Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

353 https://internationalpubls.com 

On the Structure of 𝝈𝟏 Near - Rings 
 

1V.Alies Anbukani, 2G.Sugantha, 3P.Sivagami 
1Research Scholar of Mathematics (Part time),  Reg No: 21122102092008, Email:alieslivingston@gmail.com 

PG and Research Department of Mathematics, Kamaraj College, Thoothukudi – 628003. (Affiliated to Manonmaniam 

Sundaranar University, Abishekapatti, Tirunelveli – 627012) 

2Assistant Professor of Mathematics, Pope’s College (Autonomous), Sawyerpuram, Tamil Nadu - 627 251, India. 

E.mail:sugi.trini@gmail.com (Affiliated to Manonmaniam Sundaranar University, Abishekapatti, Tirunelveli – 627012) 

3Associate Professor of Mathematics, PG and Research Department of Mathematics, Kamaraj College, Thoothukudi – 

628003. E.mail:sivagamimuthu75@gmail.com (Affiliated to Manonmaniam Sundaranar University, Abishekapatti, 

Tirunelveli – 627012) 

 

Article History: 

Received: 15-11-2024 

Revised: 26-12-2024 

Accepted:10-01-2025 

Abstract:  

If, in a ring (N, +, ∙) we ignore the commutativity of ‘+’ and one of the distributive laws, (N, +, 

∙) becomes a Near-Ring. If we do not stipulate the left distributive law, (N, +, ∙) is a right near-

ring. This research aims to introduce the concept of σ_1near-ring. N is called σ_1near - ring if 

N is a right near-ring and xy^2=yxyfor all x,y∈N .The element wise characterization for 

σ_1near-ring will be investigated and shall establish theorems and properties in this near - ring. 

Mathematics Subject Classification: 16Y30. 

Keywords: σ_1  near -ring, commutativity, near-field. 

 

1 Introduction 

A right near-ring is a non-empty set N together with two binary operations “+”and “.” such that (1) (N, 

+) is a group. (2) (N, ∙) is a semi-group and (3) (𝑛1  +  𝑛2)𝑛3  =  𝑛1𝑛3 +  𝑛2𝑛3 for all 𝑛1 , 𝑛2, 𝑛3 ∈  𝑁. 

Throughout this paper N stands for a right near-ring. (𝑁, +, . ) with at least two elements  1 and ′0′ 

denotes the identity element of the group (N, +). Obviously, 0n = 0 for all n in N. N is said to be zero-

symmetric if n0 = 0 for all n in N. As in [2], a subgroup of (𝑀, +) of (𝑁, +) is called an N-subgroup 

of N if 𝑁𝑀 ⊂  𝑀  and an invariant N subgroup of N if, in 

addition, 𝑀𝑁 ⊂  𝑀. In [6], N is defined to be pseudo commutative if 𝑥𝑦𝑧 =  𝑧𝑦𝑥 for all 𝑥, 𝑦, 𝑧 in N. 

The concept of a mate function in N has been introduced in [4] with a view to handling the regularity 

structure with considerable case. A map ′f′ from N into N is called (i) a mate function for 𝑁 if 𝑥 =

 𝑥𝑓 (𝑥)𝑥, (ii) a P3 mate function, if, in addition, 𝑥𝑓 (𝑥)  =  𝑓 (𝑥)𝑥 for all 𝑥 in 𝑁. By identity 1 of 𝑁, 

we mean only the multiplicative identity of 𝑁. 

Basic concepts and terms used but left undefined in this paper can be found in [2]. 

2 Notations 

In this section, we furnish the notations which are used frequently throughout this paper. 

(i) E denotes the set of all idempotent of N. 

    (e in N is called an idempotent if 𝑒2  =  𝑒) 

mailto:alieslivingston@gmail.com


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(ii) L denotes the set of all nilpotent of N. 

    (a in N is nilpotent if  ak = 0 for some positive integer k.) and 𝑁 is said to be reduced if  L={0}. 

(iii) 𝑁0  =  {𝑛 ∈  𝑁 / 𝑛0 =  0} - zero-symmetric part of N and 𝑁 is called zero symmetric if        𝑁 =

𝑁0. 

(iv) 𝑁𝑑  =  {𝑛 ∈  𝑁 / 𝑛(𝑥 + 𝑦)  =  𝑛𝑥 + 𝑛𝑦 for all 𝑥, 𝑦 in 𝑁}  – set of all distributive 

      element of N and 𝑁 is called distributive if 𝑁 = 𝑁𝑑. 

(v)   𝐶(𝑁)  =  {𝑛 ∈  𝑁 / 𝑛𝑥 =  𝑥𝑛 for all 𝑥 in 𝑁} - centre of N. 

(vi)  If A is any non – empty subset of 𝑁, then 

i) 𝐴∗ = 𝐴 − {0} 

ii) 𝐶(𝐴) = {𝑛 ∈ 𝑁/𝑛𝑎 = 𝑎𝑛 for all 𝑎 ∈ 𝐴} 

iii) When 𝐴 = 𝑁, 𝐶(𝑁) = {𝑛𝑎 = 𝑎𝑛 for all 𝑎 ∈ 𝑁} −called the centre of 𝑁. 

(vii)  When 𝐸 ⊆ 𝐶(𝑁), we say that the idempotent are central. 

3 Preliminary Results 

We freely make use of the following results and designate them as R (1), R (2)....etc. 

R (1) N is sub directly irreducible if and only if the intersection of any family of non-zero ideals  

of N is again non-zero (Theorem 1.60, p.25 of [2]) 

R (2) N has no non-zero nilpotent elements if and only if 𝑥2  =  0 ⇒  𝑥 =  0 for all 𝑥 in 𝑁 

(Problem 14, p.9 of [3]). 

R (3) If f is a mate function for N, then for every 𝑥 in 𝑁, 𝑥𝑓 (𝑥), 𝑓 (𝑥)𝑥 ∈  𝐸 and 

𝑁𝑥 = 𝑁𝑓 (𝑥)𝑥 , 𝑥𝑁 =  𝑥𝑓 (𝑥)𝑁 (Lemma 3.2 of [5]). 

R (4) If 𝐿 =  {0} and 𝑁 =  𝑁0, then (i) 𝑥𝑦 =  0 ⇒  𝑦𝑥 =  0 for all 𝑥, 𝑦 in 𝑁.(ii) N has  

Insertion of factors property- IFP for short- i.e for 𝑥, 𝑦 in 𝑁, 𝑥𝑦 = 0 ⇒  𝑥𝑛𝑦 =  0  . for all n in N. If 

N satisfies (i) and (ii) then N is said to have (∗, IFP) (Lemma 2.3 of [5]) 

R (5) Any pseudo commutative near-ring with a right identity is weak commutative 

 ((i.e).  𝑥𝑦𝑧 =  𝑥𝑦𝑧 for all 𝑥, 𝑦, 𝑧 in 𝑁 [2]) (Proposition 2.9 of [6]) 

R (6) A zero-symmetric near-ring N is a near- field if 𝑁𝑑 ≠  {0} and for all 𝑛 ∈  𝑁 − {0}, 

𝑁𝑛 = 𝑁 (Theorem 8.3, p.249 of [2]). 

R (7) Let N be a right near-ring. If for every 𝑥, 𝑦 in N, 𝑥𝑁 𝑦 =  𝑁𝑥𝑦 then we say N is a β1 

near-ring. (Definition 3.1.1 of [4]) 

4. Definition and Examples of 𝜎1 Near – Rings 

 In this section we define 𝜎1near-ring and give certain examples of this new concept. 



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Definition 4.1 

 Let N be a right near- ring. Then N is said to be an 𝜎1near-ring if 𝑥𝑦2 = 𝑦𝑥𝑦  

for all 𝑥,   𝑦 ∈ 𝑁. 

Example 4.2(a) 

 The near-ring (𝑁, +, . ) defined on Klein’s four group (𝑁, +) with 𝑁 = {0, 𝑎, 𝑏, 𝑐} where ‘∙’ is 

defined as per scheme 12, 𝑃 408 of Pilz [2]. 

    ∙ 0 a  b c 

    0 0 0 0 0 

    a 0 a 0 a 

    b 0 0 0 0 

    c 0 a 0 a 

is a 𝜎1near-ring 

b) Let (𝑁, +) be the Klein’s four group as in (a) above. If multiplication is defined as per scheme  

22, P.408 of pilz[2] 

    ∙ 0 a  b c 

    0 0 0 0 0 

    a a a a a 

    b 0 0 0 0 

    c a a a a 

Then N is not a 𝜎1near-ring since 𝑎𝑏2 ≠ 𝑏𝑎𝑏. 

5. Properties of 𝜎1near-ring 

In this section we prove certain important properties of 𝜎1 near-ring and give a complete 

characterization of such near-ring. 

Proposition 5.1 

Let 𝑁 = 𝑁𝑑 be a 𝜎1near-ring with identity. Then N is commutative. 

Proof 

Let N be a 𝜎1near-ring. 

Then for all 𝑥, 𝑦 ∈ 𝑁, 𝑥𝑦2 = 𝑦𝑥𝑦…………... (1) 

Replace the element 𝑦 by 𝑦 + 𝑒. in (1) 

                𝑥(𝑦 + 𝑒)2                             =  (𝑦 + 𝑒)𝑥(𝑦 + 𝑒) 

𝑥(𝑦 + 𝑒)(𝑦 + 𝑒)                   =  (𝑦 + 𝑒)𝑥(𝑦 + 𝑒)  



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𝑥[(𝑦 + 𝑒). 𝑦 + (𝑦 + 𝑒). 𝑒]  =  (𝑦 + 𝑒)𝑥(𝑦 + 𝑒)  

𝑥[𝑦. 𝑦 + 𝑦 + 𝑦 + 𝑒]             =  (𝑦𝑥 + 𝑥)(𝑦 + 𝑒)  

(𝑥𝑦)𝑦 + 𝑥𝑦 + 𝑥𝑦 + 𝑥        =  (𝑦𝑥)𝑦 + 𝑦𝑥 + 𝑥𝑦 + 𝑥  

 (𝑥𝑦)𝑦 + 𝑥𝑦                         =  (𝑦𝑥)𝑦 + 𝑦𝑥 (By right Cancellation law) 

𝑥(𝑦. 𝑦) + 𝑥. 𝑦                      =  𝑦(𝑥. 𝑦) + 𝑦. 𝑥  (By Associative Law)   

⇒ 𝑥𝑦2 + 𝑥𝑦 =  𝑥𝑦2 + 𝑦𝑥   [By equation 1] 

𝑥𝑦 =  𝑦𝑥  ∀  𝑥, 𝑦 ∈ 𝑁  

Thus 𝑁 is commutative. 

Remark 5.2 

  A quasi weak commutative near-ring can become a 𝜎1near-ring 

Proof 

 Let 𝑁 be a quasi-weak commutative near-ring 

Then 𝑥𝑦𝑧 = 𝑦𝑥𝑧 for all 𝑥, 𝑦, 𝑧 ∈ 𝑁. 

If 𝑦 = 𝑧, then 𝑥𝑦𝑦 = 𝑦𝑥𝑦, (𝑖𝑒) 𝑥𝑦2 = 𝑦𝑥𝑦. 

Consequently  𝑁 becomes a 𝜎1near-ring. 

Proposition 5.3 

 𝜎1near-ring is always zero symmetric. 

Proof 

 Suppose 𝑁 is a 𝜎1near-ring Then 𝑥𝑦2 = 𝑦𝑥𝑦  𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦 ∈ 𝑁. 

When 𝑦 = 0, 𝑥0 = 0 𝑥 0 = 0 

It follows that 𝑁 is zero symmetric. 

Proposition 5.4 

 Let 𝑁 be a 𝜎1near-ring. If 𝑁 is weak commutative then 𝑥2𝑦 = 𝑦2𝑥. for all 𝑥, 𝑦 ∈ 𝑁 

Proof 

 Let 𝑁 be a weak Commutative near-ring 

Then 𝑥𝑦𝑧 = 𝑥𝑧𝑦 … … … . . … (1) 

 Let 𝑁 be a 𝜎1near-ring 

Then 𝑥𝑦2 = 𝑦𝑥𝑦 ………….(2) 

Now, 𝑥2𝑦 = 𝑥𝑥𝑦 = 𝑥𝑦𝑥 [By equation (1)] = 𝑦𝑥2 [By equation (2)] 

Hence  𝑥2𝑦 = 𝑦𝑥2for all 𝑥, 𝑦 ∈ 𝑁. 



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Proposition 5.5 

 Homomorphic image of a 𝜎1near-ring is also a 𝜎1near-ring. 

Proof 

 The proof is straight forward. 

Proposition 5.6 

 If  I is an ideal of the 𝜎1near-ring N then N / I is also an 𝜎1near-ring 

Proof 

 The function ∅: 𝑁 →  𝑁/𝐼 defined by ∅(𝑥) = 𝐼 + 𝑥 is an epimorphism. The rest of the proof 

is taken care of by the above Proposition 5.5. 

Proposition 5.7 

Every 𝜎1near-ring N is isomorphic to a sub direct product of sub directly irreducible 𝜎1near-ring. 

Proof 

 By theorem 1.62, P 26 of Pilz [2], N is isomorphic to a sub direct product of sub directly 

irreducible 𝜎1– near-ring.𝑁𝑖’s and each 𝑁𝑖  is a homorphic image of 𝑁 under the multiplication 𝜋𝑖. 

Now the desired result follows from the above Proposition 5.5. 

Proposition 5.8 

 Let N be a 𝜎1near-ring with a mate function 𝑓. 

 Then we have 

i) 𝐿 = {0} 

ii) 𝑁 has (∗, IFP) 

iii) 𝐸 ⊆ 𝐶(𝑁) 

Proof 

              Let N be a 𝜎1near-ring 

Then 𝑥𝑦2 = 𝑦𝑥𝑦 ∀ 𝑥, 𝑦 ∈ 𝑁. ……………… (1) 

Since 𝑓 is a mate function for N then 𝑥 = 𝑥𝑓(𝑥)𝑥 ∈ 𝑥𝑁𝑥 for all  𝑥 ∈ 𝑁 

 ∴ 𝑥 = 𝑥𝑛𝑥 for some n. ………………… (2) 

  i)For 𝑛, 𝑥 ∈ 𝑁,   𝑛𝑥2 = 𝑥𝑛𝑥 [By equation (1)] = 𝑥 [By equation (2)] 

Suppose 𝑥2 = 0 

Clearly then 𝑥 = 0. [Since N is zero symmetric]. Then 𝑅(2) guarantees that 𝐿 = {0} 

(ii)  By i) 𝐿 = {0}. Now R(4) guarantees that N has (∗, IFP) 

(iii)Let 𝑒 ∈ 𝐸. Since N is a 𝜎1near − ring, 𝑛𝑒2 = 𝑒𝑛𝑒 ⟹ 𝑛𝑒 = 𝑒𝑛𝑒for all 𝑛 in 𝑁 … … . . … … (3) 



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Also we have  (𝑒𝑛𝑒 − 𝑒𝑛)𝑒 = 0 

This implies 𝑒(𝑒𝑛𝑒 − 𝑒𝑛) = 0 and 𝑒𝑛(𝑒𝑛𝑒 − 𝑒𝑛) = 0 [𝑏𝑦(𝑖𝑖)] 

Also 𝑒𝑛𝑒(𝑒𝑛𝑒 − 𝑒𝑛) = 𝑒𝑛. 0 = 0 [Since N is zero symmetric] 

Now 𝑒𝑛𝑒(𝑒𝑛𝑒 − 𝑒𝑛) − 𝑒𝑛(𝑒𝑛𝑒 − 𝑒𝑛) = 0.  

Consequently, (𝑒𝑛𝑒 − 𝑒𝑛)2 = 0 and (i) guarantees 𝑒𝑛𝑒 − 𝑒𝑛 = 0. 

Therefore 𝑒𝑛𝑒 = 𝑒𝑛 for all n in N …………….….(4) 

From Equations (3) and (4) we get 𝑒𝑛 = 𝑛𝑒 for all n in N. Thus 𝐸 ⊆ 𝐶(𝑁) 

Proposition 5.9 

Let 𝑁 be a pseudo commutative near -ring with right identity. Then if 𝑁 is a 𝜎1near-ring then for any 

𝑎, 𝑏 in 𝑁, 𝑎𝑏 = 0 implies 𝑏𝑎 = 0 

Proof 

 Let 𝑁 be a pseudo commutative near-ring. Then 𝑥𝑦𝑧 = 𝑧𝑦𝑥 for all 𝑥, 𝑦, 𝑧 ∈ 𝑁 …..…. (1) 

Now R (5) guarantees that 𝑁 is weak commutative. 

∴ 𝑥𝑦𝑧 = 𝑥𝑧𝑦 for all 𝑥, 𝑦, 𝑧 ∈ 𝑁 …….…..(2) 

Now,   

         ( 𝑥𝑎𝑥)(𝑦𝑏𝑦) = 𝑎𝑥2𝑏𝑦2  

          𝑥𝑎𝑥 𝑦𝑏𝑦      = 𝑎𝑥𝑥𝑏𝑦𝑦       

        𝑥𝑎(𝑥𝑦𝑏)𝑦     = 𝑎(𝑥𝑥𝑏)𝑦𝑦  

        𝑥𝑎(𝑏𝑦𝑥)𝑦     = 𝑎(𝑏𝑥𝑥)𝑦𝑦  [By Equation (1)]  

        (𝑥𝑎𝑏)𝑦𝑥𝑦     = 𝑎𝑏 𝑥𝑥𝑦 𝑦  

          𝑏𝑎𝑥𝑦𝑥𝑦       = 𝑎𝑏(𝑥𝑥𝑦)𝑦  [By Equation (1)] 

           𝑏𝑎 𝑥𝑦 𝑥𝑦    = 𝑎𝑏 𝑥 𝑦𝑥𝑦    [By Equation (2)]   

                           𝑏𝑎 = 𝑎𝑏. Since 𝑎𝑏 = 0 it follows that 𝑏𝑎 = 0. 

Proposition 5.10 

 N is a 𝜎1 near-ring if and only if every x in N can be written as 𝑥𝑦2 = 𝑢 + 𝑣 where 𝑢 𝜖 𝑁𝑜 and 

𝑣 𝜖 𝑁𝑐  and 𝑢 = 𝑦0(𝑛𝑦 + 𝑚) − 𝑦0 𝑥 𝑦𝑐, 𝑣 = 𝑦0𝑥 𝑦𝑐 + 𝑦𝑐 , 𝑦 = 𝑦0 + 𝑦𝑐 𝜖 𝑁0⨁𝑁𝑐  where 𝑦0, 𝑛 ∈

 𝑁0 , 𝑦𝑐 , 𝑚 ∈ 𝑁𝑐. Further more 𝑢 ∈ 𝑁0, 𝑣 ∈ 𝑁𝑐. 

Proof 

 For the ‘only if’ part, let 𝑦 ∈ 𝑁.  Since N is 𝜎1 there exist  𝑥 in N such that 𝑥𝑦2 = 𝑦𝑥𝑦. By 

using pierce decomposition we can write 𝑥 = 𝑛 + 𝑚 and 𝑦 = 𝑦0 + 𝑦𝑐  where 𝑥 ∈ 𝑁, 𝑛, 𝑦0 ∈ 𝑁0  and 

𝑚, 𝑦𝑐 ∈  𝑁𝑐 

Now                𝑥𝑦2 = (𝑦0 + 𝑦𝑐)(𝑛 + 𝑚)𝑦 



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                                  = (𝑦0 + 𝑦𝑐)(𝑛𝑦 + 𝑚𝑦)  

                                  = (𝑦0 + 𝑦𝑐)(𝑛𝑦 + 𝑚)   [𝑠𝑖𝑛𝑐𝑒 𝑚 ∈ 𝑁𝑐]  

                                  = 𝑦0(𝑛𝑦 + 𝑚) + 𝑦𝑐(𝑛𝑦 + 𝑚)  

                                  = 𝑦0(𝑛𝑦 + 𝑚) + 𝑦𝑐       [𝑠𝑖𝑛𝑐𝑒 𝑦𝑐 ∈ 𝑁𝑐]  

                                  = 𝑦0(𝑛𝑦 + 𝑚) − 𝑦0 𝑥 𝑦𝑐 + 𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐  

                                  =  𝑢 + 𝑣  where 𝑢 = 𝑦0(𝑛𝑦 + 𝑚) − 𝑦0 𝑥 𝑦𝑐 and  𝑣 = 𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐 

Now,              𝑢. 𝑜 = [𝑦0(𝑛𝑦 + 𝑚) − 𝑦0 𝑥 𝑦𝑐].0 = 𝑦0(𝑛𝑦 + 𝑚)0 − 𝑦0 𝑥 𝑦𝑐 .0 

                              = 𝑦0(𝑛𝑦0 + 𝑚0) − 𝑦0 𝑥 𝑦𝑐 . 0 

                                  = 𝑦0(𝑛𝑦0 + 𝑚) − 𝑦0 𝑥 𝑦𝑐       [𝑠𝑖𝑛𝑐𝑒  𝑚, 𝑦𝑐 ∈ 𝑁𝑐] 

                                  = 𝑦0(𝑛𝑦𝑐 + 𝑚𝑦𝑐) − 𝑦0 𝑥 𝑦𝑐    [𝑠𝑖𝑛𝑐𝑒 𝑦0 = 𝑦𝑐  𝑎𝑛𝑑 𝑚 ∈ 𝑁𝑐] 

                                  = 0.     

Also,               𝑣. 0 = [𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐]. 0  = 𝑦0 𝑥 𝑦𝑐0 + 𝑦𝑐  0 = 𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐  [𝑠𝑖𝑛𝑐𝑒 𝑦𝑐 ∈ 𝑁𝑐]  = 𝑣 

Thus 𝑥𝑦2 = 𝑢 + 𝑣  where 𝑢 ∈  𝑁0 and 𝑣 ∈ 𝑁𝑐. 

For the “if part” Assume for every y in N with 𝑥𝑦2 = 𝑢 + 𝑣  where 𝑢 ∈ 𝑁0  and 𝑣 ∈  𝑁𝑐                                          

with 𝑢 = 𝑦0(𝑛𝑦 + 𝑚) − 𝑦0𝑥 𝑦𝑐 , 𝑣 = 𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐 where 𝑦 = 𝑦0 + 𝑦𝑐, 𝑦0 , 𝑛 ∈ 𝑁𝑐 and  

𝑦𝑐, 𝑚 ∈  𝑁𝑐  

We shall show that N is a 𝜎1near – ring. 

Now,  𝑥𝑦2  = 𝑢 + 𝑣 

                       = 𝑦0(𝑛𝑦 + 𝑚) − 𝑦0𝑥 𝑦 𝑐 + 𝑦0 𝑥 𝑦𝑐 + 𝑦𝑐  

                       = 𝑦0(𝑛𝑦 + 𝑚𝑦) + 𝑦𝑐       [𝑠𝑖𝑛𝑐𝑒 𝑚 ∈ 𝑁𝑐] 

                       = 𝑦0(𝑛 + 𝑚)𝑦 + 𝑦𝑐  

                       = 𝑦0 𝑥𝑦 + 𝑦𝑐 𝑥𝑦              [𝑠𝑖𝑛𝑐𝑒 𝑦𝑐, ∈ 𝑁𝑐]  

                       = (𝑦0 + 𝑦𝑐)𝑥𝑦  

                       = 𝑦𝑥𝑦  

Thus for every 𝑦 ∈ 𝑁, 𝑥𝑦2 = 𝑦𝑥𝑦 for all 𝑥 in N. 

Hence N is a 𝜎1near – ring. 

Theorem 5.11 

 Let N be a zero symmetric weak commutative 𝜎1 near – ring then right cancellative near rings 

are integral. 

 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

360 https://internationalpubls.com 

Proof 

 Let 𝑎 ≠ 𝑜 and 𝑎𝑏 = 0 . Then 𝑏𝑎 = 0  [By Proposition 5.10] = 0𝑎  and therefore by right 

cancellative law we get 𝑏 = 0. And if 𝑏 ≠ 0 and 𝑎𝑏 = 0 then 𝑎𝑏 = 0𝑏. This implies 𝑎 = 0, by right 

cancellative law. Hence 𝑎𝑏 = 0 ⟹ either 𝑎 = 0 or 𝑏 = 0 and the result follows. 

Theorem 5.12 

 Let N be a 𝜎1 near –ring with mate function 𝑓. If N is regular and I is a proper ideal of N, then 

every element of I is a zero divisor. 

Proof 

 Let N be an 𝜎1near – ring 

Then 𝑥𝑦2 = 𝑦𝑥𝑦  ∀ 𝑥, 𝑦 ∈ 𝑁 ………………….. (1) 

Since 𝑓 is a mate function for 𝑁, then 𝑥 = 𝑥𝑓(𝑥)𝑥 ∈ 𝑥𝑁𝑥 

∴ 𝑥 = 𝑥𝑛𝑥 = 𝑛𝑥2 for some 𝑛 [By equation 1] 

Put 𝑥2 = 0 ⟹ 𝑛 0 = 𝑥 ⟹ 𝑥 = 0 ∴ 𝐿 = {0}. 

Let 𝑎 ∈ 𝐼. Then Na is an N – subgroup. Since N is regular 𝑎 = 𝑎𝑥𝑎 for some 𝑥 ∈ 𝑁.  

Let 𝑛 𝑎 ∈ 𝑁𝑎 for any 𝑛 ∈ 𝑁. ⟹ 𝑛𝑎 = 𝑛(𝑎𝑥𝑎) = 𝑛𝑎𝑥𝑎 ∈ 𝑁𝑎𝑁𝑎  

And if 𝑚 ∈ 𝑁𝑎𝑁𝑎 then for 𝑢, 𝑣 ∈ 𝑁 

𝑚 = 𝑢𝑎𝑣𝑎 = (𝑢𝑎𝑣)𝑎 ∈ 𝐼𝑎 (𝑁𝐼𝑁 ⊆ 𝐼) ⟹ 𝑚 ∈ 𝑁𝑎 

Consequently, 𝑁𝑎𝑁𝑎 = 𝑁𝑎 

Let 𝑛𝑎 = 𝑢𝑎𝑣𝑎 

Then 𝑛𝑎 − 𝑢𝑎𝑣𝑎 = 0 ⟹ (𝑛 − 𝑢𝑎𝑣)𝑎 = 0………………(2) 

If a is not a zero divisor, then 𝑛 − 𝑢𝑎𝑣 = 0. 

i.e.)𝑛 = 𝑢𝑎𝑣 ∈ 𝑁𝐼𝑁 ⊆ 𝐼 

⟹ 𝑁 = 𝐼 , which is a contradiction to I is a proper ideal of N.  

Hence a is a right zero divisor. 

Now equation (2) ⟹ 𝑎(𝑛 − 𝑢𝑎𝑣) = 0. [by Proposition 5.9]. 

This leads to the result that a is a left divisor too. Thus the result follows. 

Corollary 5.12 

If N is a 𝜎1 near – ring with no non- zero divisors then N contains no proper ideals of N. 

Lemma 5.13 

If N is 𝜎1 – near ring with mate function 𝑓 then for 𝑎, 𝑏 ∈ 𝑁, 𝑎𝑏 = 𝑏2 and 𝑏𝑎 = 𝑎2 imply 𝑎 = 𝑏. 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

361 https://internationalpubls.com 

Proof:  

Let N be a 𝜎1near-ring with a mate function 𝑓. 

Then we have 𝐿 = {0}……………………(1)  [By Proposition 5.8] 

         Let 𝑢1 = 𝑎 − 𝑏  𝑢2 = 𝑎𝑢1 and 𝑢3 = 𝑏𝑢1 

We can easily obtain the following equations. 

𝑢1𝑎 = (𝑎 − 𝑏)𝑎 = 𝑎2 − 𝑏𝑎 = 0……………… (2) 

𝑢1𝑏 = (𝑎 − 𝑏)𝑏 = 𝑎𝑏 − 𝑏2 = 0………………. (3) 

Also. 𝑢1
2 = (𝑎 − 𝑏)𝑢1 = 𝑎𝑢1 − 𝑏𝑢1 = 𝑢2 − 𝑢3…………… (4) 

As N is zero symmetric. 

𝑢2
2 = (𝑎𝑢1)(𝑎𝑢1) = 𝑎(𝑢1𝑎)𝑢1 = 𝑎. 0𝑢1 = 0 [By equation (2)] ………. (5) 

Also, 𝑢3
2 = (𝑏𝑢1)(𝑏𝑢1) = 𝑏(𝑢1𝑏)𝑢1 = 𝑏(0)𝑢1 = 0 [By equation (3)] ………. (6) 

Equations (5) & (6) imply 𝑢2 = 0 and 𝑢3 = 0 respectively.  [By equation (1)] 

Making use of these in equation (4), we get 𝑢1
2 = 0. 

 It follows that 𝑢1 = 0, [By equation (1)]    i.e.) 𝑎 − 𝑏 = 0. Thus 𝑎 = 𝑏 

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