EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 1, 2021, 126-134
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Near ring Multiplications on a Modified Near Module
Over a Near ring

A. V. Ramakrishna1,∗, T.V.N. Prasanna2,, D.V. Lakshmi3

1 Department of Mathematics, R.V.R and J.C College of Engineering, Chowdavaram,
Guntur-522019,Andhra Pradesh, India
2 Department of BS&H, Vignan’s Nirula Institute of Technology& Science for Women,
Guntur-522005, Andhra Pradesh, India
3 Bapatla Women’s Engineering College, Bapatla, Andhra Pradesh, India

Abstract. We introduce the notion of a modified near module M over a near ring N and explain
a method of obtaining near ring multiplications via a special type of maps from M into N called
semilinear maps.

2020 Mathematics Subject Classifications: 16Y30

Key Words and Phrases: Near ring, near module, semilinear map

1. Introduction

An interesting question that has attracted the attention of a good number of near ring
theorists includes J.R.Clay, R.E.Williams, C.J.Maxson, M.Johnson, K.D.Magill Jr.concerns
with finding a near ring multiplication on an algebraic structure over an underlying group.
In particular J.R. Clay (1992) [6] proved that a function π on a finite cyclic group (Zn,+)
generates a multiplication ‘* ’so that (Zn,+, ∗) is a near ring if π(π(p)q) = π(p)π(q).
K.D. Magill, Jr. (1995) [4] characterized that any near ring multiplication on a real finite
dimensional Euclidean space Rn is associated with a real-valued function f on Rn that
satisfies f(f(x)y) = f(x)f(y). In this paper we present methods [2], [3] [1] of finding
near ring multiplications on some algebraic structures which we call modified near mod-
ules. A right near ring [5]is a triple (N,+, .), where (N,+) is a (not necessarily abelian)
group, (N, .) is a semigroup satisfying the right distributive law: (a+ b)c = ac+ bc for all
a, b, c ∈ N .
By a near ring we mean a right near ring. When there is no scope for confusion, we write
N is a near ring instead of (N,+, ·) is a near ring. [6]

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i1.3781

Email addresses: amathi7@gmail.com (A. V. Ramakrishna),
tvnp11@gmail.com (T.V.N. Prasanna), himaja96@gmail.com (D.V. Lakshmi)

http://www.ejpam.com 126 c© 2021 EJPAM All rights reserved.



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 127

2. Modified near modules

Let (M,+) be a group and let N be a near ring and suppose ‘·’ is a mapping of N×M
into M .

Definition 1. (M,+, ·) is called a near module over N if

(i) (n1 + n2)m = n1m+ n2m for all n1, n2 ∈ N and m ∈M ;

(ii) (n1n2)m = n1(n2m) for all n1, n2 ∈ N and m ∈M .

Remark 1. Clearly our near module is the N-group introduced by Pilz.

Definition 2. (M,+, ·) is called a modified near module over N if

(i) n(m1 +m2) = nm1 + nm2 for all n ∈ N and m1,m2 ∈M ;

(ii) (n1n2)m = n1(n2m) for all n1, n2 ∈ N and m ∈M .

Definition 3. (M,+, ·) is called a strong near module over N if

(i) (n1 + n2)m = n1m+ n2m for all n1, n2 ∈ N and m ∈M ;

(ii) n(m1 +m2) = nm1 + nm2, for all m1,m2 ∈M and n ∈ N ;

(iii) (n1n2)m = n1(n2m) for all n1, n2 ∈ N and m ∈M .

Remark 2. A strong near module over a field is a vector space if ‘+’ is abelian and
1m = m for every m.

Example 1. Let (G,+) be a group. Define the function · from M(G) × G into G by
·(f, x) = f · x = f(x) for all f ∈M(G) and x ∈ G.
For any f, g ∈M(G) and x ∈ G, (f ◦ g)(x) = f(g(x)) = f(gx).
Also (f + g)(x) = f(x) + g(x) = fx+ gx.
and f(x+y) 6= f(x)+f(y). Therefore (G,+, ·) is a near module over near ring (M(G),+, ◦),
but not a modified near module.

Example 2. Let N be a nontrivial near ring with ab = a.
Let M = (N,+). Define the function � from N ×M into M as �(n,m) = n �m = m
for all n ∈ N,m ∈M .
For any n, n1, n2 ∈ N and m,m1,m2 ∈M ,
(1) (n1n2)�m = n1 �m = m and n1 � (n2 �m) = n2 �m = m
(2) n� (m1 +m2) = m1 +m2 and n�m1 + n�m2 = m1 +m2.
Therefore (M,+,�) is a modified near module over N .
However (M,+,�) is not a near module
since (n1 + n2)�m = m and n1 �m+ n2 �m = m+m.



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 128

Example 3. Let R = (R,+, ·).
Define φ : R→ R by φ(r) = r2. Define ‘�’: R× R into R as �(r,m) = r �m = r2m for
r,m ∈ R.
Then (R,+) is a modified near module over the near ring (R,+, ·) but not a near module.
That (R,+) is a modified near module can be verified easily. However R is not a near
module as is evident from the following:
Take r1 = 1, r2 = 1,m = 2. Then
(r1 + r2)�m = (1 + 1)� 2 = 2� 2 = 222 = 8 and
r1 �m+ r2 �m = 1� 2 + 1� 2 = 122 + 122 = 2 + 2 = 4.

Infact the above example is a special case of the following theorem:

Theorem 1. Let (R,+, ·) and (S,+1, ·1) be near rings and let φ : R → S be a mapping
such that φ(r1 · r2) = φ(r1) ·1 φ(r2) for all r1, r2 ∈ R. Let (M,⊕,�) be a left S-module.
Therefore (M,⊕,�1) is a modified near module over the near ring (R,+, ·) when �1 is
defined by r �1 m = φ(r)�m for all r ∈ R and m ∈M .

Proof. Since (M,⊕,�) is a left S-module,

(i) s� (m1 ⊕m2) = s�m1 ⊕ s�m2;

(ii) (s1 +1 s2)�m = s1 �m⊕ s2 �m;

(iii) s1 � (s2 �m) = (s1 ·1 s2)�m
for all s, s1, s2 ∈ S and m,m1,m2 ∈M .

For any r, r1, r2 ∈ R and m,m1,m2 ∈M ,
(1) r1 �1 (r2 �1 m) = φ(r1)� (r2 �1 m) = φ(r1)� [φ(r2)�m]
= [φ(r1) ·1 φ(r2)]�m = φ(r1r2)�m = (r1r2)�1 m and
(2) r �1 (m1 ⊕m2) = φ(r)� [m1 ⊕m2] = φ(r)�m1 ⊕ φ(r)�m2

= r �1 m1 ⊕ r �1 m2.
Therefore (M,+, ·) is a modified near module over (R,+, ·).

Definition 4. Let (M,+, ·) be a modified near module over N . A normal subgroup I of
M is called an ideal of M if

n(m+ i)− nm ∈ I for all n ∈ N, i ∈ I and m ∈M .

Definition 5. Let (M1,+1, ·1) and (M2,+2, ·2) be modified near modules over N . A
mapping φ : M1 →M2 is called a modified near module homomorphism if

(i) φ(m+1 m
′) = φ(m) +2 φ(m′);

(ii) φ(n ·1 m) = n ·2 φ(m) for all m,m′ ∈M1 and n ∈ N .

The proofs of the following theorems are similar to those of their counterparts in near
ring theory [5], hence omitted.



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 129

Theorem 2. Let M1,M2 be modified near modules over N and let φ : M1 → M2 be a
modified near module homomorphism. Then kerφ is an ideal of M1 and M1

kerφ ' φ(M1).

Theorem 3. The intersection of any family of ideals of a modified near module M is ideal
of M .

Proposition 1. Let M be a modified near module over N and let I be an ideal of M .
Let M

I = {m+ I|m ∈M}. Then (MI ,⊕,�) is a modified near module when ⊕ and � are
defined as

(m+ I)⊕ (m′ + I) = (m+m′) + I and
n� (m+ I) = nm+ I for all m+ I,m′ + I ∈ M

I and n ∈ N

and the natural projection map π : M → M
I defined by π(m) = m + I is a modified near

module homomorphism with kernel I.

3. Near ring Multiplication On a Modified Near Module

The following theorem explains a method of obtaining a near ring multiplications on
a modified near module over N via semilinear map from M into N .

Definition 6. Let (M,+, ·) be a modified near module over N . We call a mapping f :
M → N a semilinear if f(f(m1)m2) = f(m1)f(m2) for all m1,m2 ∈M .

Theorem 4. Let (M,+, ·) be a modified near module over a near ring (N,+, ·). Let f be a
semilinear map from M into N . Define the binary operation ∗ on M as m1∗m2 = f(m2)m1

for all m1,m2 ∈M . Then (M,+, ∗) is a near ring.

Proof. For any m1,m2,m3 ∈M ,
m1 ∗ (m2 ∗m3) = f(m2 ∗m3)m1 = f(f(m3)m2)m1 = [f(m3)f(m2)]m1 and
(m1 ∗m2) ∗m3 = f(m3)(m1 ∗m2) = f(m3)[f(m2)m1] = [f(m3)f(m2)]m1.
So the binary operation ∗ is associative.
Now (m1 + m2) ∗m3 = f(m3)(m1 + m2) = f(m3)m1 + f(m3)m2 = m1 ∗m3 + m2 ∗m3.
So the binary operation ∗ is right distributive and hence (M,+, ∗) is a near ring.

Examples 3.3 through 3.7 illustrate the technique of defining a near ring multiplication
on (M,+)

Example 4. Let M = {f |f : R→ R} and N = (End(R,+),+, ◦)
. Then (M,+, ◦) is a modified near module over (N,+, ◦).
Define α : M → N by α(f) = f ′

where

f ′(x) =

{
0 if x = 0

f(x) if x 6= 0.

Note that f(x) = 0 implies f ′(x) = 0 for x ∈ R.
We claim that α is semilinear.



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 130

Let f, g ∈M and x ∈ R.
Case(i): x 6= 0. Now α(α(f)og)(x) = α(f ′og)(x)
= (f ′ ◦ g)′(x) = (f ′ ◦ g)(x) = f ′(g(x))
= f ′(g′(x)) = (f ′ ◦ g′)(x) = (α(f) ◦ α(g))(x) implies α(α(f) ◦ g) = α(f) ◦ α(g).
Case(ii): x = 0.

Now α(α(f) ◦ g)(0) = α(f ′ ◦ g)(0)

= (f ′ ◦ g)′(0) = 0.

Also [α(f) ◦ α(g)](0) = (f ′ ◦ g′)(0) = f ′(g′(0)) = f ′(0) = 0.
So α(α(f) ◦ g) = α(f) ◦ α(g) when x = 0.
Hence α is semilinear; therefore (M,+, ∗) is a near ring with ∗ defined by f ∗g = α(g)◦f =
g′ ◦ f .

Example 5. Let M be the abelian group of all n× n circulant matrices with
real entries. Then (M,+) is a modified near module over N = (R,+, ·), if we define
kA as the matrix obtained by multiplying each entry of A by k. Define α : M → N by
α(A) = specA for all A ∈M
where specA = max{|λi||λi is an eigen value of A}.

Now α(α(A)B) = spec(α(A)B)

= α(A)specB

= α(A)α(B) for all A,B ∈M.

Hence α is semilinear; therefore (M,+, ∗) is a near ring with ∗ defined by A∗B = α(B)A =
specB A.

Example 6. Let (G,+) be a (not necessarily abelian) group. Let N = (End(G),+, ◦).
For f in N and a in G, define fa = f(a). Then G is a modified near module over N .
Define α : G→ N by α(a) = La
where La is the left addition by a: La(x) = a+ x for all x ∈ G.
Then α : M → N is semilinear.

Example 7. Let (C,+) be the module of complex numbers over the real field (R,+, ·) with

usual product. (i) Define f : C→ R by f(x+ iy) = (x2 + y2)
1
2 for all x+ iy ∈ C.

For any x1 + iy1, x2 + iy2 ∈ C,
f(f(x1 + iy1)(x2 + iy2)) = f((x1

2 + y1
2)

1
2 (x2 + iy2))

= f((x1
2 + y1

2)
1
2x2 + i(x1

2 + y1
2)

1
2 y2)

= [((x1
2 + y1

2)
1
2x2)

2 + ((x1
2 + y1

2)
1
2 y2)

2]
1
2

= (x1
2 + y1

2)
1
2 (x2

2 + y2
2)

1
2

= f(x1 + iy1)f(x2 + iy2). Hence f is semilinear; therefore (C,+, ∗) is a near ring with ∗
defined by

(x1 + iy1) ∗ (x2 + iy2) = f(x2 + iy2)(x1 + iy1)

= (x2
2 + y2

2)
1
2 (x1 + iy1).



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 131

(ii) Define f : C→ R by f(x+ iy) = |x| for all x+ iy ∈ C.
For any x1 + iy1, x2 + iy2 ∈ C,

f(f(x1 + iy1)(x2 + iy2)) = f(|x1|(x2 + iy2))

= f(|x1|x2 + i|x1|y2)
= ||x1|x2| = |x1||x2|
= f(x1 + iy1)f(x2 + iy2).

Hence f is semilinear; therefore (C,+, ∗) is a near ring with ∗ defined by

(x1 + iy1) ∗ (x2 + iy2) = f(x2 + iy2)(x1 + iy1)

= |x2|(x1 + iy1).

Example 8. Let M = {a + bi + cj + dk|a, b, c, d ∈ R} be the ring of real quaternions.
Then M is a modified near module over the real number field (R,+, ·).
Define f : M → R by f(a+ bi+ cj + dk) = a2 + b2 + c2 + d2 for all a+ bi+ cj + dk ∈M .
For any a1 + b1i+ c1j + d1k, a2 + b2i+ c2j + d2k ∈M ,
f(f(a1 + b1i+ c1j + d1k)(a2 + b2i+ c2j + d2k))
= f((a1

2 + b1
2 + c1

2 + d1
2)(a2 + b2i+ c2j + d2k))

= f(a1 + b1i+ c1j + d1k)f(a2 + b2i+ c2j + d2k).
Hence f is semilinear and therefore (M,+, ∗) is a near ring with
(a1 + b1i+ c1j + d1k) ∗ (a2 + b2i+ c2j + d2k)
= f(a2 + b2i+ c2j + d2k)(a1 + b1i+ c1j + d1k)
= (a2

2 + b2
2 + c2

2 + d2
2)(a1 + b1i+ c1j + d1k).

Example 9. Let M be the set of all n× n real matrices. Then (M,+, ·) is a strong near
module over the real number field (R,+, ·).
Define f : M → R by f(A) =

∑
1≤i,j≤n

(aij)
2. Then f is a semilinear map and hence

(M,+, ∗) is a near ring with A ∗B = f(B)A.

Theorem 5. Let (M,+, ·) be a modified near module over N . Define the function � from
N ×M into M as �(n,m) = n�m = f(n)m for all m ∈M and n ∈ N . Then (M,+,�)
is a modified near module over Nf , where Nf is a near ring induced by the semilinear map
f .

Proof. For any n ∈ N and m1,m2 ∈M ,
n� (m1 +m2) = f(n)(m1 +m2) = f(n)m1 + f(n)m2 = n�m1 + n�m2.
For any n1, n2 ∈ N and m ∈M ,
(n1 ∗ n2)�m = f(n1 ∗ n2)m = f(n1f(n2))m = [f(n1)f(n2)]m and
n1� (n2�m) = f(n1)(n2�m) = f(n1)[f(n2)m] = [f(n1)f(n2)]m. Therefore (M,+,�) is
a modified near module over Nf .

Theorem 6. Let M1,M2 be modified near modules over N and f : M1 → N be a semi-
linear map and φ : M2 →M1 be a near module homomorphism. Then foφ is a semilinear
map.



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 132

Proof. Let g = f ◦ φ.
For any m2,m2

′ ∈M2, g(g(m2)m2
′) = g([(f ◦ φ)(m2)]m2

′) = g((f(φ(m2))(m2
′))

= (f ◦ φ)[f(φ(m2))m2
′] = f [φ[f(φ(m2))m2

′]]
= f [f(φ(m2))φ(m2

′)] = f(φ(m2))f(φ(m2
′))

= g(m2)g(m2
′). Therefore g is a semilinear map.

Remark 3. Suppose a modified near module (M,+, ·) over a near ring (N,+, ·) is made
into a near ring (M,+, ∗) with the help of a semilinear map f . Then we know that Mk, the
k-fold product of (M,+, ∗) is also a near ring. It may be hoped that the near ring module
(Mk,⊕, ·) can be made into the near ring (Mk,⊕,⊗) directly by employing a suitable
semilinear map from Mk into N . The following example warns that not every modified
near module comes through a semilinear map.

As an illustration we present the following:

Example 10. Define x ·y = 2xy for all x, y ∈ R. Then (R,+, ·) is a modified near module
over R.
Define f : R→ R by f(m) = 2m for all m ∈ R.
Then f(f(a) · b) = f(a) · f(b) = 8ab.
So that f is semilinear.

Now m1 ∗m2 = f(m2) ·m1 = 2f(m2)m1 = 2(2m2)m1 = 4m2m1.

Consider (R2,⊕,⊗), the product of the near ring (R,+, ∗) with itself.
Suppose if possible there is a semilinear map g : R2 → R such that ‘⊗’ is induced by g.

Now (m1 ·m3,m2 ·m4) = (m1,m2)⊗ (m3,m4)

= g(m3,m4) · (m1,m2)

= (α ·m1, α ·m2) where α = g(m3,m4)

⇒ m1 ·m3 = α ·m1 and

m2 ·m4 = α ·m2

⇒ 2m1m3 = 2αm1 and

2m2m4 = 2αm2 for all m1,m2,m3,m4 ∈M.

Taking m1 = m3 = 1,m2 = m4 = 2, we get 2 = 2α and 8 = 4α
⇒ α = 1 and α = 2, which is a contradiction.

Theorem 7. Let M be a modified near module over (R,+) and f : M → R be a semilinear
map.

(i) If f is one-one, then (Mf ,+, ∗) is commutative.

(ii) Suppose M = (Rk,+). Then (Mf ,+, ∗) is commutative if and only if either M = {0}
or (Mf ,+, ∗) ' (R,+, ·), where Mf is a near ring induced by the semilinear map f .



A.V. Ramakrishna, T.V.N. Prasanna, D.V. Lakshmi / Eur. J. Pure Appl. Math, 14 (1) (2021), 126-134 133

Proof. (1) For any m1,m2 ∈M , m1 ∗m2 = f(m2)m1 and m2 ∗m1 = f(m1)m2.

Now f(m1 ∗m2) = f(f(m2)m1)

= f(m2)f(m1).

Also f(m2 ∗m1) = f(f(m1)m2)

= f(m1)f(m2).

Since (R, ·) is commutative, we have f(m1 ∗m2) = f(m2 ∗m1).
Since f is one-one, we have m1 ∗m2 = m2 ∗m1.
So ‘* ’ is commutative on M and hence (Mf ,+, ∗) is commutative.
(2) Suppose (Mf ,+, ∗) is commutative. Then

m1 ∗m2 = m2 ∗m1

⇒f(m2)m1 = f(m1)m2

⇒ The vectors m1 and m2 are parallel

⇒k = 0 or k = 1.

When k=1:
Now m1 ∗ m2 = f(m2)m1 and m2 ∗ m1 = f(m1)m2 ⇒ f(m2)m1 = f(m1)m2 for all
m1,m2 ∈M .
This equality is true for m1 = 1, we get f(m2) = f(1)m2.
Put f(1) = λ⇒ f(m2) = λm2 for some constant.
Therefore f is linear.

Now m1 ∗ (m2 +m3) = f(m2 +m3)m1

= [f(m2) + f(m3)]m1

= f(m2)m1 + f(m3)m1

= m1 ∗m2 +m1 ∗m3.

Therefore (Mf ,+, ∗) is a commutative ring.
Let 0 6= m ∈ R, then m ∗m1 = f(m1)m = λm1m = λmm1.
Put m1 = 1

λm . Then m ∗m1 = 1.
Define ψ : (M,+, ∗)→ (R,+, ·) by ψ(m) = λm for all m ∈M . Then (M,+, ∗) ' (R,+, ·).
Conversely suppose that M = {0} or (Mf ,+, ∗) ' (R,+, ·).
Since the ring {0} is commutative and since any ring isomorphic to (R,+, ·) is commutative,
the converse is clear.

Acknowledgements

The authors are thankful to Prof.I.Ramabhadra Sarma for his valuable comments and
suggestions.



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