# A Study of Nil Ideals and Kothe’s Conjecture in Neutrosophic Rings

Authors: Mohammad Abobala
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

The aim of this study is to determine the necessary and sufficient condition for any AH subset to be a full ideal in a neutrosophic ringR(I) and to be a nil ideal too. Also, this work shows the equivalence between Kothe’s conjecture in classical rings and corresponding neutrosophic rings, i.e., it proves that Kothe’s conjecture is true in the neutrosophic ring R(I) if and only if it is true in the corresponding classical ring R.


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## Body

## 1. Introduction

It was Zadeh [1] who introduced the notion of fuzzy sets to knob the uncertainties. Since then, the fuzzy set has been pragmatic in many commands such as medical diagnosis [2], decision making [3], and pattern recognition [4]. Keeping in view the prominence of fuzzy sets, many generalizations of fuzzy sets have been familiarized like rough sets [5], soft sets [6], intuitionistic fuzzy set [7], linear Diophantine fuzzy sets [8], bipolar valued fuzzy sets [9], bipolar soft sets [10, 11], picture fuzzy set [12], and spherical and T-spherical fuzzy sets [13]. Although all these generalizations have their own benefits, the notion of the neutrosophic set [14] has expanded much more responsiveness from the researchers as compared to the others [15].Neutrosophy is a new branch of philosophy founded by Smarandache [16] to deal with indeterminacy in real life and scientific actions. It became a useful tool in many areas such as topology [17, 18], statistics [19], probability [20], and algebra [21–23].Introducing neutrosophic algebras began with Smarandache and Kandasamy [24], where they presented concepts such as neutrosophic groups and neutrosophic rings and loops by inserting the indeterminacy element I into classical algebraic structures.In [25–27], we find two generalizations of neutrosophic rings, refined and n-refined neutrosophic rings, respectively. Many interesting properties of these kinds were discussed such as AH ideals, equations [28, 29], AH homomorphisms [30], and idempotency [31].Neutrosophic rings were applicable in other algebraic structures, where neutrosophic modules, matrices [32–35], and AH spaces [36] can be built over these rings in a similar way to classical cases.In [37], Abobala has proved that every neutrosophic ring is a homomorphic image of the corresponding refined neutrosophic ring. A similar relationship was proven in [38] for n-refined neutrosophic rings.Recently, the structure of maximal and minimal full AH ideals ofn-refined neutrosophic rings was presented [39]. Since, every neutrosophic ring R(I) can be understood as RI=R+RI=a+bI;a,b∈R.Then, each AH subset ofR(I) is defined to have the form M=P+SI; P and S are the two subsets of R. We call P the real part and S the neutrosophic part of M.An important question arises here. The question is “WhenM is a neutrosophic ideal of R(I)?” In other words, what conditions on the real part P and neutrosophic part S make M be an ideal?Also, it is clear that if Kothe’s conjecture about nil ideals is true inR(I), then it is true in R. What about the inverse relationship? In other words, if Kothe’s conjecture is true in R, then is it true for R(I)?These important questions motivate us to do this study, where we try to give full description of their answers.


## 2. Preliminaries

Definition 1.
LetR,+,× be a ring; then, RI=a+bI;a,b∈R is called the neutrosophic ring, where I is a neutrosophic element with the condition I2=I.
IfR is a field, then R(I) is called a neutrosophic field. A neutrosophic field is not a field by the classical meaning, since I is not invertible.Definition 2.
LetR be a ring and R(I) be the related neutrosophic ring, and P=P0+P1I=a0+a1I;a0∈P0,a1∈P1;P0,andP1 are the two subsets of R; then, P is called an AH subset.(a)
We say thatP is an AH ideal if P0andP1 are ideals in the ring R(b)
We say thatP is an AHS ideal if P0=P1Definition 3.
LetR(I) be any neutrosophic ring, and P=P0+P1I be any AH subset. We say that P is a full ideal of R(I) if it is an ideal by classical meaning.Definition 4.
The elementI can be split into two indeterminacies I1andI2 with the following conditions:(1)I12=I1,I22=I2,I1I2=I2I1=I1.Definition 5.
IfX is a set, then X(XI1,I2=a,bI1,cI2:a,b,c∈X is called the refined neutrosophic set generated by X, I1,andI2.Definition 6.
LetR,+,× be a ring, and RI1,I2,+,× is called a refined neutrosophic ring generated by R, I1,I2.Theorem 1.
LetRI1,I2,+,× be a refined neutrosophic ring; then, it is a ring.It is called a refined neutrosophic field ifR is a classical field.


### 2.1. Kothe Conjecture

IfR is a ring, then the sum of any two left nil ideals is nil again. For more equivalent forms of Kothe conjecture, see [20].


## 2.1. Kothe Conjecture

IfR is a ring, then the sum of any two left nil ideals is nil again. For more equivalent forms of Kothe conjecture, see [20].


## 3. Main Discussion

First of all, we determine the conditions of AH subsets to be ideals in the neutrosophic rings with unity.Theorem 2.
LetR(I) be a neutrosophic ring with unity 1 and M=P+SI be any AH subset of R(I); then, M is a neutrosophic ideal (full ideal) if and only if the following conditions are true:(a)
P is an ideal on R(b)
P is contained in S(c)
S is an ideal of RProof.
First, we assume that (a), (b), and (c) are true; we haveM,+ is a subgroup of RI,+, that is, because if a+bI,c+dI∈M;a,c∈P,b,d∈S, we find(2)a+bI−c+dI=a−c+b−dI∈M;a−c∈P,b−d∈S.
Now, suppose thata+bI∈Mandr=m+nI∈RI, we have r⋅a+bI=m⋅a+Im⋅b+n⋅b+n⋅a; by the assumption, we regard that m⋅b+n⋅b∈Sandn⋅a∈P≤S; thus, r⋅a+bI=m⋅a+Im⋅b+n⋅b+n⋅a∈P+SI=M, which means that M is a neutrosophic ideal of R(I).
Conversely, we suppose thatM=P+SI is a neutrosophic ideal of R(I). Let aandc be two arbitrary elements in P, and bandd be two arbitrary elements in S; we have a+bI,c+dI∈M; by using the assumption, we have M as an ideal; hence, a+bI−c+dI=a−c+b−dI∈M=P+SI; thus,(3)a−c∈P,b−d∈S,and P,+andS,+are the two subgroups of R,+.
For everyr∈R, we have r=r+0I∈RIandr⋅a+bI=r⋅a+r⋅bI∈M=P+SI; thus, r⋅a∈Pandr⋅b∈S; this means that PandS are the ideals in the classical ring R.
Now, we prove thatP is contained in S. We have 1−I∈RI, that is, because R(I) has a unity 1. On the other hand, we can write 1−Ia+bI=a−aI∈M=P+SI, that is, because M is an ideal of R(I); hence, −a∈S, and thus, a∈S; by regarding that a is an arbitrary element of P, we get that P≤S.Example 1.
LetR=Z be the ring of integers, and RI=ZI=a+bI;a,b∈Z be the corresponding neutrosophic ring, we have(a)
P=2,Q=4,andS=3 are the three ideals of R, with Q≤P(b)
M=Q+PI=4m+2nI;m,n∈Z is an ideal of R(I)(c)
N=P+SI=2m+3nI;m,n∈Z is not a neutrosophic ideal, that is, because P is not contained in SExample 2.
LetR=Z8be the ring of integers modulo8, and RI=a+bI;a,b∈Z8 be the corresponding neutrosophic ring. Consider the set M=0,4,2I,4I,6I,4+2I,4+6I,4+4I. We have M as an ideal of R(I), that is, because M=4+2Iand4≤2.
Now, we show that Theorem2 is not true in the case of neutrosophic rings with no unity by the following counter example.Example 3.
Consider the ringR=2Z=2m;m∈Z, and it is clear that it has no unity.
LetQ0=R and Q1=4n;n∈Z be an ideal in R. It is clear that Q0 is not contained in Q1. Now, we should check the set Q=Q0+Q1I if it makes an ideal of R(I) or not.
First of all, it is easy to see thatQ is a subgroup with respect to addition of the additive group (R(I), +).
On the other hand, we assume thatx=2n+4mI∈Q and r=2d+2cI is an arbitrary element of R(I) = 2Z(I). We have(4)r⋅x=4nd+I8md+8mc+4nc=22nd+42md+2mc+ncI∈Q.
Hence,Q is an ideal of R(I).
Now, we show that the necessary and sufficient conditions on any AH subsetP=P0+P1I to be an ideal in R(I) if R(I) has no unity.Theorem 3.
LetR be any ring without unity and R(I) be its corresponding neutrosophic ring. Then, the AH subset P=P0+P1I is an ideal of R(I) if and only if(a)
P0andP1 are ideals of R(b)
For everyx∈P0 and r∈R, we have rx∈P1Proof.
We suppose thatP is a full ideal of R(I); hence, P0andP1 are the ideals of R by a similar proof of Theorem 2. On the other hand, for every x0+y0I∈Pandr+rI∈RI,wherer∈R, we have r+rIx0+y0I∈P; thus rx0+Iry0+ry0+rx0∈P. This means that rx0∈P1, that is, because y0∈P1.
For the converse, we assume that conditions (a) and (b) are true; hence, it is clear that (P,+) is a subgroup of (R(I),+). Now, consider an arbitrary element x0+y0I∈P and any r=m+nI∈RI; we have rx0+y0I=m+nIx0+y0I=mx0+Imy0+nx0+ny0; by the assumption that (a) and (b) are true, we get nx0∈P1; hence, mx0+Imy0+nx0+ny0∈P, and P is an ideal of R(I).
The following theorem describes the structure of nil AH ideals inR(I).Theorem 4.
LetR(I) be any neutrosophic ring, we have(a)
x+yI is nilpotent in R(I) if and only if x,x+y are nilpotent elements in R(b)
IfP=P0+P1I is a full ideal of R(I), then P is nilpotent if and only if P0,P1, and P0+P1 are nilpotent(c)
IfP=P0+P1I is a right/left full ideal of R(I), then P is nil if and only if P0,P1,P0+P1 are nilProof.
(a)
First of all, we prove thatx+yIn=xn+x+yn−xnI, where n is any positive integerForn = 1, it is clear. We suppose that it is true for n = k, we shall prove it for k + 1.(5)x+yIk+1=x+yIkx+yI=xk+x+yk−xkI⋅x+yI=xk+1+Ixk⋅y+x+yk⋅y+x+yk⋅x−xk⋅x−xk⋅y=xk+1+Ix+yk+1−xk+1.Thus, it is true by inductionNow, we suppose thatx+yI is nilpotent in R(I); hence, there is a positive integer n, such that x+yIn=0. By the previous statement, we get xn=0andx+yn−xn=0;thus,x+yn=0. Thus, xandx+y are the nilpotent elements in R. The converse is clear.(b)
LetP=P0+P1I be a nilpotent ideal of R(I); then, there is a positive integer n, such that Pn=0. For any element x∈P0, we have x∈P; hence, xn=0. On the other hand, for any element y∈P1, we have yI∈P1I≤P; hence, yIn=ynI=0,andyn=0. This means that P0andP1 are nilpotent. Also, we have that x+yn=0, as a direct result of the equation x+yIn=xn+x+yn−xnI; hence, P0+P1 is nilpotent.For the converse, we assume thatP0,P1,andP0+P1 are nilpotent ideals of R; then, there are two positive integers mandn, such that P0n=P0+P1m=0. Hence, for every element x+yI∈P, we have x+yIn+m=xn+m+x+yn+m−xn+mI=0. This implies that Pn+m=0, that is, because xn+m=x+yn+m=0, and P must be a nilpotent ideal.(c)
First of all, we assume thatP=P0+P1I is nil in R(I); hence, P0andP1I are nil as a direct result from the inclusion P0≤PandP1I≤P; thus, P0andP1 are nil in R. To prove that the sum P0+P1 is nil, we consider two arbitrary elements x∈P0andy∈P1; hence, x+yI∈P, but P is nil; thus, there is a positive integer n, such that x+yIn=0; this implies that xn=0andx+yn−xn=0;thus,x+yn=0, so we get P0+P1 is nil. For the converse, we suppose that P0,P1,andP0+P1 are nil ideals in R, and we shall prove that P is nil.
Letx+yI be an arbitrary element of P; we have x∈P0andy∈P1. Under the assumption of theorem, we can find two positive integers mandn, such that xn=x+ym=0. According to (a), we have x+yIn+m=xn+m+x+yn+m−xn+mI=x+yn+mI=x+ynx+ymI=0. Thus, P is a nil ideal of R(I).
The following theorem shows the equivalence between Kothe’s conjecture in the classical ringR and the corresponding neutrosophic ring R(I).Theorem 5.
Kothe’s conjecture is true inR(I) if and only if it is true in R.Proof.
According to Theorem4, we have P=P0+P1I is nil in R(I) if and only if P0,P1,andP0+P1 are nil in R. Thus, the sum of two left nil ideals P=P0+P1IandQ=Q0+Q1I, say it is S=P0+Q0+P1+Q1I in R(I) is nil if and only if the following three sums P0+Q0,P1+Q1,andP0+Q0+P1+Q1 are nil in R; hence, Kothe’s conjecture is true in R(I) if and only if it is true in R.Theorem 6.
If Kothe’s conjecture is true in the refined neutrosophic ringRI1,I2 or in the n-refined neutrosophic ring RnI, then it is true in the ring R.Proof.
According to [37, 38], every neutrosophic ring R is a homomorphic image of its corresponding refined neutrosophic ring RI1,I2 and n-refined neutrosophic ring RnI; thus, if Kothe’s conjecture holds in RI1,I2,RnI, then it holds in R.Remark 1.
It is still unknown that if Kothe’s conjecture is true inR, then it is true in RI1,I2,RnI.


### 3.1. Open Questions

Two interesting open questions are coming to light according to this work. The first can be asked as follows. If Kothe’s conjecture is true in the ringR, then is it true in the corresponding refined neutrosophic ring RI1,I2. The second is the following. If Kothe’s conjecture is true in R, then is it true in the corresponding n-refined neutrosophic ring RnI. These open questions maybe the future of the study of Kothe conjecture in neutrosophic ring theory.


## 3.1. Open Questions

Two interesting open questions are coming to light according to this work. The first can be asked as follows. If Kothe’s conjecture is true in the ringR, then is it true in the corresponding refined neutrosophic ring RI1,I2. The second is the following. If Kothe’s conjecture is true in R, then is it true in the corresponding n-refined neutrosophic ring RnI. These open questions maybe the future of the study of Kothe conjecture in neutrosophic ring theory.



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