Content uploaded by Pushpalatha Kolluru
Author content
All content in this area was uploaded by Pushpalatha Kolluru on Mar 29, 2019
Content may be subject to copyright.
On subdirectly irreducible
Boolean Near Rings
K. Pushpalatha
Andhra Loyola Institute of Engineering and Technology,
Vijaywada, A.P., India
kpushpamphil@gmail.com
February 15, 2017
Abstract
It is well known that every sub directly irreducible
Boolean ring R6= (0) is a two-element field. In this paper
we generalize the result to near-rings. For this we introduce
the concept of an almost trivial near-ring. We prove that a
non constant Boolean near-ring Nis sub directly irreducible
if and only if Nis an almost trivial near-ring such that its
constant part Ncdoes not contain any non zero ideal of N.
Examples of sub directly irreducible Boolean near-rings are
given. As a consequence of the above result we show that a
zero-symmetric Boolean near-ring is subdirectly irreducible
if and only if its multiplication is trivial in the sense that
nm =nif m6= 0 and nm = 0 if m= 0, for all n, m ∈N. We
also prove that if a sub directly irreducible Boolean near-
ring has a nonzero distributive element then it is a two-
element field. Some more interesting results are proved in
this paper.
AMS Classification (1985 Revision): Primary 16A76,
Secondary 06E20,16A30,16A32
Key Words Near-ring, subdirectly irreducible, almost
trivial near-ring, Boolean near-ring
International Journal of Pure and Applied Mathematics
Volume 113 No. 13 2017, 272 – 281
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
Special Issue ijpam.eu
ijpam.eu 272 2017
1 Preliminaries:
A near-ring is a system (N, +,·) where (i) (N, +) is a group, (ii)
(N, .) is a semi group and (iii) (a+b)c=ac +bc for all a, b, c ∈N.
For near-rings, we always have the law 0n= 0. If N is a near-
ring then N0={n∈N / n0 = 0}and Nc={n∈N / n0 = 0}
are called the zero-symmetric and constant parts of Nrespectively.
Further N0∩Nc= (0) and every n∈Ncan be uniquely written as
n=n0+nc, where n0∈N0,nc∈Nc. N is called a zero-symmetric
(constant) near-ring if N=N0(N=Nc) . For any n∈N, write
(0 : n) = {m∈N / mn = 0}. Clearly, (0 : 0) = N0. [2]
proved that every Boolean near-ring Nis weakly commutative, i.e.,
abc =acb for all a, b, c ∈N. As a consequence of this one can prove
that every Boolean near-ring has the IFP property [1]. Therefore,
if Nis a Boolean near-ring , then for all n∈N, (0 : n) is an ideal
of N. In particular, we have that N0is an ideal. Throughout the
paper we consider non-constant Boolean near-rings. So, henceforth,
by a near-ring we mean a non-constant near-ring, unless otherwise
specified. For notation and basic results we refer to Pilz[1].
2 Subdirectly Irreducible Boolean Near-
rings:
Let us recall that
Definition 2.1. Let Nbe a near-ring.Then
1. Nis called a Boolean near-ring if n=n2for all n∈N.
2. Nis said to be subdirectly irreducible if the intersection of
all nonzero ideals of Nis nonzero.
We now introduce the following
Definition 2.2. A near- ring Nis called an almost trivial
near-ring if for all x, y ∈N,xy =x, if y∈Ncand xy =xcif
y∈Ncwhere xc=x0is the constant part of x. Clearly every
almost trivial near-ring is a Boolean near-ring.
Before proving a characterization theorem we prove a lemma.
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 273 2017
Lemma 2.3. If Nis a subdirectly irreducible Boolean near-
ring, then
1. Nis an almost trivial near-ring.
2. N0is the smallest nonzero ideal of N.
Proof. Note that Nis non-constant ⇐⇒ N6=Nc⇐⇒ N06=
(0).
(i) let J={n∈N/(0 : n)6= (0)}.JΦ , since N0= (0 : 0) 6={0}.
Since Nis subdirectly irreducible,
∩n∈J(0n)6= (0)
. Let 0e∈ ∩n∈J(0n) . Since e /∈(0 : e) , it follows that e /∈J
and hence (0 : e) = (0). For any a∈N, (aae)e= 0 ⇒aae ∈
(0 : e) = (0) ⇒a=ae. Thus eis a right identity of N. On
the other hand, if j∈Jthen e∈(0 : j) and ej = 0. Therefore
j=je = (je)2=jeje =j0⇒j∈Nc. Thus J⊆Nc. Moreover,
if n /∈Nc, then n /∈Jand (0 : n) = (0). By the above argument,
we can prove that nis a right identity. Also if nc∈Ncthen
xnc=xnc0 = x0nc=x0 = xc, for any x∈N. Therefore we get
that
xy =xif y /∈Nc
=xcif y∈Nc
(ii) Let Pbe the intersection of all nonzero ideals in N. Then
P6= (0). Since N0is a nonzero ideal, P⊆N0. Let 0 6=e∈P. Then
e /∈Ncand by (i) eis right identity. If x∈N0, then x=xe ∈P
and hence N0⊆P. Thus P=N0and N0is the smallest nonzero
ideal of N.
We now prove a characterization theorem for subdirectly irre-
ducible Boolean near-rings
Theorem 2.4. A Boolean near-ring Nis subdirectly irre-
ducible if and only if Nis an almost trivial near-ring such that its
constant part Ncdoes not contain any nonzero ideal of N.
Proof. Suppose Nis subdirectly irreducible. By lemma 2.3 , N
is an almost trivial near-ring and N0is the smallest nonzero ideal of
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 274 2017
N. If Iis any ideal of Nsuch that I⊆Nc, then N0∩I⊆N0∩Nc=
(0) and hence N0∩I= (0) . Since Nis subdirectly irreducible
and N0is a nonzero ideal , we have that I= (0) . Therefore Nc
does not contain any nonzero ideal of N. Conversely, suppose that
Nis almost trivial near-ring such that Ncdoes not contain any
nonzero ideal of N. Since Nis a non constant Boolean near-ring,
N0is a nonzero ideal of N. Let Ibe any non zero ideal of N. By
assumption, I"Nc. Then there exists an element a∈Isuch that
a /∈Nc. Since Nis almost trivial , xa =xfor all x∈N. Since
Iis an ideal, by (1) , N0I⊆I. So, for all n0∈N0, n0=n0a∈I
and N0⊆I. Thus N0is the smallest nonzero ideal of Nand Nis
subdirectly irreducible.
Example 2.5. : Let N=D8={0, a, 2a, 3a, b, a +b, 2a+
b, 3a+b}be the dihedral group of order 8. Then Nis a Boolean
near-ring with the following addition and multiplication.
+ 0 a 2a 3a b a+b 2a+b 3a+b
0 0 a 2a 3a b a+b 2a+b 3a+b
a a 2a 3a 0 a+b 2a+b 3a+b b
2a 2a 3a 0 a 2a+b 3a+b b a+b
3a 3a 0 a 2a 3a+b b a+b 2a+b
b b 3a+b 2a+ b a+b 0 3a 2a a
a+b a+b b 3a+b 2a+b a 0 3a 2a
2a+b 2a+b a+b b 3a+b 2a a 0 3a
3a+b 3a+b 2a+b a+b b 3a 2a a 0
·0 a 2a 3a b a+b 2a+b 3a+b
0 0 0 0 0 0 0 0 0
a 0 a a a 0 a a a
2a 0 2a 2a 2a 0 2a 2a 2a
3a 0 3a 3a 3a 0 3a 3a 3a
b b b b b b b b b
a+b b a+b a+b a+b b a+b a+b a+b
2a+b b 2a+b 2a+b 2a+b b 2a+b 2a+b 2a+b
3a+b b 3a+b 3a+b 3a+b b 3a+b 3a+b 3a+b
Nis a Boolean near-ring. (N , +) is non-abelian N0={0, a, 2a, 3a}
is an ideal. Write R={0,2a}.Ris a normal subgroup (since it
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 275 2017
is the center of (N, +) and RN ⊆N. Therefore Ris a right ideal.
But Ris not an ideal, since (3a+b)(0+2a)−(3a+b)0 = 3a+b−b=
3a /∈R.Nis subdirectly irreducible.
Nis an almost trivial near-ring. Nc={0, b}.Ncis not an
ideal of N. since Ncis not a normal subgroup by theorem 2.4, N
is a subdirectly irreducible Boolean near-ring. Further Nis not a
simple near-ring, since N0={0, a, 2a, 3a}is a nontrivial ideal of N.
Example 2.6. : Let N={0, a, b, c}be the kleins 4-group.
Then the addition and multiplication tables are as follows:
+ 0 a b c
0 0 a b c
a a 0 c b
b b c 0 a
c c b a 0
·0 a b c
0 0 0 0 0
a a a c b
b 0 0 b b
c a a c c
Nis a special Boolean near-ring, which is neither a Boolean
ring nor a constant near-ring. Nis an almost trivial near-ring.
N0={0, b}and Nc={0, a}are nonzero ideals of Nsuch that
N0∩Nc= (0). Therefore Nis not subdirectly irreducible . Thus
the condition that Ncdoes not contain any nonzero ideal of Nin
theorem 2.4 is essential.
Corollary 2.7. Let Nbe a Boolean near-ring. Then Nis
subdirectly irreducible if and only if N0is the smallest nonzero
ideal of N.
.
Proof. By lemma 2.3 , if Nis subdirectly irreducible, then N0
is the smallest nonzero ideal of N. Conversely if N0is the smallest
nonzero ideal of N, then the intersection of all nonzero ideals of N
is N0, which is nonzero. Hence Nis subdirectly irreducible.
Corollary 2.8. Let Nbe a zero-symmetric Boolean near-ring
. Then Nis subdirectly irreducible if and only if Nis a trivial near-
ring in the sense that nm =nif m6= 0 and nm = 0 if m= 0 , for
all n, m ∈N.
Proof. Note that Nis zero-symmetric if and only if Nc= (0) .
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 276 2017
Since Nc(0), Nccannot contain any nonzero ideal of N. Therefore
Nis subdirectly irreducible if and only if Nis almost trivial. Nis
almost trivial if and only if for n, m ∈N,nm =n, if m6= 0 ( ie.,
m /∈Nc)nm = 0 , if m= 0 ( ie., m∈Nc) , since nc =n0 = 0 .
ie., Nis trivial.
Corollary 2.9. Let Nbe a zero-symmetric boolean near-ring.
Then Nis subdirectly irreducible if and only if Nis simple.
Proof. By corollary 2.8, if Nis subdirectly irreducible then for
all n, m ∈N,nm =nif m6= 0 and nm = 0 if m= 0 , Let Ibe
any nonzero ideal of N. Then N I ⊆N. Let 0 6=a∈I. For any
n∈N, n =na ∈I. Hence I=Nand Nis simple. Converse is
trivial.
Note Corollary 2.8 is not true for arbitrary boolean near-rings.
This can be seen from the example 2.5.
Theorem 2.10. Let Nbe a subdirectly irreducible boolean
near-ring. Then Ncontains a nonzero distributive element if and
only if Nis a two element field (i.e., N≈Z2).
Proof. Let dbe a nonzero distributive element in N. Clearly
d∈N0. Then dr={n∈N / dn = 0}is an ideal of N[3, p ·34] .
We claim that N0∩dr= (0). If 0 6=x∈N0∩drthen x∈Ncand
dx = 0 . Since Nis subdirectly irreducible, by theorem 2.4, Nis an
almost trivial near-ring. Since x∈Nc,yx =yfor all y∈N. Hence
d=dx = 0 and this is a contradiction. Therefore N0∩dr= (0)
. Since N0and drare ideals and N06= (0), dr= (0). If nc ∈Nc
then dnc =dnc0 = d0nc = 0, by [2] and hence Nc⊆dr. Thus
Nc⊆dr= (0) ⇒Nc= (0) ⇒Nis zero-symmetric. By corollary
2.8, nm =nif m6= 0 and nm = 0 if m= 0 for all m, n ∈N.
Since d6= 0, nd =nfor all n∈N. For 0 6=m∈N, dm =dand
d(dm −md) = ddm −dmd =dm −dm = 0 , by [2]. Therefore
dmmd ∈dr= (0) and dm =md. Hence d=dm =md =mfor all
0∈m∈N. Thus N={0, d}is a two element field.
By theorem 2.10, we obtain the following
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 277 2017
Corollary 2.11. If R6= (0) is a subdirectly irreducible Boolean
ring, then Ris a two element field.
3 Subdirect Product Representation of
Boolean Near-Rings:
Theorem 3.1. Every Boolean near-ring is a subdirect product
of near-rings {Ni}, where each Niis either a constant near-ring or
an almost trivial near ring such that its constant part (Ni)cdoes
not contain any nonzero ideal of Ni.
Proof. Let Nbe the subdirect product of a family of subdirectly
irreducible near-rings {Ni}[1]. Each Niis a subdirectly irreducible
boolean near-ring. If Niis constant, then it is o.k. Otherwise, by
theorem 2.4, Niis an almost trivial near-ring such that its constant
part (Ni)cdoes not contain any ideal of Ni. This completes the
proof.
Corollary 3.2. Every zero-symmetric boolean near-ring Nis
a subdirect product of near-rings having trivial multiplication.
Proof. If Nis the subdirect product of a family of subdirectly
irreducible near-rings {Ni}[1], then each Niis a zero-symmetric
subdirectly irreducible boolean near-ring. By corollary 2.8 Nihas
trivial multiplication. This completes the proof.
Corollary 3.3. Let Nbe a boolean near-ring such that N0=
Nd, where Ndis the set of all distributive elements of N. Then N
is a subdirect product of near-rings {Ni}, where each Niis either a
constant near-ring or a two-element field.
Proof. Let Nbe the subdirect product of a family of subdirectly
irreducible near-rings {Ni}. For each i, since Niis a homomorphic
image of N,Niis a subdirectly irreducible boolean near-ring.
If Niis a constant near-ring then it is o.k. Otherwise, Niis a non-
constant near-ring and so (Ni)0= (Ni)dand hence (Ni)d6= 0 .
Thus Nihas non-zero distributive elements. By theorem 2.8 , Niis
a two element field. Therefore each Niis either a constant near-ring
or a two element field.
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 278 2017
Corollary 3.4. Let Nbe a boolean near-ring such that N=
Ndwhere Ndis the set of all distributive elements in N. Then Nis
a subdirect product of two element fields. Therefore Nis a Boolean
ring.
Proof. Since N=Nd, it is easy to show that N=N0=Ndand
hence Nis zero-symmetric. Therefore every homomorphic image
of Nis also zero-symmetric and hence it cannot be constant. Now
the result follows from corollary 3.3.
Corollary 3.5. Every boolean ring R6= 0 is a subdirect
product of two-element fields.
Proof. Note that every boolean ring is a boolean near-ring. Let
Rbe a subdirect product of subdirectly irreducible rings {Ri}[4].
Each Ri6= (0) is a subdirectly irreducible boolean ring. By theorem
2.10, Riis a two-element field. This completes the proof.
Acknowledgment
The author extends her gratitude to Prof.Y.V.Reddy,
Rtd professor,ANU,Guntur and also to the management of Andhra
Loyola Institute of Engineering and Technology, Vijayawada and to
Mrs.Sangeetha Joshi who designed this file in latex.
References
[1] G.Pilz, Near rings, the Theory and its Applications, North Hol-
land Publishing Company,1983.
[2] Hansen,D.J. and Jiang Luh, Boolean near rings and weak com-
mutativity,J.Aust. Math.Soc. (series A) 103-107 (1989).
[3] K.Pushpalatha ,On Special Boolean Near Rings and Boolean
Like Near Rings, Ph.D Thesis,Acharya Nagarjuna University,
Guntur, A.P., India (2015).
[4] Thomas W.Hungerford,Algebra, Holt, Rinehart and Winston
Inc; Newyork(1974).
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 279 2017
[5] Groenewald, Nico.J.and divier, werner A., on regularities in
near-rings, Acta.Math.Hungar.74, 177-190 (1997).
International Journal of Pure and Applied Mathematics Special Issue
ijpam.eu 280 2017
281