397 

© 2025 The Author(s). Published by College of Education for Pure Science (Ibn Al-Haitham), 

University of Baghdad. This is an open-access article distributed under the terms of the Creative Commons 

Attribution 4.0 International License 

 

 
Centralizer on Lie-ideal of Semi-prime Inverse Semi-ring 

 

Ali JA. Abass1*  , Abdulahman H. Majeed 2 , Mohammed Yasin 3  and  

Shrooq Bahjat Smeein4    
1 Department of Mathematics, College, of Science, University of Baghdad, Baghdad, Iraq. 

2 Department of Mathematic , Al-Mamoun University College, Baghdad, Iraq. 
3 Department of Mathematics, An-Najah National University, Nablus P400, Palestine. 

4 Information Department, Section Mathematics, University of Technology and Applied Science - 

Muscat, Sultanate of Oman. 

*Corresponding Author. 

 

 Received: 10 May 2023                        Accepted: 27 August 2023                   Published: 20 January 2025  

 doi.org/10.30526/38.1.3482 

 
 Abstract   

     The summary purpose of this work: We extending certain results on α-centralizer of inverse 

semiring under specific conditions, achieve new results on lie ideal of inverse semiring with some 

consequent collieries, generalize assorted α-centralizer for lie ideal of inverse semiring with some 

collieries, investigate significant theorems on jordan α-centralizer of prime inverse semiring and 

we extend certain results of 𝛼 −centralizers and jordan 𝛼 −centralizers on lie-ideals of prime 

semi-rings to prime inverse semi-ring, we generalizing the results of Mary in to α-centralizer on 

semiring, Also we generalize our results on lie ideals of inverse semiring. We extending the results 

of Shafiq,  Aslam,  Javed to 𝛼 − centralizer of Inverse semiring. 𝑠𝑖𝑛𝑐𝑒 𝑅 is left (right) Jordan 𝛼 −

centralizer on”V, we get the output R is a left (right) 𝛼 − centralizer on 𝑉.”If it where 

𝛼 is an automorphism of  V,𝑅(𝑢)  ∈  𝑉, for any  𝑢 ∈  𝑉, and 𝛼(𝑍(𝑉)) =  𝑍(𝑉). We also get the 

following output  R is 𝑎 𝛼 − centralizer on 𝑉.  

Keywords: Lie-ideal, prime inverse semi-ring, semi-prime inverse semi-ring, 𝛼 −centralizer, 

jordan α-centralizer. 

 

1. Introduction 

     Let 𝑀 be a non-empty set with binary operation (•) defined on 𝑀, then (𝑀,•) is named  semi −

group iff  𝑘 •  (𝑠 •  𝑡)  =  (𝑘 •  𝑠)  •  𝑡 for any  𝑘, 𝑠, 𝑡 ∈  𝑀(1), a semi − group 𝑀 is named 

commutative semi − group if 𝑘 •  𝑠 =  𝑠 •  𝑘, holds   𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑠 ∈  𝑀 (1), A non − empty set 

with two − binary operations(+) and (•) is named  semi-ring iff the following requirements 

hold: 

i)   (𝑀, +) is commutative semi − group.  

https://creativecommons.org/licenses/by/4.0/
https://creativecommons.org/licenses/by/4.0/
https://doi.org/10.30526/38.1.3501
https://orcid.org/0009-0005-8166-2106
mailto:ali.jaafar1603b@sc.uobaghdad.edu.iq
https://orcid.org/0000-0001-8534-0749
mailto:dulrahman.h.majeed@almamonuc.edu.iq
https://orcid.org/0009-0009-9394-5698
mailto:m.yasin@najah.edu
https://orcid.org/0009-0002-9351-4176
mailto:shrooq.smeein@hct.edu.om


IHJPAS. 2025, 38 (1) 
 

398 
 

ii)   (𝑀,•) semi − group. 

iii) 𝑎 •  (𝑘 +  𝑠)  =  𝑎 •  𝑘 +  𝑎 •  𝑠 and(𝑘 +  𝑠)  •  𝑎 = 𝑘 •  𝑎 +  𝑠 •  𝑎𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎, 𝑘, 𝑠 ∈

 𝑀 (2), (𝑀, +) is named additive commutative with neutral element 0. ( i. e. 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈ 𝑀, 𝑘 +

0 = 0 + 𝑘 = 𝑘)  iff  𝑘 +  𝑠 =   𝑘 +  𝑛  holds for any  𝑘, 𝑠 ∈  𝑀, and (𝑀,•) is a semi − group with 

zero 0, 𝑖. 𝑒. , 0. 𝑎 =  𝑎. 0 =  0 for any  𝑎 ∈  𝑀. A semi − ring (𝑀, +,•) is named commutative 

iff 𝑘 •  𝑠 =  𝑠 • 𝑘  holds for any 𝑘, 𝑠 ∈  𝑀 (2), Let (M, +, •) be an additively commutative semi-

ring. Then M is named inverse semi-ring, if (M, +) is an inverse semi-group (i.e) for each 𝑘 ∈

 𝑀 there are a unique 𝑘′ ∈  𝑀 such that, 𝑘 =  𝑘 +  𝑘′ +  𝑘      and    𝑘′ +  𝑘 +  𝑘′ =   𝑘′  (2), 

and is  called  cancellative semi − ring iff for any  𝑘, 𝑠, 𝑚 ∈  𝑀, such that  𝑘 +  𝑠  =   𝑘 +

 𝑚, then 𝑠 =  𝑚.A semi-ring𝑀 is named prime semi-ring if for any  𝑘, 𝑠  ∈  𝑀, 𝑘 𝑀 𝑠 =

 0 implies that either 𝑘 =  0 𝑜𝑟 𝑠 =  0. A semi − ring 𝑀  is named a semi-prime if  for any  𝑘 ∈

 𝑀, 𝑘 𝑀 𝑘 =  0  mplies that 𝑘 =  0. (3), A semi-ring M is named 𝑞 − torsion free where 𝑞 ≠  0 

is an integer if whenever q𝑘 =  0 with 𝑘 ∈  𝑀, then 𝑘 =  0 . A 

commutator [. , . ] in inverse semi − rings defines as [𝑘, 𝑠]  =  𝑘𝑠 +  𝑘𝑠´ 𝑎𝑛𝑑, 𝑘 𝑜 𝑠 =  𝑘𝑠 +  𝑘𝑠 

(3). In (4) Albas presented the 𝛼 − centralizer concept and the Jordan α −centralizer concept, 

which could be a generalization of Jordan centralizer and centralizer and tried beneath particular 

requirements on a 2 −torsion free semi − prime ring, each Jordan α-centralizer is α centralizer, 

where α could be a surjective homomorphism. Inverse semi-rings considered in different directions 

by numerous authors, see (5-12). In this work our aim is to consider the results of Majeed and 

Meften (13) in the inverse semi-ring. In this article, M will represent additive inverse semi-ring that 

satisfies the requirement that for any r ∈ M, 𝑘 + �́� is located in the center Z(M) of M . 

 

2. Preliminaries 

     We recalled the definitions of  lie − ideal, square closed Lie − ideal of a semiring  𝑀,  and 

some definitions, lemmas that will be used later. 

Definition (2.1):(14) 

      An  additive sub semi − group  of inverse semi − ring 𝑀 satisfies[𝑛, q]  =  𝑛q +  q′𝑘 ∈

 𝑉 for any  𝑘 ∈  𝑉, q ∈  𝑀, is named a Lie-ideal of M . 

Definition (2.2):(14) 

     Let V be a lie − ideal of a ring, "then 𝑉 is named a squane closed Lie − ideal" of 𝑀 if  𝑘2  ∈

 𝑉 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈  𝑉. 

Note that if V is a square closed Lie-idealof 𝑀, then 2𝑘q ∈  𝑉 for any  𝑘, q ∈  𝑉. 

Definition (2.3):(2), (15) 

     Let I be a nonzero ideal of 𝑀, the set 𝑍(𝐼)  =  {𝑘 ∈  𝐼, 𝑘q =  q𝑘, for any  q ∈  𝐼} is named 

the center of  𝐼. 

Definition (2.4):(2), (16) 

     Let  𝑞 ∈  𝑀 , the set 𝑍(𝑀)  =  {𝑘 ∈  𝑀 , 𝑘𝑞 =  𝑞𝑘, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑞 ∈  𝑀} is named the center 

of the semi − ring M. Clearly that 𝑍(𝑀) is a subsemi − ring of 𝑀. 

Note that if M is multiplicatively commutative then 𝑍(𝑀)  =  𝑀. 

 

 



IHJPAS. 2025, 38 (1) 
 

399 
 

Lemma (2.5):(10), (17) 

     Let M be an additive inverse semi-ring, for any  k, 𝑞 ∈  𝑀, 𝑖𝑓 𝑘 +  𝑞 =  0 then 𝑘 =  𝑞′. Note 

that in general 𝑘 + 𝑘 ′ ≠ 0, 𝑘 + 𝑘 ′ = 0, iff there are some 𝑞 𝜖 𝑀 with 𝑘 + 𝑞 = 0 [2] 

Proposition (2.6):(12),(18) 

     For any r, s ∈ M, the following are holds: 

i. (𝑘 +  𝑞)′ =  𝑘′ +  𝑞′ 

ii. (𝑘 𝑞)′′ =  𝑘′𝑞 =  𝑘𝑞′ 

iii. 𝑘′′ =  𝑘 

iv. 𝑘′𝑞′ =  (𝑘′𝑞)′ =  (𝑘𝑞)′′ =  𝑘𝑞. 

Lemma (2.7):(12),(19) 

     Let M be ring and k, 𝑞, 𝑤 ∈ 𝑀 then 

i. [𝑘, 𝑘]  =  0 

ii. [𝑘 +  𝑞, 𝑤]  =  [𝑘, 𝑤]  +  [𝑞, 𝑤] 

iii. [𝑘𝑞, 𝑤]  =  𝑘[𝑞, 𝑤]  + [𝑘, 𝑤]𝑞 

iv. [𝑘, 𝑞𝑤]  =  𝑞[𝑘, 𝑤]  +  [𝑘, 𝑞]𝑤. 

Definition (2.8):(15),(20) 

     Let M be a semi-ring,an  additive  mapping 𝑅: 𝑀 →  𝑀 is nameda (𝛼, 𝛼) −

derivation" 𝑖𝑓 𝑅(𝑘𝑞) = 𝑅(𝑘)𝛼(𝑞) +  𝛼(𝑘)𝑅(𝑞) for any 𝑘, 𝑞 ∈  𝑀, and we say that R is 

Jordan (𝛼, 𝛼) − derivation" if  𝑅(𝑘2) = 𝑅(𝑘)𝛼(𝑘) +  𝛼(𝑘)𝑅(𝑘) for any 𝑘 ∈  𝑀,where 𝛼 be 

additive mapping on 𝑀.  

Every derivation is (𝛼, 𝛼) − derivation is Jordan (𝛼, 𝛼) − derivation, but the converse in general 

is not true. 

Definition (2.9):(3),(21) 

     A left (right)𝛼 − centralizer"" of a semi-ring 𝑀 is an “additive mapping” 𝑅: 𝑀 →  𝑀  which 

satisfies 𝑅(𝑘𝑞) +  𝑅(𝑘)𝛼(𝑞)′ = 0, (𝑅(𝑘𝑞) +  𝛼(𝑘)′𝑅(𝑞) = 0) for any  𝑘, 𝑞 ∈

 𝑀.  "𝛼 −centralizer of a ring 𝑀 is “both left and right”   𝛼 − centralizer", "where  𝛼  is an additive 

mapping” on M . 

Definition (2.10):(3),(22) 

     A left (right) Jordan” 𝛼 − "centralizer"" of a semi-ring M is an “addittive mapping” 𝑅: 𝑀 →

 𝑀 which satisfy 𝑅(𝑘2) +  𝑅(𝑘) 𝛼(𝑘)′ = 0, (𝑅(𝑘2) +  𝛼(𝑘)′𝑅(𝑘) = 0)  for any  𝑘 ∈ 𝑀, 𝛼 −

""Jordan"  centralizer of a ring M is both “left and right Jordan”  𝛼 −  centralizer, where  𝛼  be 

“additive mapping” on M . 

 

3. Main Results 

     To verify our main results, we must utilize the following. 

Lemma (3.1):(4),(23) 

     If  𝑉 ⊄ 𝑍(𝑀) is a Lie-ideal of a 2 − tortion free prime" semirig 𝑀 𝑎𝑛𝑑 𝑘, 𝑞 ∈  𝑀 such that 

𝑘 𝑉 𝑞 =  0, then 𝑘 =  0 or 𝑚 =  0. 

From this we mean by V is a square closed lie − ideal of 𝑀. " 

Lemma (3.2) 



IHJPAS. 2025, 38 (1) 
 

400 
 

     Let M be a 2 − tortion free prime semi-ring. Suppose that 𝐹, 𝐺 ∶  𝑉𝑥𝑉 →  𝑉 biadditive 

mappings. If 𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑘, 𝑞)  =  0 for any  𝑘, 𝑞, 𝑤 ∈  𝑉, then 𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑣)  =  0 for any  

𝑘, 𝑞, 𝑢, 𝑣, 𝑤 ∈  𝑉. 

Proof: 

                 𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑘, 𝑞)  =  0                                     𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑤 ∈  𝑉                              (*) 

Replace 𝑘 with 𝑘 +  𝑢, we have 

                𝐹(𝑘 +  𝑢, 𝑞) 𝑤 𝐺(𝑘 +  𝑢, 𝑞)  =  0                 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑤, 𝑢  ∈  𝑉 

By using the additive of F and G 

                𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑞)  =  𝐹(𝑢, 𝑞)′ 𝑤 𝐺(𝑘, 𝑞)  

Replace w by 24 𝐹(𝑘, 𝑞) 𝑧 𝐺(𝑢, 𝑞) 

               (𝐹(𝑘, 𝑞)𝑤 24𝐺(𝑢, 𝑞)) 𝑧 𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑞)  = 

                𝐹(𝑢, 𝑞)′ 𝑤 24𝐺(𝑢, 𝑞) 𝑧 𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑘, 𝑞)  =  0 

by (*), we get 

               24𝐹(𝑘, 𝑞)𝑤𝐺(𝑢, 𝑞)𝑧𝐹(𝑘, 𝑞)𝑤𝐺(𝑢, 𝑞) = 0     𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑢, 𝑧 ∈ 𝑉                         (**) 

If  𝑉 ⊄  𝑍(𝑀), by Lemma(3.1), we get” 

                𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑞)  =  0                                      𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑢, 𝑤 ∈  𝑉” 

If  𝑉 ⊂  𝑍(𝑀), multiply the relation (**) from the right by 𝑧𝑡, where 𝑡 ∈  𝑀, we get 

      24𝐹(𝑘, 𝑞)𝑤 𝐺(𝑢, 𝑞) 𝑧 𝑡 𝐹(𝑘, 𝑞)𝑤𝐺(𝑢, 𝑞)𝑧 =  0,    𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑢, 𝑧, 𝑤 ∈ 𝑉, 𝑡 ∈  𝑀 

Since M is 2 − tortion free prime semi-ring, we have 

               𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑞) 𝑧 =  0                                  𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑢, 𝑧, 𝑤 ∈  𝑉 

If we multiply the relation by t an element of M, which is prime, and do a right multiplication, the 

result is                 

               𝐹(𝑘, 𝑞) 𝑤 𝐺(𝑢, 𝑞)  =  0                                    𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑢, 𝑤 ∈  𝑉 

We can acquire the lemma's claim by exchanging 𝑞  𝑓𝑜𝑟  𝑞 + 𝑣,  in a way analogous to the one 

used above. 

Theorem (3.3) 

     Let 𝑀  be 2 − tortion free prime semi-ring. If 𝑅 is left (right) Jordan 𝛼 − centralizer on”V, 

then R is a left (right) 𝛼 − centralizer on 𝑉.” 

Proof: 

𝑅(𝑘2)  +  𝑅(𝑘)′𝛼(𝑘)  =  0                               𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈  𝑉                                                         (1) 

we replace  𝑘  by 𝑘 +  𝑞 when 𝑘, 𝑞 in 𝑈, we get 

                      𝑅((𝑘 + 𝑞)2)  =  𝑅(𝑘 +  𝑞 )𝛼(𝑘 +  𝑞) 

 𝑅(𝑘2 + 𝑘𝑞 + 𝑞𝑘 + 𝑞2) = 𝑅(𝑘2)  +  𝑅(𝑘𝑞 +  𝑞𝑘)  +  𝑅(𝑞2)                     

                                       =  𝑅(𝑘)𝛼(𝑘) +  𝑅(𝑘𝑞 +  𝑞𝑘) +  𝑅(𝑞)𝛼(𝑞) 

“𝑅(𝑘 + 𝑞)𝛼(𝑘 + 𝑞) = 𝑅(𝑘)𝛼(𝑘) + 𝑅(𝑘)𝛼(𝑞) + 𝑅(𝑞)𝛼(𝑘) + 𝑅(𝑞)𝛼(𝑞)” 

We get 

𝑅(𝑘𝑞 + 𝑞𝑘) + 𝑅(𝑘)𝛼(𝑞)′ + 𝑅(𝑞)𝛼(𝑘 )′ = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈ 𝑉                                                     (2) 

By replacing   𝑞 with 2(𝑘𝑞 +  𝑞𝑘) and using (2), we get 

2𝑅(𝑘(𝑘𝑞 +  𝑞𝑘) +  (𝑘𝑞 +  𝑞𝑘)𝑘) +  2𝑅(𝑘)𝛼(𝑘𝑞 )′ +  2𝑅(𝑘)𝛼(𝑞𝑘)′ +  𝑅(𝑘𝑞 +  𝑞𝑘)𝛼(𝑘)′   

=  0 

 2𝑅(𝑘(𝑘𝑞 +  𝑞𝑘) + (𝑘𝑞 +  𝑞𝑘)𝑘) = 2𝑅(𝑘)𝛼(𝑘𝑞 ) + 2𝑅(𝑘)𝛼(𝑞𝑘) + 2 𝑅(𝑘𝑞 + 𝑞𝑘)𝛼(𝑘)  



IHJPAS. 2025, 38 (1) 
 

401 
 

                                                                                                                                                          (3) 

This can also be computed using an alternate way 

2𝑅(𝑘2𝑞 + 𝑞𝑘2) +  4𝑅(𝑘𝑞𝑘) + 2 𝑅(𝑘)𝛼(𝑘𝑞)′ + 2𝑅(𝑞)𝛼(𝑘2)′ =  0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈ 𝑉                 (4) 

From (3) and (4), we obtain  

𝑅(𝑘𝑞𝑘)  +  𝑅(𝑘)𝛼(𝑞𝑘)′ =  0                       𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉                                                         (5) 

If we linearize (5), we get   

𝑅(𝑘𝑞𝑡 + 𝑡𝑞𝑘) + 𝑅(𝑘)𝛼(𝑞𝑡)′ + 𝑅(𝑡)𝛼(𝑞𝑘)′ = 0    𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑡 ∈  𝑉                                         (6) 

Since V is a square closed Lie-ideal, we have 

                                      24(𝑘𝑞𝑡𝑞𝑘 + 𝑞𝑘𝑡𝑘𝑞) ∈ 𝑉.  

Now we shall compute  𝑓 =  24𝑅(𝑘𝑞𝑡𝑞𝑘 +  𝑞𝑘𝑡𝑘𝑞) in two different ways, using (5) we have 

𝑓 + 24𝑅(𝑘)𝛼(𝑞𝑡𝑞𝑘)′ + 𝑅(𝑞)𝛼(𝑘𝑡𝑘𝑞)′ = 0          𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑡 ∈ 𝑉                                            (7) 

Using (6) we have  

𝑓 + 24𝑅(𝑘𝑞)𝛼(𝑡𝑞𝑘)′ + 𝑅(𝑞𝑘)𝛼(𝑡𝑘𝑞)′ = 0          𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑡 ∈ 𝑉                                           (8) 

Comparing (7) and (8)  

 𝑅(𝑘)𝛼(𝑞𝑡𝑞𝑘)′ + 𝑅(𝑞)𝛼(𝑘𝑡𝑘𝑞)′ + 𝑅(𝑘𝑞)𝛼(𝑡𝑞𝑘) + 𝑅(𝑞𝑘)𝛼(𝑡𝑞𝑘) = 0  

 (𝑅(𝑘𝑞) + 𝑅(𝑘)𝛼(𝑞)′)𝛼(𝑡𝑞𝑘) + (𝑅(𝑞𝑘) +  𝑅(𝑞)𝛼(𝑘)′) 𝛼(𝑡𝑘𝑞) = 0 

Introducing a additive mapping,    

                           𝐺(𝑘, 𝑞)  =  𝑅(𝑘𝑞)  +  𝑅(𝑘)𝛼(𝑞)′,  

we arrive at 

                           𝐺(𝑘, 𝑞)𝛼(𝑡𝑞𝑘)  +  𝐺(𝑞, 𝑘)(𝑡𝑘𝑞)  =  0 

By Lemma (2.5) 

𝐺(𝑘, 𝑞)𝛼(𝑡𝑞𝑘)  =  𝐺(𝑞, 𝑘)′𝛼(𝑡𝑘𝑞)                                                                                                 (9) 

 We can be rewritten  equality (2)in this notation as  

                           𝐺(𝑘, 𝑞)  +  𝐺(𝑞, 𝑘)′ =  0.  

 Using equality (9) and this fact, we obtain         

𝐺(𝑘, 𝑞)𝛼( 𝑡 [𝑘, 𝑞])  =  0       𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑡, 𝑧 ∈ 𝑉                                                                       (10) 

Now using Lemma (3.2), we have 

𝐺(𝑘, 𝑞)𝛼( 𝑧 [𝑢, 𝑣]) =  0                       𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧, 𝑢, 𝑣 ∈  𝑉                                                    (11)  

     (i)      If 𝑉  is non commutative” 

        Since 𝛼 is surjective and using Lemma (3.1), we have 

                          𝐺(𝑘, 𝑞) =  0            𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉 

     (ii)     If 𝑉  is commutative and 𝑉 ⊄  𝑍(𝑀)” 

Compute 𝑁 = 24 𝑅(𝑘𝑞𝑧𝑞𝑘) in two different ways. Using (5), we have  

𝑁 + 24 𝑅(𝑘)′𝛼(𝑞𝑧𝑞𝑘) = 0            𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧  ∈  𝑉                                                                (12) 

𝑁 + 24 𝑅(𝑘𝑚)′𝛼(𝑧𝑚𝑘) = 0         𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧  ∈  𝑉                                                                (13) 

From (12) and (13), we arrive at 

                           𝑅(𝑘𝑞)𝛼(𝑧𝑞𝑘) +  𝑅(𝑘)′𝛼(𝑞𝑧𝑞𝑘) =  0 

                         (𝑅(𝑘𝑞)  +  𝑅(𝑘)′𝛼(𝑞))𝛼(𝑧𝑞𝑘)  =  0 

𝐺(𝑘, 𝑞)𝛼(𝑧𝑞𝑘)  =  0                        𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧 ∈  𝑉                                                                 (14) 

Let 𝜓 (𝑘, 𝑞) = 𝛼(𝑞𝑘),“it's clear that 𝜓 is additive mapping,  therefore 



IHJPAS. 2025, 38 (1) 
 

402 
 

                         𝐺(𝑘, 𝑞)𝛼(𝑧)𝜓(𝑘, 𝑞) =  0                𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧 ∈  𝑉 

Using Lemma (3.2), we have 

                         𝐺(𝑘, 𝑞)𝛼(𝑧)𝜓(𝑢, 𝑣)  =  0              𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧, 𝑢, 𝑣 ∈  𝑉 

Implies that 

𝐺(𝑘, 𝑞)𝛼(𝑧𝑢𝑣)  =  0                      𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧, 𝑢, 𝑣 ∈  𝑉                                                           (15) 

Replacing 𝛼(𝑣) with 2𝐺(𝑘, 𝑞)𝛼(𝑧), 𝑢sing Lemma (3.1) and M  is prime semi-ring, we have 

                         𝐺(𝑘, 𝑞)𝛼(𝑧) =  0                           𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞, 𝑧 ∈  𝑉 

Using Lemma (3.1) 

                         𝐺(𝑘, 𝑞)  =  0                                  𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉 

(i)  If 𝑉 ⊂  𝑍(𝑀) 

Multiplying relation (15) on the right by t, where t ∈  𝑀 and since M is a prime, we can obtain the 

result.    

                         𝐺(𝑘, 𝑞)  =  0                               𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉 

If 𝑅(𝑘2) + 𝛼(𝑘)′𝑅(𝑘) = 0, reaching the conclusion of the theorem with the same procedure as 

before completes the proof. 

Lemma (3.4) 

     Let M be a 2 − tortion free prime semi − ring, 𝐻, 𝛼: 𝑀 →  𝑀,  H is (𝛼, 𝛼) − derivation on 

𝑉 𝑎𝑛𝑑 𝑎 ∈  𝑉 some fixed element, where 𝛼   is automorphism of  𝑉, such that 𝛼(𝑉) = 𝑉  then  

(ii) 𝐻(𝑘)𝐻(𝑞)  =  0 for any  𝑘, 𝑞 ∈  𝑈 implies 𝐻 =  0 on 𝑉. 

(iii)𝑎𝛼(𝑘)  +  𝛼(𝑘)′𝑎 ∈  𝑍(𝑉) for any  𝑘 ∈  𝑉 implies 𝑎 ∈  𝑍(𝑉). 

Proof: 

(i) 𝐻(𝑘)𝛼(𝑞)𝐻(𝑘)  =  𝐻(𝑘)𝐻(𝑞𝑘)  +  𝐻(𝑘)′𝐻(𝑞)𝛼(𝑘)  

      𝐻(𝑘)(𝐻(𝑞)𝛼(𝑘)  +  𝛼(𝑞)𝐻(𝑘))  +  𝐻(𝑘)′𝐻(𝑞)𝛼(𝑘) = 0                        

𝐻(𝑘)𝐻(𝑞)𝛼(𝑘) + 𝐻(𝑘)𝛼(𝑞)𝐻(𝑘) + 𝐻(𝑘)′𝐻(𝑞)𝛼(𝑘) =  0 

By hypothesis, and M is inverse semi-ring, we get 

                            𝐻(𝑘)𝛼(𝑞)𝐻(𝑘)  =  0                

Since 𝛼  is automorphism of  𝑉, such that 𝛼(𝑉) = 𝑉,  we get                  

                            𝐻(𝑘) 𝑉 𝐻(𝑘)  =  0                              𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈  𝑉 

If 𝑉 ⊄  𝑍(𝑀), and 𝛼  is automorphism of  𝑉, Lemma (3.2) we have 𝐻 =  0 𝑜𝑛 𝑉. 

If V ⊂ Z(M)  

                              𝐻(𝑘)𝑡𝐻(𝑘)  =  0                                𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈  𝑉, 𝑡 ∈  𝑀 

 So, by primness of M, we have 

                              𝐻 =  0 𝑜𝑛 𝑉 

(ii)   Define 𝐻(𝑘)  =  𝑎𝛼(𝑘)  +  𝛼(𝑘)𝑎′                                                                                                  

 “It is easy to see that” H is a (𝛼, 𝛼) − derivations, since 𝐻(𝑘) ∈  𝑍(𝑉)  for any  𝑘 ∈  𝑉, we have 

𝐻(𝑞)𝛼(𝑘)  =  𝛼(𝑘)𝐻(𝑞) and also 2𝐻(𝑞𝑧)𝛼(𝑘)  =  2 𝛼(𝑘)𝐻(𝑞𝑧) 

Since M is prime, we get 

                        𝐻(𝑞)𝛼(𝑧𝑘) +  𝛼(𝑞)𝐻(𝑧)𝛼(𝑘) 

                        =  𝛼(𝑘)𝐻(𝑞)𝛼(𝑧)  +  𝛼(𝑘𝑞)𝐻(𝑧) 

𝐻(𝑞)(𝛼(𝑧)𝛼(𝑘) + 𝛼(𝑘)𝛼(𝑧)′)  =  𝐻(𝑧)(𝛼(𝑞)𝛼(𝑘)′ +  𝛼(𝑘)(𝑞)) 

                      𝐻(𝑞)[𝛼(𝑧), 𝛼(𝑘)] = 𝐻(𝑧)[𝛼(𝑞), 𝛼(𝑘)] 



IHJPAS. 2025, 38 (1) 
 

403 
 

 Since 𝛼  is automorphism, take 𝛼(𝑧)  =  𝑎. Obviously 𝐻(𝑎)  =  0, so, we obtain by (i) 

                                   𝐻(𝑞)𝐻(𝑘)  =  0 

“By virtue of (i) we get” H = 0 and hence a ∈  Z(M). 

Lemma (3.5) 

     Let M be a 2 − tortion free prime semi − ring, R and α are additive mappings on M, and 𝑎 ∈

 𝑉 some fixed element. If 𝑅(𝑘) = 𝑎 𝛼(𝑘) + 𝛼(𝑘)𝑎and𝑅(𝑘 𝑜 𝑞) + 𝑅(𝑘)𝑜 𝛼(𝑞)′ =

0 and 𝑅(𝑘 𝑜 𝑞) + 𝛼(𝑘)′𝑜𝑅(𝑞) = 0 for any  𝑘, 𝑞 ∈  𝑉 then “𝑎 ∈  𝑍(𝑉), “where  𝛼 is a 

surjective” endomorphism of  𝑉 .                                                                

Proof: 

By hypothesis 

 “ 𝑅(𝑘𝑞 + 𝑞𝑘) = 𝑅(𝑘)𝛼(𝑞)  +  𝛼(𝑞)𝑅(𝑘)                        𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈ 𝑉” 

 𝑅(𝑘𝑞) + 𝑅(𝑞𝑘) =  𝑅(𝑘)𝛼(𝑞)  +  𝛼(𝑞)𝑅(𝑘)                    𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈ 𝑉 

 𝑅(𝑘𝑞) + 𝑅(𝑞𝑘) = 𝑎𝛼(𝑘𝑞) + 𝛼(𝑘𝑞)𝑎 +  𝑎𝛼(𝑞𝑘)  +  𝛼(𝑞𝑘)𝑎 

 =  𝑎𝛼(𝑘)𝛼(𝑞)  +  𝛼(𝑘)𝛼(𝑞)𝑎 +  𝑎𝛼(𝑞)𝛼(𝑘)  +  𝛼(𝑞)𝛼(𝑘)𝑎 

 𝑅(𝑘)𝛼(𝑞) +  𝛼(𝑞)𝑅(𝑘) = 𝑎𝛼(𝑘)𝛼(𝑞) +  𝛼(𝑘)𝑎𝛼(𝑞) + 𝛼(𝑞)𝑎𝛼(𝑘)  +  𝛼(𝑞)𝛼(𝑘)𝑎 

 + 𝑎𝛼(𝑘)𝛼(𝑞)  +  𝛼(𝑘)𝛼(𝑞)𝑎 +  𝑎𝛼(𝑞)𝛼(𝑘)  +  𝛼(𝑞)𝛼(𝑘)𝑎 

   = 𝑎𝛼(𝑘)𝛼(𝑞) +  𝛼(𝑘)𝑎𝛼(𝑞) +  𝛼(𝑞)𝑎𝛼(𝑘) +  𝛼(𝑞)𝛼(𝑘)𝑎  

 (𝑎 +  𝑎′)𝛼(𝑘)𝛼(𝑞)  +  𝛼(𝑞)𝛼(𝑘)(𝑎 +  𝑎′)  +  𝛼(𝑘)𝛼(𝑞)𝑎 + 𝑎𝛼(𝑞)𝛼(𝑘)   +  𝛼(𝑘)𝑎′𝛼(𝑞)  +

 𝛼(𝑞)𝑎′𝛼(𝑘)  =  0 

Since 𝑎 +  𝑎′ ∈  𝑍(𝑉) 

     𝛼(𝑘)𝛼(𝑞)(𝑎 + 𝑎′ + 𝑎) + (𝑎 + 𝑎′ + 𝑎)𝛼(𝑞)𝛼(𝑘) + 𝑎𝛼(𝑞)𝛼(𝑘) + 𝛼(𝑘)𝑎′𝛼(𝑞) = 0  

     𝛼(𝑘)𝛼(𝑞)𝑎 + 𝛼(𝑘)𝑎′𝛼(𝑞) +𝑎𝛼(𝑞)𝛼(𝑘) + 𝛼(𝑞)𝑎′𝛼(𝑘) = 0  

     𝛼(𝑘)(𝛼(𝑞)𝑎 + 𝑎′𝛼(𝑞)) + (𝛼(𝑞)𝑎 + 𝑎′𝛼(𝑞))𝛼(𝑘)′ =  0  

But 𝛼 is a surjective   

𝑎𝛼(𝑘)  +  𝛼(𝑘)𝑎′ ∈  𝑍(𝑉)   

By Lemma (3.4) (ii), we get 𝑎 ∈  𝑍(𝑉) 

            𝑅(𝑘 𝑜 𝑞) + 𝑅(𝑘)𝑜 𝛼(𝑞)′ = 0 and 𝑅(𝑘 𝑜 𝑞) +  𝛼(𝑘)′𝑜 𝑅(𝑞) = 0. 

Lemma (3.6) 

   Let  𝑀 be 𝑎 2 − tortion free prime semi-ring, and 𝑅, 𝛼  are additive mappings on 𝑀, 𝑅 satisfies 

𝑅(𝑘 o 𝑞) + 𝑅(𝑘)o 𝛼(𝑞)′ = 0 and 𝑅(𝑘 𝑜 𝑞) +  𝛼(𝑘)′ 𝑜  𝑅(𝑞) = 0 for any𝑘, 𝑞 ∈  𝑉, then 𝑅(𝑧)  ∈

 𝑍(𝑉) for any  𝑧 ∈  𝑍(𝑉), where 𝛼 is a surjective endomorphism of 𝑉.. 

Proof: 

               𝑅(𝑘𝑞 +  𝑞𝑘)  +  𝑅(𝑘)𝛼(𝑞)′ + 𝛼(𝑞)′𝑅(𝑘)  =  0 

              𝑅(𝑘𝑞 +  𝑞𝑘)  +  𝛼(𝑘)′𝑅(𝑞)  +  𝑅(𝑞)𝛼(𝑘)′ =  0.          

because 𝑅(𝑧)  ∈  𝑍(𝑉) 

Take any t ∈ Z(U)  and denote a = R(t)” 

          2𝑅(𝑡𝑘) =  𝑅(𝑡𝑘 +  𝑘𝑡) = 𝑅(𝑡)𝛼(𝑘) +  𝛼(𝑘)𝑅(𝑡)     

                                                 =  𝑎𝛼(𝑘)  +  𝛼(𝑘)𝑎 

A simple check reveals that 𝑀(𝑘)  =  2𝑅(𝑡𝑘) is satisfies  

𝑀(𝑘 𝑜 𝑞)  =  2𝑅(𝑡(𝑘𝑞 + 𝑞𝑘)  =  2𝑅(𝑡𝑘𝑞 +  𝑞𝑡𝑘) 

                 =  2𝑅(𝑡𝑘)𝛼(𝑞)  + 2 𝛼(𝑞)𝑅(𝑡𝑘) 



IHJPAS. 2025, 38 (1) 
 

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                 =  𝑀(𝑘)𝛼(𝑞)  +  𝛼(𝑞)𝑀(𝑘) 

                 =  𝑀(𝑘)𝑜𝛼(𝑞) 

𝑀(𝑘 𝑜𝑞)  =  2𝑅(𝑡(𝑘𝑞 + 𝑞𝑘)  = 2𝑅(𝑘(𝑡𝑞)  +  (𝑡𝑞)𝑘) 

                 =  2𝛼(𝑘)𝑅((𝑡𝑞)  +  2𝑅(𝑡𝑞)𝛼(𝑘) 

                 =  𝛼(𝑘)𝑀(𝑞)  +  𝑀(𝑞)𝛼(𝑘) 

                 =  𝛼(𝑘) 𝑜 𝑀(𝑞) 

𝑀(𝑘 𝑜 𝑞)  =  𝑀(𝑘) 𝑜 𝛼(𝑞)  =  𝛼(𝑘) 𝑜 𝑀(𝑞)  𝑓𝑜𝑛 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑀 

By Lemma (3.5), we have 𝑅(𝑡) ∈  𝑍(𝑀).  

Theorem (3.7) 

    Let 𝑀 be 2 − tortion free prime semi − ring and 𝑅, 𝛼: 𝑀 →  𝑀  additive mappings, R 

satisfies 𝑅(𝑘 𝑜 𝑞) + 𝑅(𝑘)𝑜𝛼(𝑞)′ = 0 𝑎𝑛𝑑  𝑅(𝑘 𝑜 𝑞) + 𝛼(𝑘)′𝑜𝑅(𝑞) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈ 𝑉 

then 𝑅 is 𝑎 𝛼 − centralizer on 𝑉, where 𝛼 is an automorphism of 𝑉, 𝑅(𝑢)  ∈  𝑉, for any  𝑢 ∈  𝑉,  

and 𝛼(𝑍(𝑉)) =  𝑍(𝑉). 

Proof :  

Since U is a square closed Lie − ideal of 𝑀, and by Lemma (2.5), we get 

                          2𝑅(𝑘𝑞 +  𝑞𝑘) = 2𝑅(𝑘)𝛼(𝑞) +  2𝛼(𝑞)𝑅(𝑘) 

                                                   = 2𝛼(𝑘)𝑅(𝑞) + 2𝑅(𝑞)𝛼(𝑘)                                                  

If V is a commutative, we have 

                           R(r2) = R(r)α(r) = α(r)R(r)                                                                              

If V is a non-commutative 

Replace  𝑞 by 2𝑘𝑞 +  2𝑞𝑘  in (2), we get,    

           4𝑅(𝑘)𝛼(𝑘𝑞 +  𝑞𝑘) + 4𝛼(𝑘𝑞 +  𝑞𝑘)𝑅(𝑘)  

           = 4𝛼(𝑘)𝑅(𝑘𝑞 +  𝑞𝑘) +  4𝑅(𝑘𝑞 +  𝑞𝑘)𝛼(𝑘) 

4𝑅(𝑘)𝛼(𝑘)𝛼(𝑞) +  4𝑅(𝑘)𝛼(𝑞)𝛼(𝑘) + 4𝛼(𝑘)𝛼(𝑞)𝑅(𝑘) + 4𝛼(𝑞)𝛼(𝑘)𝑅(𝑘)  = 

4𝛼(𝑘)𝑅(𝑘)𝛼(𝑞)  + 4𝛼(𝑘)𝛼(𝑞)𝑅(𝑘) + 4𝑅(𝑘)𝛼(𝑞)𝛼(𝑘)  +  4𝛼(𝑞)𝑅(𝑘)𝛼(𝑘)   

By using the property of 2 − tortion free semi − ring, we obtain 

𝑅(𝑘)𝛼(𝑘)𝛼(𝑞) +  𝛼(𝑞)𝛼(𝑘)𝑅(𝑘) + 𝛼(𝑘)′𝑅(𝑘)𝛼(𝑞) + 𝛼(𝑞)𝑅(𝑘)𝛼(𝑘)′  = 0 

Now it follows that  

       [𝑅(𝑘), 𝛼(𝑘)]𝛼(𝑞) =   𝛼(𝑞)[𝑅(𝑘), 𝛼(𝑘)]                         𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉 

but α is surjective, then we get  

[𝑅(𝑘), 𝛼(𝑘)]  ∈  𝑍(𝑉)                                    𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑞 ∈  𝑉 

The next goal is to show that [𝑅(𝑘), 𝛼(𝑘)] =  0  𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈  𝑉.  

Take any t ∈ Z(U) 

              4𝑅(𝑡𝑘)  =  2𝑅(𝑡𝑘 +  𝑘𝑡)  =  2𝑅(𝑡)𝛼(𝑘)  +  2𝛼(𝑘)𝑅(𝑡)  

                            =  2𝑅(𝑘)𝛼(𝑡)  +  2𝛼(𝑡)𝑅(𝑘) 

Using Lemma (3.6), we get  

                            𝑅(𝑡𝑘) =  𝑅(𝑘)𝛼(𝑡) =  𝑅(𝑡)𝛼(𝑘)                𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑡 ∈  𝑉                                      

     4[𝑅(𝑘), 𝛼(𝑘)]𝛼(𝑡)  =  4𝑅(𝑘)𝛼(𝑘𝑡)  +  4𝛼(𝑘)′𝑅(𝑘)𝛼(𝑡)                       

                                      =  4𝑅(𝑘)𝛼(𝑡𝑘)   +  4𝑅(𝑘)𝛼(𝑡)𝛼(𝑘)′ = 0 

Since 𝛼(𝑍(𝑉))  =  𝑍(𝑉), and [𝑅(𝑘), 𝛼(𝑘)] itself is central element, By Lemma (3.1), we get our 

goal. 



IHJPAS. 2025, 38 (1) 
 

405 
 

                        2𝑅(𝑘2) =  𝑅(𝑘𝑘 +  𝑘𝑘) =  𝑅(𝑘)𝛼(𝑘) +  𝛼(𝑘)𝑅(𝑘) 

                                     =  2𝑅(𝑘)𝛼(𝑘)  =  2𝛼(𝑘)𝑅(𝑘). 

By Theorem 3.3, we get our result. 

 

4. Conclusion 

     In this work, we extend certain results of 𝛼-centralizers and Jordan 𝛼-centralizers on lie ideals 

of prime rings to prime inverse semirings. We got the output R is a left (right) 𝛼 −

centralizer on 𝑉.”If it where 𝛼 is an automorphism of  V,𝑅(𝑢)  ∈  𝑉, for any  𝑢 ∈  𝑉, and 

𝛼(𝑍(𝑉)) =  𝑍(𝑉). We also get the following output  R is 𝑎 𝛼 − centralizer on 𝑉 

 

Acknowledgment 

      Our researcher extends his Sincere thanks to the editor and members of the preparatory 

committee of the Ibn AL-Haitham Journal of Pure and Applied Sciences. 

 

Conflict of Interest  

      There are no conflicts of interest. 

 

Funding  

     There is no funding for the article. 

 

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