, B12) +Qn(u,Ψ(u), . . . , u, α1 − α2, B12) + · · ·+
Qn(u, u, . . . ,Ψ(u), α1 − α2, B12) +Qn(u, u, . . . , u,Ψ(α1 − α2), B12) +Qn(u, u,

. . . , u, α1 − α2,Ψ(B12)) +Qn(Ψ(u), u, . . . , u, α1 − α2, B21) +

Qn(u,Ψ(u), . . . , u, α1 − α2, B21) + · · ·+Qn(u, u, . . . ,Ψ(u), α1 − α2, B21)

+Qn(u, u, . . . , u,Ψ(α1 − α1), B21) +Qn(u, u, . . . , u, α1 − α2,Ψ(B21))

= Qn(Ψ(u), u, . . . , u, α1 − α2, B12 +B21) +Qn(u,Ψ(u), . . . , u, α1 − α2, B12 +B21)

+ · · ·+Qn(u, u, . . . ,Ψ(u), α1 − α2, B12 +B21) +Qn(u, u, . . . , u,Ψ(α1 − α2), B12

+B21) +Qn(u, u, . . . , u, α1 − α2,Ψ(B12) + Ψ(B21)).

Alternatively, we have

Ψ(Qn(u, u, . . . , u, α1 − α2, B12 +B21))

= Qn(Ψ(u), u, . . . , u, α1 − α2, B12 +B21) +Qn(u,Ψ(u), . . . , u, α1 − α2, B12 +B21) +

· · ·+Qn(u, u, . . . ,Ψ(u), α1 − α2, B12 +B21) +Qn(u, u, . . . , u,Ψ(α1 − α2), B12 +B21)

+Qn(u, u, . . . , u, α1 − α2,Ψ(B12 +B21).

From the above, Qn(u, u, . . . , u, α1 − α2,H) = 0 that implies H11 = 0 and H22 = 0.
Therefore, H = 0, i.e.,

Ψ(B12 +B21) = Ψ(B12) + Ψ(B21).



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 5 of 16

Lemma 3. For any B11 ∈ B11, B12 ∈ B12, B21 ∈ B21, and B22 ∈ B22, we have

Ψ(B11 +B12 +B21) = Ψ(B11) + Ψ(B12) + Ψ(B21)

and
Ψ(B12 +B21 +B22) = Ψ(B12) + Ψ(B21) + Ψ(B22).

Proof. Let H = Ψ(B11+B12+B21)−Ψ(B11)−Ψ(B12)−Ψ(B21). The aim to prove that
H = 0. Applying Qn(u, u, . . . , B11, α1, α2) = Qn(u, u, . . . , B21, α1, α2) = 0 and Lemma 1,
we find

Ψ(Qn(u, u, . . . , B11 +B12 +B21, α1, α2))

= Ψ(Qn(u, u, . . . , B11, α1, α2)) + Ψ(Qn(u, u, . . . , B12, α1, α2)) + Ψ(Qn(u, u, . . . , B21, α1,

α2))

= Qn(Ψ(u), u, . . . , B11, α1, α2) +Qn(u,Ψ(u), . . . , B11, α1, α2) + · · ·+Qn(u, u, . . . ,

Ψ(B11), α1, α2) +Qn(u, u, . . . , B11,Ψ(α1), α2) +Qn(u, u, . . . , B11, α1,Ψ(α2)) +

Qn(Ψ(u), u, . . . , B12, α1, α2) +Qn(u,Ψ(u), . . . , B12, α1, α2) + · · ·+Qn(u, u, . . . ,

Ψ(B12), α1, α2) +Qn(u, u, . . . , B12,Ψ(α1), α2) +Qn(u, u, . . . , B12, α1,Ψ(α2)) +

Qn(Ψ(u), u, . . . , B21, α1, α2) +Qn(u,Ψ(u), . . . , B21, α1, α2) + · · ·+Qn(u, u, . . . ,

Ψ(B21), α1, α2) +Qn(u, u, . . . , B21,Ψ(α1), α2) +Qn(u, u, . . . , B21, α1,Ψ(α2))

= Qn((Ψ(u), u, . . . , B11 +B12 +B21, α1, α2) +Qn((u,Ψ(u), . . . , B11 +B12 +B21, α1, α2)

+ · · ·+Qn((u, u, . . . ,Ψ(B11) + Ψ(B12) + Ψ(B21), α1, α2) +Qn((u, u, . . . , B11 +

B12 +B21,Ψ(α1), α2) +Qn(u, u, . . . , B11 +B12 +B21, α1,Ψ(α2)).

Alternatively, we obtain

Ψ(Qn(u, u, . . . , B11 +B12 +B21, α1, α2))

= Qn(Ψ(u), u, . . . , B11 +B12 +B21, α1, α2) +Qn(u,Ψ(u), . . . , B11 +B12 +B21, α1, α2)

+ · · ·++Qn(u, u, . . . ,Ψ(B11 +B12 +B21), α1, α2) +Qn(u, u, . . . , B11 +B12 +B21,

Ψ(α1), α2) +Qn(u, u, . . . , B11 +B12 +B21, α1,Ψ(α2)).

Comparing the last expressions for Ψ(Qn(u, u, . . . , B11+B12+B21, α1, α2)) to get Qn(u, u,
. . . ,H, α1, α2) = 0. That impliesH12 = 0, andH21 = 0. Based on the facts Qn(u, u . . . , (α1

− α2), B12) = Qn(u, u, . . . (α1 − α2), B21) = 0 and use Lemma 1 to get that

Ψ(Qn(u, u, . . . , (α1 − α2), (B11 +B12 +B21)))

= Ψ(Qn(u, u, . . . , (α1 − α2), B11)) + Ψ(Qn(u, u, . . . , (α1 − α2), B12)

+Ψ(Qn(u, u, . . . , (α1 − α2), B21))

= Qn(Ψ(u), u, . . . , (α1 − α2), B11) +Qn(u,Ψ(u), . . . , (α1 − α2), B11)

+ · · ·+Qn(u, u, . . . ,Ψ(α1 − α2), B11) +Qn(u, u, . . . , (α1 − α2),Ψ(B11))

+Qn(Ψ(u), u, . . . , (α1 − α2), B12) +Qn(u,Ψ(u), . . . , (α1 − α2), B12)



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 6 of 16

+ · · ·+Qn(u, u, . . . ,Ψ(α1 − α2), B12) +Qn(u, u, . . . , (α1 − α2),Ψ(B12))

+Qn(Ψ(u), u . . . , (α1 − α2), B21) +Qn(u,Ψ(u), . . . , (α1 − α2), B21)

+ · · ·+Qn(u, u, . . . ,Ψ(α1 − α2), B21) +Qn(u, u, . . . , (α1 − α2),Ψ(B21))

= Qn(Ψ(u), u, . . . , (α1 − α2), (B11 +B12 +B21)) +Qn(u,Ψ(u), . . . , (α1 − α2), (B11

+B12 +B21)) + · · ·+Qn(u, u, . . . ,Ψ(α1 − α2), (B11 +B12

+B21)) +Qn(u, u, . . . , (α1 − α2),Ψ(B11) + Ψ(B12) + Ψ(B21)).

However, we also have

Ψ(Qn(u, u, . . . , (α1 − α2), B11 +B12 +B21))

= Qn(Ψ(u), u, . . . , (α1 − α2), (B11 +B12 +B21)) +Qn(u,Ψ(u), . . . , (α1 − α2), (B11 +

B12 +B21)) + · · ·+Qn(u, u, . . . ,Ψ(α1 − α2), (B11 +B12 +B21))

+Qn(u, u, . . . , (α1 − α2),Ψ(B11 +B12 +B21)).

Examine the two relations for Ψ(Qn(u, u, . . . , (α1 − α2), B11 +B12 +B21)) to have
Qn(u, u, . . . , (α1 − α2),H) = 0 that gives H11 = 0 and H22 = 0. Hence H = 0, that is,

Ψ(B11 +B12 +B21) = Ψ(B11) + Ψ(B12) + Ψ(B21).

Similarly, we can show that Ψ(B12 +B21 +B22) = Ψ(B12) + Ψ(B21) + Ψ(B22).

Lemma 4. For any B11 ∈ B11, B12 ∈ B12, B21 ∈ B21, and B22 ∈ B22, we have

Ψ(B11 +B12 +B21 +B22) = Ψ(B11) + Ψ(B12) + Ψ(B21) + Ψ(B22).

Proof. Let H = Ψ(B11 + B12 + B21 + B22) − Ψ(B11) − Ψ(B12) − Ψ(B21) − Ψ(B22).
Applying Qn(u, u, . . . , α1, B22) = 0 with the above lemmas, we get the following

Ψ(Qn(u, u, . . . , α1, (B11 +B12 +B21 +B22)))

= Ψ(Qn(u, u, . . . , α1, (B11 +B12 +B21))) + Ψ(Qn(u, u, . . . , α1, B22))

= Qn(Ψ(u), u, . . . , α1, (B11 +B12 +B21)) +Qn(u,Ψ(u), . . . , α1, (B11 +B12 +B21))

+ · · ·+Qn(u, u, . . . ,Ψ(α1), (B11 +B12 +B21)) +Qn(u, u, . . . , α1, (Ψ(B11) +

Ψ(B12) + Ψ(B21))) +Qn(Ψ(u), u, . . . , α1, B22) +Qn(u,Ψ(u), . . . , α1, B22)

+ · · ·+Qn(u, u, . . . ,Ψ(α1), B22) +Qn(u, u, . . . , α1,Ψ(B22))

= Qn(Ψ(u), u, . . . , α1, (B11 +B12 +B21 +B22)) +Qn(u,Ψ(u), . . . , α1, (B11 +B12

+B21 +B22)) + · · ·+Qn(u, u, . . . ,Ψ(α1), (B11 +B12 +B21 +B22))

+Qn(u, u, . . . , α1, (Ψ(B11) + Ψ(B12) + Ψ(B21) + Ψ(B22))).

On the other hand, we have

Ψ(Qn(u, u, . . . , α1, (B11 +B12 +B21 +B22)))

= Qn(Ψ(u), u, . . . , α1, (B11 +B12 +B21 +B22)) +Qn(u,Ψ(u), . . . , α1, (B11 +B12



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 7 of 16

+B21 +B22)) + · · ·+Qn(u, u, . . . ,Ψ(α1), (B11 +B12 +B21 +B22))

+Qn(u, u, . . . , α1, (Ψ(B11 +B12 +B21 +B22))).

Similarly, for Ψ(Qn(u, u, . . . , α1, (B11+B12+B21+B22))), we get Qn(u, u, . . . , α1,H) = 0
This leads to the conclusion that H11 = H12 = H21 = 0. Similarly, we can show that
H22 = 0. Thus H = 0, that is,

Ψ(B11 +B12 +B21 +B22) = Ψ(B11) + Ψ(B12) + Ψ(B21) + Ψ(B22).

Lemma 5. For any B12, B
′
12 ∈ B12 and B21, B

′
21 ∈ B21 we have

Ψ(B12 +B
′
12) = Ψ(B12) + Ψ(B′

12) and Ψ(B21 +B
′
21) = Ψ(B21) + Ψ(B′

21).

Proof. Using the fact that Qn(
u
2 ,

u
2 , . . . ,

u
2 , (α1+B12), (α2+B

′
12)) = B12+B

′
12+B∅

12+

B
′
12B

∅
12 and Lemma 4, we have

Ψ(B12 +B
′
12) + Ψ(B∅

12) + Ψ(B
′
12B

∅
12)

= Ψ
(
B12 +B

′
12 +B∅

12 +B
′
12B

∅
12)

= Ψ(Qn(
u

2
,
u

2
, . . . ,

u

2
, (α1 +B12), (α2 +B12

′)))

= Qn(Ψ(
u

2
),
u

2
, . . . ,

u

2
, (α1 +B12), (α2 +B

′
12)) +Qn(

u

2
,Ψ(

u

2
), . . . ,

u

2
, (α1 +B12),

(α2 +B
′
12)) + · · ·+Qn(

u

2
,
u

2
, . . . ,Ψ(

u

2
), (α1 +B12), (α2 +B

′
12)) +Qn(

u

2
,
u

2
, . . . ,

u

2
,

Ψ(α1 +B12), (α2 +B
′
12)) +Qn(

u

2
,
u

2
, . . . ,

u

2
, (α1 +B12),Ψ(α2 +B

′
12))

= Ψ(Qn(
u

2
,
u

2
, . . . ,

u

2
, α1, α2)) + Ψ(Qn(

u

2
,
u

2
, . . . ,

u

2
, α1, B

′
12)) + Ψ(Qn(

u

2
,
u

2
, . . . ,

u

2
,

B12, α2)) + Ψ(Qn(
u

2
,
u

2
, . . . ,

u

2
, B12, B

′
12))

= Ψ(B12) + Ψ(B
′
12) + Ψ(B∅

12) + Ψ(B
′
12B

∅
12).

Hence Ψ(B12 + B
′
12) = Ψ(B12) + Ψ(B

′
12) for any B12, B

′
12 ∈ B12. The proof for the other

part follows similarly.

Lemma 6. For any Bii, B
′
ii ∈ Bii for (i = 1, 2), we have

Ψ(B11 +B
′
11) = Ψ(B11) + Ψ(B

′
11) and Ψ(B22 +B

′
22) = Ψ(B22) + Ψ(B

′
22).

Proof. LetH = Ψ(B11+B
′
11)−Ψ(B11)−Ψ(B

′
11). Using the fact thatQn(u, u, . . . , u, α2,

B11) = Qn(u, u, . . . , u, α2, B
′
11) = 0. Lemma 1 gives

Ψ(Qn(u, u, . . . , u, α2, (B11 +B
′
11))).

= Ψ(Qn(u, u, . . . , u, α2, B11)) + Ψ(Qn(u, u, . . . , u, α2, B
′
11))



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 8 of 16

= Qn(Ψ(u), u, . . . , u, α2, B11) +Qn(u,Ψ(u), . . . , u, α2, B11) + · · ·+
Qn(u, u, . . . ,Ψ(u), α2, B11) +Qn(u, u, . . . , u,Ψ(α2), B11) +Qn(u, u, . . . , u, α2,

Ψ(B11)) +Qn(Ψ(u), u, . . . , u, α2, B
′
11) +Qn(u,Ψ(u), . . . , u, α2, B

′
11) + · · ·+

Qn(u, u, . . . ,Ψ(u), α2, B
′
11) +Qn(u, u, . . . , u,Ψ(α2), B

′
11) +Qn(u, u, . . . , u, α2,

Ψ(B
′
11))

= Qn(Ψ(u), u, . . . , u, α2, (B11 +B
′
11)) +Qn(u,Ψ(u), . . . , u, α2, (B11 +B

′
11)) + · · ·+

Qn(u, u, . . . ,Ψ(u), α2, (B11 +B
′
11)) +Qn(u, u, . . . , u,Ψ(α2), (B11 +B

′
11))

+Qn(u, u, . . . , u, α2, (Ψ(B11) + Ψ(B
′
11))).

Alternatively, we have

Ψ(Qn(u, u, . . . , u, α2, (B11 +B
′
11))).

= Qn(Ψ(u), u, . . . , u, α2, (B11 +B
′
11)) +Qn(u,Ψ(u), . . . , u, α2, (B11 +B

′
11)) + · · ·+

Qn(u, u, . . . ,Ψ(u), α2, (B11 +B
′
11)) +Qn(u, u, . . . , u,Ψ(α2), (B11 +B

′
11))

+Qn(u, u, . . . , u, α2, (Ψ(B11 +B
′
11))).

Compare the above for Ψ(Qn(u, u, . . . , u, α2, (B11+B
′
11))), we find that Qn(u, u, . . . , u, α2,

H) = 0, This suggests that H12 = H21 = H22 = 0. Next, we show that H11 = 0. Let
X12 ∈ B12 and it is straightforward to observe that Qn(u, u, . . . , α1, B11, X12),
Qn(u, u, . . . , α1, B

′
11, X12) ∈ B12. Hence, apply Lemma 5 to get

Ψ(Qn(u, u, . . . , α1, (B11 +B
′
11), X12)).

= Ψ(Qn(u, u, . . . , α1, B11, X12)) + Ψ(Qn(u, u, . . . , α1, B
′
11, X12))

= Qn(Ψ(u), u, . . . , α1, B11, X12) +Qn(u,Ψ(u), . . . , α1, B11, X12)

+ · · ·+Qn(u, u, . . . ,Ψ(α1), B11, X12) +Qn(u, u, . . . , α1,Ψ(B11), X12)

+Qn(u, u, . . . , α1, B11,Ψ(X12)) +Qn(Ψ(u), u, . . . , α1, B
′
11, X12)

+Qn(u,Ψ(u), . . . , α1, B
′
11, X12) + · · ·+Qn(u, u, . . . ,Ψ(α1), B

′
11, X12)

+Qn(u, u, . . . , α1,Ψ(B
′
11), X12) +Qn(u, u, . . . , α1, B

′
11,Ψ(X12))

= Qn(Ψ(u), u, . . . , α1, (B11 +B
′
11), X12) +Qn(u,Ψ(u), . . . , α1, (B11 +B

′
11), X12)

+ · · ·+Qn(u, u, . . . ,Ψ(α1), (B11 +B
′
11), X12)

+Qn(u, u, . . . , α1, (Ψ(B11) + Ψ(B
′
11)), X12) +Qn(u, u, . . . , α1, (B11 +B

′
11),Ψ(X12)).

However, in contrast, we get

Ψ(Qn(u, u, . . . , α1, (B11 +B
′
11), X12)).

= Qn(Ψ(u), u, . . . , α1, (B11 +B
′
11), X12) +Qn(u,Ψ(u), . . . , α1, (B11 +B

′
11), X12)

+ · · ·+Qn(u, u, . . . ,Ψ(α1), (B11 +B
′
11), X12)

+Qn(u, u, . . . , α1, (Ψ(B11 +B
′
11)), X12) +Qn(u, u, . . . , α1, (B11 +B

′
11),Ψ(X12)).

Examine just the last two statements for Ψ(Qn(u, u, . . . , α1, (B11 + B
′
11), X12)) to get

Qn(u, u, . . . , α1,H, X12) = 0, implies that α1TX12+Hα1X12+X12H∅α1 = 0. Multiplying



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 9 of 16

it both sides by α1 and α2 from left and right respectively, we obtain α1Tα1Xα2 = 0 for
all Y ∈ B. Utilizing condition (♠) yields H11 = 0. Hence H = 0, that is, Ψ(B11 +B

′
11) =

Ψ(B11) + Ψ(B
′
11). Symmetrically, one can prove that Ψ(B22 +B

′
22) = Ψ(B22) + Ψ(B

′
22).

Lemma 7. The mapping Ψ is additive on B.

Proof. For any element L,R ∈ B, we get L = L11 + L12 + L21 + L22 and R =
B11 +B12 +B21 +B22. Applying the Lemmas, we get

Ψ(L+R) = Ψ(L11 + L12 + L21 + L22 +B11 +B12 +B21 +B22)

= Ψ(L11 +B11) + Ψ(L12 +B12) + Ψ(L21 +B21) + Ψ(L22 +B22)

= Ψ(L11) + Ψ(B11) + Ψ(L12) + Ψ(B12) + Ψ(L21) + Ψ(B21) + Ψ(L22) +

Ψ(B22)

= Ψ(L11 + L12 + L21 + L22) + Ψ(B11 +B12 +B21 +B22)

= Ψ(L) + Ψ(R).

Lemma 8. (1) α1Ψ(α1)α2 = −α1Ψ(α2)α2.
(2) α2Ψ(α1)α1 = −α2Ψ(α2)α1.
(3) α1Ψ(α2)α1 = α2Ψ(α1)α2 = 0.

Proof. Using the fact that Qn(α1, α1, α1, . . . , α1, α2) = 0 and Lemma 1 , we obtain

0 = Ψ(Qn(α1, α1, α1, . . . , α1, α2))

= Qn(Ψ(α1), α1, α1, . . . , α1, α2) +Qn(α1,Ψ(α1), α1, . . . , α1, α2) +Qn(α1, α1,Ψ(α1),

. . . , α1, α2) + · · ·+Qn(α1, α1, α1, . . . ,Ψ(α1), α2) +Qn(α1, α1, α1, . . . , α1,Ψ(α2))

= α1Ψ(α1)α2 + α2Ψ(α1)
∅α1 + α1Ψ(α1)α2 + α2Ψ(α1)

∅α1 + 2α1Ψ(α1)α2 + 2α2Ψ(α1)
∅

α1 + · · ·+ 2n−3α1Ψ(α1)α2 + 2n−3α2Ψ(α1)
∅α1 + 2n−2α1Ψ(α2) + 2n−2Ψ(α2)α1.

By multiplying the above relation by α1 on the left and α2 on the right, we obtain

α1Ψ(α1)α2 = −α1Ψ(α2)α2.

(2) Since Qn(α2, α2, α2, . . . , α2, α1) = 0 and Lemma 1 , we obtain

0 = Ψ(Qn(α2, α2, α2, . . . , α2, α1))

= Qn(Ψ(α2), α2, α2, . . . , α2, α1) +Qn(α2,Ψ(α2), α2, . . . , α2, α1) +Qn(α2, α2,Ψ(α2),

. . . , α2, α1) + · · ·+Qn(α2, α2, α2, . . . ,Ψ(α2), α1) +Qn(α2, α2, α2, . . . , α2,Ψ(α1))

= α2Ψ(α2)α1 + α1Ψ(α2)
∅α2 + α2Ψ(α2)α1 + α1Ψ(α2)

∅α2 + 2α2Ψ(α2)α1 + 2α1Ψ(α2)
∅

α2 + · · ·+ 2n−3α2Ψ(α2)α1 + 2n−3α1Ψ(α2)
∅α2 + 2n−2α2Ψ(α1) + 2n−2Ψ(α1)α2.



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 10 of 16

After multiplying the final relation by α2 from the left and by α1 from the right, we get

α2Ψ(α1)α1 = −α2Ψ(α2)α1.

(3) un (1), we have

0 = α1Ψ(α1)α2 + α2Ψ(α1)
∅α1 + α1Ψ(α1)α2 + α2Ψ(α1)

∅α1 + 2α1Ψ(α1)α2 + 2α2Ψ(α1)
∅

α1 + · · ·+ 2n−3α1Ψ(α1)α2 + 2n−3α2Ψ(α1)
∅α1 + 2n−2α1Ψ(α2) + 2n−2Ψ(α2)α1.

Taking the left and right sides and multiplying them by α1, respectively, yields

α1Ψ(α2)α1 = 0.

Similarly, in (2), we have

0 = α2Ψ(α2)α1 + α1Ψ(α2)
∅α2 + α2Ψ(α2)α1 + α1Ψ(α2)

∅α2 + 2α2Ψ(α2)α1 + 2α1Ψ(α2)
∅

α2 + · · ·+ 2n−3α2Ψ(α2)α1 + 2n−3α1Ψ(α2)
∅α2 + 2n−2α2Ψ(α1) + 2n−2Ψ(α1)α2.

When we take the left and right sides and multiply them by α2, respectively, we get

α2Ψ(α1)α2 = 0.

Lemma 9. α1Ψ(α1)α1 = α2Ψ(α2)α2 = 0.

Proof. For B12 ∈ B12, we have 2n−2B12 = Qn(α1, α1, α1, . . . , α1, B12) and Using
Lemma 7, we obtain

2n−2Ψ(B12) = Ψ(Qn(α1, α1, α1, . . . , α1, α1, B12))

= Qn(Ψ(α1), α1, α1, . . . , α1, α1, B12) +Qn(α1,Ψ(α1), α1, . . . , α1, α1, B12)

+Qn(α1, α1,Ψ(α1), . . . , α1, α1, B12) + · · ·+Qn(α1, α1, α1, . . . ,Ψ(α1), α1,

B12) +Qn(α1, α1, α1, . . . , α1,Ψ(α1), B12) +Qn(α1, α1, α1, . . . , α1, α1,

Ψ(B12))

= Ψ(α1)α1B12 + (2n−2 − 2){α1Ψ(α1)α1B12}+ α1Ψ(α1)B12 +B12Ψ(α1)
∅

α1 +Ψ(α1)α1B12 + (2n−2 − 2){α1Ψ(α1)α1B12}+ α1Ψ(α1)B12 +

B12Ψ(α1)
∅α1 + 2Ψ(α1)α1B12 + (2n−2 − 4){α1Ψ(α1)α1B12}+ 2α1Ψ(α1)

B12 + 2B12Ψ(α1)
∅α1 + · · ·+ 2n−4Ψ(α1)α1B12 + (2n−2 − 2n−3)

{α1Ψ(α1)α1B12}+ 2n−4α1Ψ(α1)B12 + 2n−4B12Ψ(α1)
∅α1 +

2n−3α1Ψ(α1)B12 + 2n−3Ψ(α1)α1B12 + 2n−3B12Ψ(α1)
∅α1 +

2n−2α1Ψ(B12) + 2n−2Ψ(B12)α1.

Multiplying both sides by α1 and α2 from left and right, respectively and using 2-torsion
freeness of B. We obtain α1Ψ(α1)α1B12 = 0, implies α1Ψ(α1)α1Sα2 = 0 for all S ∈ B. It
follows from (♠) that α1Ψ(α1)α1 = 0. Similarly, we can prove that α2Ψ(α2)α2 = 0.



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 11 of 16

Lemma 10. Ψ(u) = 0.

Proof. Applying the last three lemmas to get the desired outcome.

Lemma 11. For every S ∈ B, Ψ(S∅) = Ψ(S)∅.

Proof. Notice that Qn(u, u, . . . , S, u) = 2n−2(S +S∅), for all S ∈ B. Apply Lemmas 7,
10 and using the condition that rings B is 2-torsion free to get

2n−2(Ψ(S) + Ψ(S∅)) = Ψ(Qn(u, u, . . . , S, u, u))

= Qn(u, u, . . . ,Ψ(S), u, u)

= 2n−2(Ψ(S) + Ψ(S)∅)

which implies
Ψ(S∅) = Ψ(S)∅.

Now, let M = α1Ψ(α1)α2 − α2Ψ(α1)α1, then M∅ = −M . Define a mapping ζ : B →
B by ζ(L) = Ψ(L) − (LM − ML) for every L ∈ B. It is simple to confirm that
ζ(Qn(L1, L2, . . . , Ln)) =

∑n
i=1Qn(L1, . . . , Li−1, ζ(Li), Li+1, . . . , Ln) for all L1, L2, . . . , Ln ∈

B,

Remark 2.1. ζ possesses the following behaviors
(a) ζ(L∅) and ζ(L)∅ identical.
(b) ζ is additive.
(c) ζ(α1) and ζ(α2) are vanish.
(d) ζ(u) is zero.
(e) ζ is a ∅-derivation iff Ψ is a ∅-derivation.

Lemma 12. ζ(Aij) ⊆ Aij i,j=1,2.

Proof. From Qn(u, u, . . . , u, α1, A12) = 2n−2A12 and the above remark, we get

2n−2ζ(A12) = ζ(Qn(u, u, . . . , u, α1, A12))

= Qn(u, u, . . . , u, α1, ζ(A12)

= 2n−2{α1ζ(A12) + ζ(A12)α1}.

This implies that α1ζ(A12)α1 = 0 and α2ζ(A12)α2 = 0, applying Qn(u, u, . . . , u, A12, α1) =
0 and the last remark 2.1, we find

0 = ζ(Qn(u, u, . . . , u, A12, α1))

= Qn(u, u, . . . , u, ζ(A12), α1)

= 2n−2{ζ(A12)α1 + α1ζ(A12)
∅}.

This implies that α2ζ(A12)α1 = 0, thus ζ(A12) ⊆ A12. Similarly, we can show that
ζ(A21) ⊆ A21. Similar as last two steps,

0 = ζ(Qn(u, u, . . . , u, α2, A11))



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 12 of 16

= Qn(u, u, . . . , u, α2, ζ(A11))

= 2n−2{α2ζ(A11) + ζ(A11)α2}.

This implies that α2ζ(A11)α2 = α1ζ(A11)α2 = α2ζ(A11)α1 = 0, thus ζ(A11) ⊆ A11. Simi-
larly, we can show that ζ(A22) ⊆ A22.

Lemma 13. For any Aij , Bij ∈ Bij , 1 ≤ i, j ≤ 2, we get
(1) ζ(A11B12) = ζ(A11)B12 +A11ζ(B12) and ζ(A22B21) = ζ(A22)B21 +A22ζ(B21).
(2) ζ(A12B21) = ζ(A12)B21 +A12ζ(B21) and ζ(A21B12) = ζ(A21)B12 +A21ζ(B12).
(3) ζ(A11B11) = ζ(A11)B11 +A11ζ(B11) and ζ(A22B22) = ζ(A22)B22 +A22ζ(B22).
(4) ζ(A12B22) = ζ(A12)B22 +A12ζ(B22) and ζ(A21B11) = ζ(A21)B11 +A21ζ(B11).

Proof. (1) From Qn(u, u, . . . , A11, B12) = 2n−2(A11B12) and Remark 2.1, we find

2n−2ζ(A11B12) = ζ(Qn(u, u, . . . , A11, B12))

= Qn(u, u, . . . , ζ(A11), B12) +Qn(u, u, . . . , A11, ζ(B12)).

Using the last lemma, we find

ζ(A11B12) = ζ(A11)B12 +A11ζ(B12).

Similarly, we can prove that ζ(A22B21) = ζ(A22)B21 +A22ζ(B21).

(2) Again, Qn(u, u, . . . , A12, B21) = 2n−2A12B21 and last remark, we get

2n−2ζ(A12B21) = ζ(Qn(u, u, . . . , A12, B21))

= Qn(u, u, . . . , ζ(A12), B21) +Qn(u, u, . . . , A12, ζ(B21))

Applying the last Lemma, we have

ζ(A12B21) = ζ(A12)B21 +A12ζ(B21).

Similarly, we can prove that ζ(A21B12) = ζ(A21)B12 +A21ζ(B12).

(3) For any X12 ∈ B12 and Qn(u, u, . . . , A11B11, X12) = 2n−2(A11B11X12) and making use
of the last remark, we get

2n−2ζ(A11B11X12) = ζ(Qn(u, u, . . . , A11B11, X12)

= Qn(u, u, . . . , ζ(A11B11), X12) +Qn(u, u, . . . , A11B11, ζ(X12))

= 2n−2{ζ(A11B11)X12 +X12ζ(A11B11) +A11B11ζ(X12) + ζ(X12)

A11B11}.



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 13 of 16

Lemma 12 yields that

ζ(A11B11X12) = ζ(A11B11)X12 +A11B11ζ(X12).

Further, Qn(u, u, . . . , A11, B11X12) = 2n−2(A11B11X12) and making use of Remark 2.1
implies the following

2n−2ζ(A11B11X12) = ζ(Qn(u, u, . . . , A11, B11X12)

= Qn(u, u, . . . , ζ(A11), B11X12) +Qn(u, u, . . . , A11, ζ(B11X12))

= 2n−2{ζ(A11)B11X12 +A11ζ(B11X12)}.

Using Lemma 13(1), we have

ζ(A11B11X12) = ζ(A11)B11X12 +A11ζ(B11)X12 +A11B11ζ(X12).

Comparing the above two expressions for ζ(A11B11X12), we get (ζ(A11B11)−ζ(A11)B11−
A11ζ(B11))X12 = 0, implies (ζ(A11B11)− ζ(A11)B11 − A11ζ(B11))Y α2 = 0 for all Y ∈ B.
it follows from (♠) that ζ(A11B11) = ζ(A11)B11+A11ζ(B11). Similarly, we can prove that

ζ(A22B22) = ζ(A22)B22 +A22ζ(B22).

(4) Apply Qn(u, u, . . . , u, α1, A12, B22) = 2n−3(A12B22 + B22A
∅
12) and making use of Re-

mark 2.1 to get

2n−3{ζ(A12B22) + ζ(B22A
∅
12)}

= ζ(Qn(u, u, . . . , u, α1, A12, B22))

= Qn(u, u, . . . , u, α1, ζ(A12), B22) +Qn(u, u, . . . , u, α1, A12, ζ(B22)).

Lemma 12 yields

ζ(A12B22) + ζ(B22A
∅
12) = ζ(A12)B22 +B22ζ(A12)

∅ +A12ζ(B22) + ζ(B22)A
∅
12.

From Lemmas 11 and 13(1), we obtain

ζ(A12B22) + ζ(B22)A
∅
12 +B22ζ(A

∅
12) = ζ(A12)B22 +B22ζ(A12)

∅ +A12ζ(B22)

+ζ(B22)A
∅
12.

Hence
ζ(A12B22) = ζ(A12)B22 +A12ζ(B22).

Similarly, we can prove that ζ(A21B11) = ζ(A21)B11 +A21ζ(B11).

Lemma 14. ζ(LR) = ζ(L)R+ Lζ(R), for all L,R ∈ B.



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 14 of 16

Proof. For any L,R ∈ B, write L = L11+L12+L21+L22 and R = B11+B12+B21+B22.
Use the additivity of ζ Lemma 13 to get

ζ(LR) = ζ(L11B11 + L11B12 + L12B21 + L12B22

+L21B11 + L21B12 + L22B21 + L22B22)

= ζ(L11B11) + ζ(L11B12) + ζ(L12B21)ζ(L12B22)

+ζ(L21B11) + ζ(L21B12) + ζ(L22B21) + ζ(L22B22)

= ζ(L11 + L12 + L21 + L22)(B11 +B12 +B21 +B22)

+(L11 + L12 + L21 + L22)ζ(B11 +B12 +B21 +B22).

= ζ(L)R+ Lζ(R).

Lemma 14 and Remark 2.1 show that ζ is an additive ∅-derivation. As a result, it deduces
that Ψ will be an additive ∅- derivation. Which completes the proof of the Theorem 1.

3. Corollaries

Remember that a ring B is said to be prime if, B1, B2 ∈ B, B1BB2 = {0} implies that
either B1 = 0 or B2 = 0. it’s easy to observed that every prime ∅-rings satisfies property
(♠). Therefore, Theorem 1 directly leads to the following conclusion:

Corollary 3.1. Let B be a 2-torsion free unital prime ∅-rings having a non-trivial sym-
metric idempotent. Then a map Ψ : B → B (not necessarily additive) satisfies

Ψ(Qn(B1, B2, . . . , Bn)) =
n∑

i=1

Qn(B1, . . . , Bi−1,Ψ(Bi), Bi+1, . . . , Bn) (3.1)

for all B1, B2, . . . , Bn ∈ B iff Ψ is an additive ∅-derivation.

Remember that a algebra B is said to be prime if, B1, B2 ∈ B, B1BB2 = {0} implies
that either B1 = 0 or B2 = 0. it’s easy to observed that every prime ∅-algebra satisfies
property (♠). Thus, we can infer the following outcome as ac immediate implication of
Theorem 1:

Corollary 3.2. If B is a unital prime ∅-algebra containing a non-trivial projection α1

and α2 = u− α1, then a mapping Ψ : B → B (not necessarily additive) satisfies

Ψ(Qn(B1, B2, . . . , Bn)) =

n∑
i=1

Qn(B1, . . . , Bi−1,Ψ(Bi), Bi+1, . . . , Bn) (3.2)

for all B1, B2, . . . , Bn ∈ B if and only if Ψ is an additive ∅-derivation.

Since a factor von Neumann algebra is also prime, it always satisfies (♠). The following
result follows as a consequence this corollary 3.2:



Md Arshad Madni et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6084 15 of 16

Corollary 3.3. If B is a factor von Neumann algebra having a dimension greater than or
equal to 2, then a mapping Ψ : B → B (not necessarily additive) fulfills

Ψ(Qn(B1, B2, . . . , Bn)) =
n∑

i=1

Qn(B1, . . . , Bi−1,Ψ(Bi), Bi+1, . . . , Bn) (3.3)

for all B1, B2, . . . , Bn ∈ B iff Ψ is an additive ∅-derivation.

Acknowledgments

The authors of the present article extend their gratitude to the Deanship of Gradu-
ate Studies and Scientific Research of the Islamic University of Madinah for the support
provided to the Post-Publication Program 4.

Author Contributions

Conceptualization by M.A. Madni and A.Z. Ansari; validation by M.R. Modumder and
F. Shujat; investigation by M.A. Madni and F. Shujat; writing-original draft by M.A.
Madni and M.R. Modumder; writing–review and editing by A.Z. Ansari and F. Shujat;
supervision by A.Z. Ansari. All authors have read and agreed for the publication of the
present manuscript.

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