# Solutions of Some Kandasamy–Smarandache Problems about Neutrosophic Complex Numbers and Group of Units’ Problem

Authors: Mohammad Abobala; Ahmed Hatip; Necati Olgun
Journal: Journal of Mathematics
 (2022)

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## Abstract

Neutrosophic complex numbers are a novel interesting generalization of classical complex numbers. This study aims to study the neutrosophic complex infinite rings, where we classify those rings as direct products by using classical complex fieldC. Also, this work introduces full solutions for 18 Kandasamy–Smarandache open problems concerning these structures, as well as the group of unit’s famous problem, where the group of units (invertible elements) of a neutrosophic complex ring will be classified as a direct product of two classical groups of units.


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## Body

## 1. Introduction

Neutrosophy, as a new kind of generalized logic that deals with indeterminacy in nature, reality, and the human mind, found its way into pure mathematics. A lot of neutrosophic algebraic structures were defined and studied in a wide range such as neutrosophic topological groups and sets [1, 2], neutrosophic rings, neutrosophic graphs and homomorphisms [3, 4], and neutrosophic vector spaces [5, 6].In [7], Smarandache and Kandasamy introduced the neutrosophic complex numbers as an interesting generalized structure of the classical complex numbers, by adding an element with logical property (Indeterminate I) to the field C.Their work has generalized the concept of classical complex numbers into neutrosophic rings [8], and they proposed 150 open problems concerning substructures and factorization properties in these complex neutrosophic structures.In this study, we aim to continue their efforts and to suggest a classification of neutrosophic complex infinite rings by finding isomorphisms between these neutrosophic structures and the corresponding classical structures. Also, we use these isomorphisms to suggest solutions for 18 problems of Kandasamy–Smarandache problems introduced in [7].


## 2. Preliminaries

In this section, we present the basic definitions that are useful in this research.Definition 1 (see [7]).
LetCZ∪I=a+bI+ci+idI|a,b,c,d∈Z; then,(1)
An integer complex neutrosophic group under addition(2)
An integer neutrosophic complex monoid commutative monoid under multiplication(3)
An integer neutrosophic complex commutative ring with a unit of infinite order under addition+ and multiplication ×Definition 2 (see [7]).
LetS=aij|aij∈CZ∪I;1≤i,j≤n be a collection of n×n complex neutrosophic integer matrices. S is a ring of n×n integer complex neutrosophic ring of infinite order and is noncommutative. S has zero divisors, units, idempotents, subrings, and ideals.Definition 3 (see [7]).
LetCQ∪I=a+bi+cI+idI|a,b,c,d∈Q; then,(1)
CQ∪I is a rational complex neutrosophic ring that has no zero divisors(2)
CQ∪I is a rational complex neutrosophic fieldDefinition 4 (see [7]).
LetCZn = {a+biF|a,b∈Zn,iF be the finite complex modulo number such that iF2=n–1,n<∞}; we define iF as the finite complex modulo number. CZn is the finite complex modulo integer numbers.Definition 5 (see [8]).
A Smarandache ring (S-ring) is defined to be a ringA, such that a proper subset of A is a field with respect to the operations induced. By proper subset, we understand a set included in A different from the empty set, from the unit element if they were existed and from A.


## 3. Results

### 3.1. Problem1 and Problem 2

Problem 1.
CanCZ∪I be a Smarandache ring?
The answer is no. We give a proof.
We suppose thatCZ∪I is a Smarandache ring; then, there is a proper subset A⊆CZ∪I, such that A has a field structure with respect to multiplication on CZ∪I. Consider an arbitrary element n=a+bI+c+dIi,a,b,c,d∈Z, in A; since A is an Abelian group under addition, we can see that r⋅x∈A for every r∈Z; thus, A has infinite cardinality.
It is well known that the minimal field of infinite cardinality is the field of rationalesQ; hence, the field A has a characteristic zero (A contains an isomorphic image of Q).
The fieldA has only two principal ideals 0 and A; hence, A=a+bI+c+dIi. It is probable that A has a unity different from 1, and we will prove that the identity of A must be 1.
For this goal, we suppose thatm=x+yI+z+tIi is the identity of A and different from 1.
We havem⋅n=n; thus, a+bI+c+dIix+yI+z+tIi=a+bI+c+dIi, and we get a⋅x−c⋅z+Ia⋅y+b⋅x+b⋅y+a⋅zi+Ii(a⋅t+b⋅z+b⋅t+c⋅xi+Iic⋅y+d⋅x+d⋅y+I−c⋅t−d⋅z−d⋅t=a+bI+c+dIi; this implies a⋅x−c⋅z+a⋅z+c⋅xi+Ia⋅y+b⋅x+b⋅y−c⋅t−d⋅z−d⋅t+Iia⋅t+b⋅z+b⋅t+c⋅y+d⋅x+d⋅y=a+ci+bI+dIi; hence,(1)
a+ci=a⋅x−c⋅z+a⋅z+c⋅xi(2)
a⋅y+b⋅x+b⋅y−c⋅t−d⋅t−d⋅z=b(3)
a⋅t+b⋅z+b⋅t+c⋅y+d⋅x+d⋅y=d
From equation (1), we get the following two equivalent equations.(i)
a=a⋅x−c⋅z(ii)
c=a⋅z+c⋅x
We multiply (i) byc and (ii) by a to obtain(i)
a⋅c=a⋅c⋅x−c2z.(ii)
a⋅c=a2z+a⋅c⋅x; we compute (ii)−(i); 0=a2z+c2z=za2+c2; thus, z=0, that is because a2+c2≠0 so that x=1. By putting z=0 and x=1 in equations (2) and (3), we obtain the result.(iii)
a⋅y+b+b⋅y−c⋅t−d⋅t=b; hence, ya+b−tc+d=0.(iv)
a⋅t+b⋅t+c⋅y+d+d⋅y=d; hence, yc+d+ta+b=0.
We multiply equation (iii) bya+b and equation (iv) by c+d, and then, we add them to obtain y⋅a+b2+c+d2=0; hence, y=0, and then, t=0 so that m=1.
The previous discussion implies thatQ⊆A; thus, Q⊆CZ∪I which is a contradiction; thus, A cannot be a field and CZ∪I is not a Smarandache ring.Problem 2.
IsM=a1,a2;a1,a2∈CZ∪I under × a Smarandache semigroup?
The answer is yes, and we give a proof.
We have to search for a proper subsetA of M, where A has a group structure.
It is easy to see thatZI⊆CZ∪I, so if we take the group of units in the ring ZI, which is equal to UZI=1,−1,1−2I,−1+2I, it will be a subgroup of the semigroup CZ∪I; hence, the direct product UZI×UZI is a subgroup contained in the semigroup M; thus, M is a Smarandache semigroup.


## 3.1. Problem1 and Problem 2

Problem 1.
CanCZ∪I be a Smarandache ring?
The answer is no. We give a proof.
We suppose thatCZ∪I is a Smarandache ring; then, there is a proper subset A⊆CZ∪I, such that A has a field structure with respect to multiplication on CZ∪I. Consider an arbitrary element n=a+bI+c+dIi,a,b,c,d∈Z, in A; since A is an Abelian group under addition, we can see that r⋅x∈A for every r∈Z; thus, A has infinite cardinality.
It is well known that the minimal field of infinite cardinality is the field of rationalesQ; hence, the field A has a characteristic zero (A contains an isomorphic image of Q).
The fieldA has only two principal ideals 0 and A; hence, A=a+bI+c+dIi. It is probable that A has a unity different from 1, and we will prove that the identity of A must be 1.
For this goal, we suppose thatm=x+yI+z+tIi is the identity of A and different from 1.
We havem⋅n=n; thus, a+bI+c+dIix+yI+z+tIi=a+bI+c+dIi, and we get a⋅x−c⋅z+Ia⋅y+b⋅x+b⋅y+a⋅zi+Ii(a⋅t+b⋅z+b⋅t+c⋅xi+Iic⋅y+d⋅x+d⋅y+I−c⋅t−d⋅z−d⋅t=a+bI+c+dIi; this implies a⋅x−c⋅z+a⋅z+c⋅xi+Ia⋅y+b⋅x+b⋅y−c⋅t−d⋅z−d⋅t+Iia⋅t+b⋅z+b⋅t+c⋅y+d⋅x+d⋅y=a+ci+bI+dIi; hence,(1)
a+ci=a⋅x−c⋅z+a⋅z+c⋅xi(2)
a⋅y+b⋅x+b⋅y−c⋅t−d⋅t−d⋅z=b(3)
a⋅t+b⋅z+b⋅t+c⋅y+d⋅x+d⋅y=d
From equation (1), we get the following two equivalent equations.(i)
a=a⋅x−c⋅z(ii)
c=a⋅z+c⋅x
We multiply (i) byc and (ii) by a to obtain(i)
a⋅c=a⋅c⋅x−c2z.(ii)
a⋅c=a2z+a⋅c⋅x; we compute (ii)−(i); 0=a2z+c2z=za2+c2; thus, z=0, that is because a2+c2≠0 so that x=1. By putting z=0 and x=1 in equations (2) and (3), we obtain the result.(iii)
a⋅y+b+b⋅y−c⋅t−d⋅t=b; hence, ya+b−tc+d=0.(iv)
a⋅t+b⋅t+c⋅y+d+d⋅y=d; hence, yc+d+ta+b=0.
We multiply equation (iii) bya+b and equation (iv) by c+d, and then, we add them to obtain y⋅a+b2+c+d2=0; hence, y=0, and then, t=0 so that m=1.
The previous discussion implies thatQ⊆A; thus, Q⊆CZ∪I which is a contradiction; thus, A cannot be a field and CZ∪I is not a Smarandache ring.Problem 2.
IsM=a1,a2;a1,a2∈CZ∪I under × a Smarandache semigroup?
The answer is yes, and we give a proof.
We have to search for a proper subsetA of M, where A has a group structure.
It is easy to see thatZI⊆CZ∪I, so if we take the group of units in the ring ZI, which is equal to UZI=1,−1,1−2I,−1+2I, it will be a subgroup of the semigroup CZ∪I; hence, the direct product UZI×UZI is a subgroup contained in the semigroup M; thus, M is a Smarandache semigroup.


## 4. Neutrosophic Complex Rings as Direct Products

This section is devoted to solve Problems3–11.Problem 3.
Obtain some interesting results enjoyed by(a)
Neutrosophic complex reals(b)
Neutrosophic complex modulo integers(c)
neutrosophic complex rationales
The best answer we can obtain is to classify those rings as direct products of classical algebraic rings. This desired classification can help in many other problems.Theorem 1.
LetCR∪I be the neutrosophic complex ring of reals; then, CR∪I=CI.Proof.
It is clear that the neutrosophic ringCI is contained in CR∪I. Conversely, suppose that x=a+bI+ci+diI∈CR∪I; then, x=a+ci+Ib+di∈CI.
Thus, our proof is complete.Theorem 2.
CR∪I≅C×C.Proof.
We shall prove that there is a ring isomorphism betweenCIandC×C.
We definef:CI⟶C×C,fx+yI=x,x+y,x,y∈C.(a)
f is well defined: suppose that a+bI=x+yI;a,b,x,y∈C; this implies a=xandb=y. Hence, a,a+b=x,x+y,i.e.,fa+bI=fx+yI.(b)
f is bijective: it is clear that f is a surjective function. On the contrary, we assume that fx+yI=fa+bI; this meansa,a+b=x,x+y; thus, a=xandb=y.(c)
f is a ring homomorphism: we take m=a+bI,n=c+dI∈CI; then, m+n=a+c+b+dI and m⋅n=a⋅c+Ia⋅d+b⋅c+b⋅d so that  fm+n=a+c,a+c+b+d=a,a+b+c,c+d=fm+fn and fm⋅n=a⋅c,a⋅c+a⋅d+b⋅c+b⋅d=a⋅c,a+bc+d=a,a+b⋅c,c+d=fm⋅fn.
Thus,f is an isomorphism and the proof holds.Problem 4.
CanCR∪I have irreducible polynomials?
The answer is no, and we give a proof.
SinceCR∪I≅C×C, then CR∪Ix≅C×Cx, i.e., for each polynomial Px in CR∪Ix, there is a corresponding polynomial with form (gx,hx) in C×Cx; if px is not irreducible in CR∪Ix, then one of gx,hx at least is irreducible over the field of complex numbers C, which is not possible, that is because C is an algebraically closed field.
The following theorem classifiesCQ∪I.Theorem 3.
LetCQ∪I be the neutrosophic complex ring of rationales; then, CQ∪I≅Qi×Qi, where Qi=a+bi;a,b∈Q is the algebraic extension of the field Q by i.Proof.
We definef:CQ∪I⟶Qi×Qi;fa+bi+c+diI=a+bi,a+bi+c+di, where a,c,b,d∈Q. We have(a)
f is well defined: suppose that a+bi+c+diI=x+yi+z+tiI; then, a=x,b=y,c=z,andd=t.(b)
f is a bijective map: it is similar to that of Theorem 2.(c)
f is a ring homomorphism: consider two arbitrary elements m=a+bi+c+diIandn=x+yi+z+tiI, and we have(1)fm+n=a+bi+a+yi,c+di+z+ti+a+bi+x+yi=fm+fnfm⋅n=fa+bi⋅x+yi+Ia+biz+ti+c+dix+yi+c+diz+tia+bix+yi,a+bi⋅x+yi+a+biz+ti+c+dix+yi+c+diz+ti=fm⋅fnRemark 1.
By the same argument, we can writeCZ∪I≅Zi×Zi, i.e., CZ∪I can be classified as a direct product of the ring Zi=a+bi;a,b∈Z with itself.
By our classification results, we can answer many other open problems.Problem 5.
IsCQ∪I a field? Is it a prime field?
The answer is no, that is because, sinceCQ∪I≅Qi×Qi, we can find that it is not a field since the element x=1,0∈Qi×Qi and it is not invertible; hence, its inverse isomorphic image 1−I is not invertible in CQ∪I. Thus, CQ∪I cannot be a field.Problem 6.
Determine the irreducible polynomials overCQ∪I.
It is really a hard problem, but by using the factCQ∪I≅Qi×Qi, we can find all irreducible polynomial over CQ∪I.
LetPx be any polynomial defined over CQ∪I; then, it has a corresponding polynomial g,h in (Qi×Qi)x. It is sufficient to compute its isomorphic image g,h.
If one ofg,h is irreducible at least over Qi; then, px is irreducible over CQ∪I.Example 1.
Letpx=X2+1+1+iIX+1+5−iI, where X=x1+x2I;x1,x2∈Qi, be a polynomial defined over CQ∪I.
The corresponding isomorphic polynomial ofpx is(2)nx1,x2=fpx=fX2+f1+1+iI⋅fX+f1+5−iI=x12,x1+x22+1,2+i⋅x1,x1+x2+1,6−i=x12+x1+1,x1+x22+2+ix1+x2+6−i.
We get the following two equivalent polynomials:gx1=x12+x1+1,hx1,x2=x1+x22+2+ix1+x2+6−i; it is clear that g is irreducible over Qi; thus, px is irreducible over CQ∪I.Problem 7.
Find irreducible polynomials inCZ∪Ix? Is every ideal in CZ∪I principal?
The first part of Problem7 can be solved in a similar way of Problem 6, just by taking the isomorphic corresponding polynomial since CZ∪I≅Zi×Zi.Example 2.
Letpx=X2+1+1+iIX+1+5−iI, where X=x1+x2I,x1,x2∈Zi, be a polynomial defined over CZ∪I.
The corresponding isomorphic polynomial ofpx is(3)nx1,x2=fpx=fX2+f1+1+iI⋅fX+f1+5−iIx12,x1+x22+1,2+i⋅x1,x1+x2+1,6−i=x12+x1+1,x1+x22+2+ix1+x2+6−i.
We get the following two equivalent polynomials:(4)gx1=x12+x1+1,hx1,x2=x1+x22+2+ix1+x2+6−i.
It is clear thatg is irreducible over Zi; thus, px is irreducible over CZ∪I.Problem 8.
Can one say for all polynomials with complex neutrosophic coefficientsCR∪I is algebraically closed?
The answer is yes. That is becauseCR∪I≅C×C, and C is algebraically closed; thus, CR∪I is an algebraically closed ring, i.e., each root of any polynomial with coefficients from CR∪I is from CR∪I.Problem 9.
IsG=abcd;a⋅d−c⋅b≠0anda,b,c,d∈CZ∪I a group? Is G simple?
The answer is no.G is not even a group; we take 1237∈G; its inverse is not in G; thus, G is not a group.Problem 10.
IsCZ∪I a unique factorization domain?
The answer is no. We clarify our claim in the following discussion.
We haveCZ∪I≅Zi×Zi; it is easy to see that 1,0=1,21,0=1,3⋅1⋅0, i.e., Zi×Zi is not a unique factorization domain; hence, CZ∪I is not a unique factorization domain.Problem 11.
CanCR∪I be a principal ideal domain?
The answer is no. It is sufficient to prove thatCR∪I has zero divisors.
We haveCR∪I≅C×C and 1,0⋅0,1=0,0 so that C×C is not a principal ideal domain because it has zero divisors; thus, CR∪I is not a principal ideal domain.


## 5. Other Open Problems

This section is devoted to study Problems12–18.Problem 12.
Can any geometrical interpretation be given to the field of neutrosophic complex numbersCQ∪I?
The answer is no. However, there is an algebraic interpretation ofCQ∪I We describe it by the following theorem.Theorem 4.
LetCQ∪I be the complex neutrosophic ring of rationales. Then, it can be considered as an algebraic extension of the neutrosophic ring QI with degree two.Proof.
We havePx=x2+1 is a monic polynomial over QI; we shall prove that it is irreducible over QI.
Suppose thatpx=x+a+bIx+c+dI,a,b,c,d∈Q; then, Px=x2+xa+bI+c+dI+a⋅c+Ia⋅d+b⋅c+b⋅d; hence, a+bI+c+dI=0anda⋅d+b⋅c+b⋅d=0anda⋅c=1 so that a+c=0∗andb+d=0.
We get from∗a=−c; thus, −a2=1, which is a contradiction since a∈Q. Thus, px is irreducible. Px has a root m=i; hence, the ring QIi is an algebraic extension of QIQI with a degree equal to degP=2. It is clear that QIi=x+yi;x,y∈QI=a+bI+ci+diI;a,b,c,d∈Q=CQ∪I.Problem 13.
LetV=a1,a2,a3,a4;ai∈CQ∪I,+ be a group.(i)
Define an automorphismη:V⟶V so that kerη is a nontrivial subgroup.(ii)
IsV≅CQ∪I×CQ∪I×CQ∪I×CQ∪I?
(i) is not possible since every group automorphism needs to be a bijective map; hence, its kernel will be trivial.
Question (ii) is easy and clear, that is becauseV is defined to be the direct product of CQ∪I with itself four times.Problem 14.
LetM=a1a2a3a4;ai∈CQ∪I be a semigroup under multiplication.(i)
ProveM is an S-semigroup.(ii)
IsM commutative?(iii)
Find at least three zero divisors inM.(iiv)
DoesM have ideals?(v)
Give sub-semigroups inM which are not ideals.(a)
We defineA=x00x;x∈CQ∪I, where A is a proper subset of M, and it is, clearly, an abelian group. Thus, M is an S-semigroup.(b)
No, it is not. Since matrices overQ do not commute, and Q is contained in CQ∪I.(c)
Takex=0001,y=0002,andz=2300. It is easy to see that x⋅z=y⋅z=0; thus, x,y,andz are zero divisors.(d)
TakeS=aIbIcIdI;a,b,c,d∈CQ∪I, where S is a subgroup with respect to addition.Letm=xyzt∈M and =aIbIcIdI∈S; we have m⋅n∈S; thus, S is an ideal and M has ideals.(e)
We defineS1=abcd;a,b,c,d∈Qi,S2=abcd;a,b,c,d∈Q. These two sets are semigroups with respect to multiplication, but they are not ideals clearly.Problem 15.
LetS=a1a2a3a4a5a6a7a8a9a10a11a12;ai∈CQ∪I,+.(i)
Find subgroups ofS.(ii)
CanS have ideals?(iii)
CanS have idempotents?(iv)
CanS have zero divisors?
We summarize the answer as follows:(i)
Consider thatHi,+ is a subgroup of CQ∪I,+ and M=a1a2a3a4a5a6a7a8a9a10a11a12;ai∈Hi;1≤i≤12; then, S has the following property:For everyx,y∈M, we have x−y∈M; hence, M is a subgroup of S. All subgroups will have the same form.(ii)
The answer is no. That is because the multiplication is not defined onS; thus, it is not even a ring.(iii)
No, for the same reason.(iv)
No, for the same reason.Problem 16.
LetV=a1a2a3a4;ai∈CQ∪I be a semigroup under product.(i)
IsV commutative?(ii)
CanV have idempotents?(iii)
DoesV have a semigroup which is not an ideal?(iv)
CanV have zero divisors?(v)
Give an ideal inV.(vi)
IsV a Smarandache semigroup?(vii)
IsV a Smarandache semigroup?
The answer is(i)
No, it is not, since matrices overQ do not commute and Q is contained in CQ∪I(ii)
Yes, for example, takex=I00I(iii)
Yes, the setS=abcd;a,b,c,d∈Q is a semigroup that is not an ideal(iv)
Yes, we haveIIII⋅−I−III=0000(v)
ConsiderS=aIbIcIdI;a,b,c,d∈CQ∪I, and it is an ideal in V(vi)
Yes,V contains the set S=abcd;a⋅d−c⋅b≠0anda,b,c,d∈Q, which is a group with respect to multiplication(vii)
Yes,V contains the set S=a00a;a≠0anda∈Q, which is an Abelian group with respect to multiplicationProblem 17.
What are the advantages of using the algebraic structureCR∪I?Problem 18.
Give some uses of this complete algebraic structureCR∪I.
Problems17 and 18 are solved partially, that is because CR∪I is being classified as a direct product of the complex field C with itself according to Theorem 3.


## 6. Group of Units’ Problem

A well-known problem in the theory of rings is to describe the group of units under multiplication for a ringR. We will solve this famous problem in the case of CR∪I,CQ∪I,CZ∪I, by using classification properties.Theorem 5.
The group of units of the ringCR∪I is U=C∗×C∗.Proof.
SinceCR∪I≅C×C, then U=UC×UC, but C is a field; hence, UC=C∗. Thus, the proof is complete.Theorem 6.
The group of units of the ringCQ∪I is U=(Qi)∗×(Qi)∗.Proof.
SinceCQ∪I≅Qi×Qi, then U=UQi×UQi, but Qi is a field; hence, UQi=(Qi)∗. Thus, the proof is complete.Theorem 7.
The group of units of the ringCZ∪I is U =  Z2×Z2×Z2×Z2.Proof.
SinceCZ∪I≅Zi×Zi, then U=UZi×UZi, but UZi=1,−1,i,−i≅Z2×Z2; thus, U =  Z2×Z2×Z2×Z2.


## 7. Conclusions

In this study, we have built some algebraic isomorphisms between neutrosophic complex structures and their corresponding classical structures. Also, we have used these isomorphisms to present a full solution to 18 open problems proposed by Smarandache and Kandasamy concerning the algebraic structure of infinite rings of neutrosophic complex numbers.On the contrary, we have applied these isomorphisms to determine the classification of the group with units of these rings as a direct product of well-known classical groups.As a future research direction, we aim to solve many other Smarandache–Kandasamy open problems about neutrosophic complex rings.



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