# The Source of Semiprimeness of Semigroups

Authors: Barış Albayrak; Didem Yeşil; Didem Karalarlioğlu Camci
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this study, we define new semigroup structures using the setSS=a∈S|aSa=0 which is called the source of semiprimeness for a semigroup S with zero element. SS−idempotent semigroup, SS−regular semigroup, SS−reduced semigroup, and SS−nonzero divisor semigroup which are generalizations of idempotent, regular, reduced, and nonzero divisor semigroups in semigroup theory are investigated, and their basic properties are determined. In addition, we adapt some well-known results in semigroup theory to these new semigroups.


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## Body

## 1. Introduction

Semiprime ideals play a very important role in semigroups. Since every ring can be considered as a semigroup under multiplication, we have more generalized theorems of [1]. The aim of this study is to obtain new semigroup structures by using the definition of semiprimeness in semigroups in the sense of the study by Aydın et al. [1]. There are different equivalent definitions of semiprimeness. One of these is, if aSa=0 with a∈S implies a=0, then S is called a semiprime semigroup. In [1], the authors have defined the set SR=a∈R∣aRa=0 called source of semiprimeness of ring R by using the ring version of semiprimeness definition. In addition, they have obtained new ring and field structures. In this study, new semigroup structures have been obtained by using the source set of semiprimeness of semigroups. These structures are investigated, and some results of the semigroup theory are adapted to the new semigroups.


## 2. Preliminaries

In semigroup theory, there are many studies on different semigroup types. Adams examined the properties of semiprime semigroups in [2]. In [3], Van Rooyen worked on the ideals of semigroups. In [4–6], the authors have worked on reduced, regular, and zero-divisor semigroups. This article will bring a different perspective to these various semigroups.First, let us give the definitions mentioned in this article which are frequently used in semigroup theory. The studies by Clifford and Preston and Grillet [7, 8] are used for definitions.LetS,⋅ be a semigroup. A zero element of a semigroup S is an element 0∈S, such that 0⋅a=a⋅0=0 for all a∈S. Throughout this study, S will be taken as a semigroup with zero element. The e∈S element, such that e2=e, is called an idempotent element. The semigroup whose all elements are idempotent is called an idempotent semigroup. An element xof a semigroup S is called regular element if there exist at least one y∈S, such that xyx=x. The semigroup whose all elements are regular is called a regular semigroup. For a∈S, if there exist n∈ℤ+, such that an=0 and an−1≠0, then a is called nilpotent. A semigroup without nilpotent elements other than zero is called a reduced semigroup. An element x of a semigroup S is called zero divisor element if there exist 0≠y,0≠z∈S, such that xy=0 (left zero divisor) and zx=0 (right zero divisor). S is called the nonzero divisor semigroup, if there is no nonzero zero divisor element in S.Now, let us give the definitions of ideal and semiprime semigroup which are the basis of our article. From [2, 9], the subset I of semigroup S is called an ideal if IS⊆I (right ideal) and SI⊆I (left ideal). One of the equivalent definition of the semiprime semigroup, which using this study, is given as follows: if aSa=0 with a∈S implies a=0, then S is called the semiprime semigroup.


## 3. Results of Different Types ofSS−Semigroups

Definition 1.
LetS be a semigroup with zero and A be a nonempty subset of S. The subset of S,(1)SSA=a∈S∣aAa=0,is called the source of semiprimeness of A in S. For semigroup S, the notation SS is used instead of SSS. Then, the source of semiprimeness of the S semigroup is defined as follows:(2)SS=a∈S∣aSa=0.Now, let us examine the basic properties of the source of semiprimeness of semigroupS. In [1], the set of source of semiprimeness was investigated for the rings. The properties that can be exactly provided for semigroups are given with reference without proof in this study. First of all, let us mention some facts that can be easily seen but useful for understanding the set.(1)
For every semigroup with zero, since0∈SSA, SSA≠∅ is provided for every subset A of S(2)
It is easy to see thatSA⊆SSA for the subsemigroup A of S(3)
For subsetsA and B, A⊆B implies SSB⊆SSA ([1], Proposition 2.2)Now, let us investigate the properties of idempotent, regular, nilpotent, and zero divisor elements. These properties form the basis for defining the new semigroup structures that we will define in the next section. By using the results obtained, we will obtain the definitions of theSS−idempotent semigroup, SS−regular semigroup, SS−reduced semigroup, and SS−nonzero divisor semigroup.Lemma 1.
LetS be a semigroup, e be an idempotent element, and x be a regular element, such that xyx=x for y∈S. Then, the following properties are provided.(1)
eSS⊆SeS(2)
xySS⊆SxyS(3)
IfS is an idempotent semigroup, then SS=0(4)
IfS is a regular semigroup, then SS=0(5)
Ifa∈SS, then a is a nilpotent element(6)
Ifa∈SS, then a is a zero divisor elementProof.
(1)
Letea∈eSS. Then, since a∈SS, we write aSa=0. From this equation, we get(3)eaeSea=eaeSea⊆eaSa=0.This gives usea∈SeS. So, eSS⊆SeS is provided.(2)
For regular elementx, such that xyx=x for y∈S, this implies that xyxy=xy, and xy is an idempotent element. From (1), it is clear that xySS⊆SxyS.(3)
LetS be an idempotent semigroup and a∈SS. Since aSa=0, the equation aaa=0 is satisfied for a∈S. Hence, using a as an idempotent element, we get a=0. Then, SS=0.(4)
LetS be a regular semigroup and x∈SS. In this case, xSx=0, and there exist y∈S, such that xyx=x. Specially, x=xyx=0 is provided. This means that SS=0.(5)
Ifa∈SS, then aSa=0. Since a3=0, a is a nilpotent element.(6)
If0≠a∈SS, then aaa=0. This equation can be written as aaa=0 and aaa=0. Since 0≠a, if aa=0, then a is a zero divisor element. On the other hand, if aa≠0, then a is also a zero divisor element specific with a2.Using the above Lemma1, it is easy to see that the following corollary.Corollary 1.
For semigroupS, the following holds true:(1)
There is no idempotent element other than zero inSS(2)
There is no regular element other than zero inSS(3)
Every element inSS is nilpotent element(4)
Every element inSS is zero divisor elementNow, we will define different type ofSS−semigroups.Definition 2.
LetS be semigroup with zero and S≠SS.(1)
S is called the SS−idempotent semigroup if every element of S−SS is idempotent(2)
S is called the SS−regular semigroup if every element of S−SS is regular(3)
S is called the SS−reduced semigroup if S−SS has no nilpotent element(4)
S is called the SS−nonzero divisor semigroup if S−SS has neither a left nor a right zero divisorFirst, let us mention the basic characteristics properties of these newly defined semigroups. The following results can be easily seen using the definitions.(1)
IfS=0, then SS=0=S. In this case, since S−SS=∅, definitions would be meaningless for the zero semigroup. Similarly, S−SS=∅ is provided in case of S=SS. So, in this case too, the definitions would be meaningless.(2)
For the elements of a semigroup, “ifa is an idempotent element, then a is regular element,” “if a is a nilpotent element, then a is a zero divisor element,” and “if a is a nilpotent element, then a is not an idempotent element” and conditions are always provided. Using these conditions, the following properties can be easily seen.(4)IfSisSS−idempotent semigroup,thenSisSS−regular semigroup,IfSisSS−idempotent semigroup,thenSisSS−reduced semigroup,IfSisSS−nonzero divisor semigroup,thenSisSS−reduced semigroup.Additional conditions for providing other directions of relations will be investigated in our study(3)
It is clear that, ifS is an idempotent (regular, reduced, and nonzero divisor) semigroup, thenS is a SS−idempotent (regular, reduced, and nonzero divisor) semigroupWe will give examples of each of these four semigroups before proceeding with conclusions. It can also be seen from these examples that the above relations are one sided.Example 1.
Let the operation table of the semigroupS be given as follows.(5)⋅0abc00000a0000b0000c000c.
Using the table, we get(6)SS=0,a,b,S−SS=c.
Sincec2=c, it is seen that c is idempotent and regular element of S−SS. So, S is a SS−idempotent and SS−regular semigroup. Also, since c is not nilpotent, S is a SS−reduced semigroup. However, c is a zero divisor element for ac=0 and ca=0. Thus, S is not a SS−nonzero divisor semigroup.Example 2.
Let the operation table of the semigroupS be given as follows.(7)⋅0abc00000a0bb0b0bb0c0000.
For the semigroupS, it is easy to see that(8)SS=0,c,S−SS=a,b.
If we investigate elements ofS−SS, we see that only b is an idempotent and regular element. So, S is not a SS−idempotent or SS−regular semigroup. Also, since ac=0 and ca=0, a and c are the zero divisor elements. Then, S is not a SS−nonzero divisor semigroup. On the other hand, it is clear that a and b are not nilpotent elements. So, S is a SS−reduced semigroup.Example 3.
Let the operation table of the semigroupS be given as follows.(9)⋅0abc00000a00aab00bbc00cc.
Now, it turns out that(10)SS=0,a,S−SS=b,c.
It is seen thatb and c are the idempotent and regular elements of S−SS. Then, S is a SS−idempotent and a SS−regular semigroup. Also, b and c are nonzero divisors. Therefore, S is a SS−nonzero divisor semigroup. On the other hand, since a and b are not nilpotent elements, S is also a SS−reduced semigroup.Example 4.
Let the operation table of the semigroupS be given as follows.(11)⋅0abc00000a0abcb0b0bc0cba.
Using the table, we get(12)SS=0,b,S−SS=a,c.
If we investigate the elements ofS−SS, we see that only a is an idempotent element. Then, S is not a SS−idempotent semigroup. However, since aaa=a and ccc=c, a and c are the regular elements. So, S is a SS−regular semigroup. Also, a and c are not zero divisor or nilpotent elements. Therefore, S is a SS−nonzero divisor and a SS−reduced semigroup.Example 5.
Consider the semigroupℕ,⋅. Since(13)Sℕ=0,we get(14)ℕ−Sℕ=1,2,3,….
In the setℕ−Sℕ, since only 1 is idempotent and regular, ℕ is not a Sℕ−idempotent or Sℕ−regular semigroup. On the other hand, since there are no zero divisor elements in ℕ−Sℕ, S is a Sℕ−nonzero divisor and Sℕ−reduced semigroup.Let us now give a proposition and an example about characterization of the subgroups of these new semigroups.Proposition 1.
LetS be a semigroup and A be a subsemigroup of S. Then, the following conditions are satisfied.(1)
IfS is a SS−idempotent (regular) semigroup, then A is a SA−idempotent (regular) semigroup(2)
IfS is a SS−reduced (nonzero divisor) semigroup, then A is a SA−reduced (nonzero divisor) semigroupProof.
(1)
Ifa∈A−SA, then a∈A and a∉SA. Thus, we write a∈S and a∉SS. This means that a∈S−SS. So, A−SA⊆S−SS. Therefore, since every element in S−SS is idempotent (regular), every element in A−SA is an idempotent (regular) element. So, A is a SA−idempotent (regular) semigroup.(2)
We showed thatA−SA⊆S−SS in 1. So, since there is no nilpotent (zero divisor) element in S−SS, there is no nilpotent (zero divisor) element in A−SA. In this case, A is a SA−reduced (nonzero divisor) semigroup.Example 6.
Consider the set(15)M=xy00x000x∣x,y∈ℝ.
M is a semigroup with zero element by multiplication operation in matrices. For this semigroup, it is not hard to see that(16)SM=0y0000000∣y∈ℝ,M−SM=xy00x000x∣x,y∈ℝ,x≠0.
Since there existB=1/x−y/x2001/x0001/x∈M−SM, such that ABA=A for each A=xy00x000x∈M−SM, M is a SM−regular semigroup. On the other hand, the elements of the set M−SM can only be nilpotent elements for x=0. So, M is a SM−reduced semigroup. Also, it is easy to see that M is a SM−nonzero divisor semigroup because there is no zero divisor element in M−SM.
Now, let define the subsemigroup,(17)A=x000x000x∣x∈ℝ,of semigroup M. From Proposition 1, A is a SA−regular, SA−reduced, and SA−nonzero divisor semigroup. Indeed, it is clear that(18)SA=000000000,A−SA=x000x000x∣x≠0.
If we investigateA−SA similar to the above, we see that all elements are regular elements, and there are no nilpotent or zero divisor elements.Obviously, the setS−SS does not have to be a subsemigroup. The following proposition shows that the S−SS set is the subsemigroup with additional conditions for SS−semigroup types. In the next Lemma, the characterization of the SS is given for the SS−nonzero divisor and SS−reduced semigroups.Proposition 2.
LetS be a semigroup. Then, the following conditions are satisfied.(1)
IfS is a SS−nonzero divisor semigroup, then S−SS is a subsemigroup(2)
IfS is a commutative and SS−idempotent (regular) semigroup, then S−SS is a subsemigroupProof.
(1)
LetS be a SS−nonzero divisor semigroup. If x,y∈S−SS, then x and y are the nonzero divisors. Then, element xy is also a nonzero divisor. This gives us xy∈S−SS. So, S−SS is a subsemigroup.(2)
Note that product of idempotent elements is an idempotent element, product of regular elements is a regular element, and product of nilpotent elements is a nilpotent element in a commutative semigroup. In this case, ifa,b∈S−SS, then ab∈S−SS for the SS−idempotent (regular) semigroup S. So, S−SS is a subsemigroup.Lemma 2.
IfS is a SS−nonzero divisor or SS−reduced semigroup, then SS=a∈S∣a3=0. In addition, if S is a monoid, then SS=a∈S∣a2=0 is provided for SS−nonzero divisor or SS−reduced semigroup S.Proof.
First, letS be a SS−nonzero divisor semigroup and A=a∈S∣a3=0. The inclusion SS⊆A is always provided. Conversely, let us take an arbitrary element a of A. Assume that a∈S−SS. Since S is a SS−nonzero divisor semigroup, a is a nonzero divisor element. From definition of the set A, we write 0=a3=aa2. Using a as a nonzero divisor, we get a2=0. Similarly, from the equation 0=a2=aa, we get a=0. However, this result leads us to the a∉S−SS contradiction. So the assumption is incorrect, and thus, a∈SS. Then, A⊆SS is provided.
Now, letS be a SS−reduced semigroup. Similar to above, SS⊆A is always provided. Conversely, if a∈A, then a3=0. This gives us a is a nilpotent element. Since S is a SS−reduced semigroup, there is no nilpotent element in S−SS. So, we get a∈SS. Hence A⊆SS is provided.
In addition, ifS is a monoid, then there exist 1S∈S, such that a1Sa=a2=0 for all a∈SS. This means that SS⊆A. Also, similar to the above proof, we obtain A⊆SS.In this part of our study, we will give the results aboutℤn,⋅ is an Sℤn−semigroup. As it is known, if monoid S has a−1∈S, such that aa−1=a−1a=1 for a∈S, a is called an unit element. Since ℤn is a monoid with inverse and unit elements, different results can be reached from other semigroups. In [1], studies on domains and reduced rings were given. In the following part, similar results are obtained by different methods for semigroups.Lemma 3.
Letℤn,⋅ be semigroup of integers modulo n. If n is p2 for prime number p, then Sℤn=0¯,p¯,2p¯,…,p−1p¯. In addition, Sℤn=0¯ for n=p.Proof.
Letn=p2 for prime number p. From Lemma 2, we write a¯2=0¯ for an arbitrary element a¯ of Sℤn. This means that p2∣a2. Since p is prime, we get p∣a2, and so, we obtain p∣a. Then, a=pk for k∈ℤ. So, we get Sℤn=0¯,p¯,2p¯,…,p−1p¯.
Letn=p be a prime number and a¯ of Sℤn. Using the same procedure as in the above paragraph, we have p∣a. Since p is a prime number, the only element that can be divided by p is itself. So, we get a=p. This result gives us a¯=0¯. So, we get Sℤn=0¯.Theorem 1.
Letℤn,⋅ be semigroup of integers modulo n. The following holds true:(1)
Forn>2, ℤn is not a Sℤn−idempotent monoid(2)
Ifn is either a prime p or p2, then ℤn is a Sℤn−regular monoid(3)
n is either a prime p or p2 if and only if ℤn is a Sℤn−nonzero divisor monoid(4)
Ifn=p is a prime number, then ℤn is a Sℤn−reduced monoidProof.
(1)
SinceSℤ2=0¯, we get ℤ2−Sℤ2=1¯. It is obvious that 1¯ is an idempotent element. Therefore, ℤ2 is a Sℤ2−idempotent monoid.Now, we consider monoidℤn for n>2. Let a¯ be an unit element of Sℤn. Then, a¯x¯a¯=0¯ for all x¯∈ℤn. Specially, a¯a¯−1a¯=0 is provided for a¯−1∈ℤn. This result leads us to the contradiction a¯=0¯. This means that there is no unit element in Sℤn. So, every unit element is element of ℤn−Sℤn. In this case, if 1¯≠a¯ is an unit element in ℤn−Sℤn, a¯ is also an idempotent element. However, an unit element different from the identity cannot be an idempotent element. Therefore, we get a¯=1¯. So, the assumption is incorrect, and thus, ℤn is not Sℤn−idempotent monoid.(2)
We recall that ifgcdn,a=1, then a¯∈ℤn is an unit element. Let n=p be a prime number. Then, we get Sℤn=0¯ from Lemma 3. In this case, every element in ℤn−Sℤn=ℤn−0¯ is an unit element. Since an unit element is also a regular element, ℤn is a Sℤn−regular monoid.Letn=p2 for prime number p. Then, we have Sℤn=0¯,p¯,2p¯,…,p−1p¯ from Lemma 3. This means that, if a¯∈ℤn−Sℤn, then gcdn,a=1. So, a¯ is an unit element. Since an unit element is also a regular element, ℤn is a Sℤn−regular monoid.(3)
Letn be either a prime p or p2. We showed that in 2, if a¯∈ℤn−Sℤn, then a¯ is an unit element. It is well-known that, an unit element cannot be a zero divisor in ℤn. Therefore, there is no zero divisor in ℤn−Sℤn. So, ℤn is a Sℤn−nonzero divisor monoid.Conversely, letℤn be a Sℤn−nonzero divisor monoid. In this case, every element a¯ in ℤn−Sℤn is a nonzero divisor. This means that a¯ is an unit element, and gcda,n=1. These results provide for n=p any prime number. Assume that n is not prime and n=pk for prime number p and for some 1<k<n. Since gcdp,n≠1, p¯ is a nonunit element. Hence, p¯ must be in Sℤn, and so, p¯3=0¯ from Lemma 2. Therefore, we get n∣p3. Since n=pk, we write pk∣p3. This means that k∣p2, and so k∣p. Since 1<k<n, we obtain k=p. So, we get n=p2.(4)
Letn=p be a prime number. We showed that in 2, if a¯∈ℤn−Sℤn, then a¯ is an unit element. If an=0, then 0=ana−1=an−1. If we continue in a similar way, then we obtain a=0 contradiction. Thus, there is no nilpotent element in ℤn−Sℤn. This means that ℤn is a Sℤn−reduced monoid.Example 7.
Consider the monoidℤ4,⋅. For this monoid, we get(19)Sℤ4=0¯,2¯,ℤ4−Sℤ4=1¯,3¯.
From Theorem1, ℤ4 is not a Sℤ4−idempotent monoid. Besides that ,ℤ4 is not Sℤ4−regular and Sℤ4−nonzero divisor monoid. Indeed, if the elements of ℤ4−Sℤ4 are investigated, it is seen that 3¯ is not an idempotent element. However, since 3¯⋅3¯⋅3¯=3¯ and 1¯⋅1¯⋅1¯=1¯, these two elements are regular elements. Then, ℤ4 is a Sℤ4−regular monoid. Also, since 1¯ and 3¯ are not zero divisors or nilpotent elements, ℤ4 is a Sℤ4−nonzero divisor and Sℤ4−reduced monoid.Example 8.
Consider the monoidℤ8,⋅. Now, it turns out that(20)Sℤ8=0¯,4¯,ℤ8−Sℤ8=1¯,2¯,3¯,5¯,6¯,7¯.
From Theorem1, ℤ8 is not a Sℤ8−idempotent monoid. Indeed, only element 1¯ is idempotent in ℤ8−Sℤ8. Also, since 2¯ is not regular element, ℤ8 is not a Sℤ8−regular monoid. On the other hand, since 2¯ is zero divisor and nilpotent, ℤ8 is a not Sℤ8−nonzero divisor or Sℤ8−reduced monoid.In [10], the relation between the regular semigroups and their ideals was investigated and characterization of the regular semigroups was given. We will now investigate the relation of SS−regular semigroups with their ideals.Lemma 4.
LetI and J are the nonempty subsemigroups of semigroup S. Then, the following properties are provided.(1)
SI∩SJ⊆SI∩J(2)
IfI is left (right, both sided) ideal, then SSI is right (left, both sided) ideal(3)
IfI is ideal, then SI is ideal. Specifically, SS is ideal of S.(4)
IfI and J are ideals, then SSI.SSJ⊆SSIJ(5)
IfI and J are ideals, then SI.SJ⊆SIJProof.
We will prove (1), (2), and (4). (3) and (5) are easily seen; similarly,(1)
Ifa∈SI∩SJ, then aIa=aJa=0. Since I∩J⊆I, we get aI∩Ja⊆aIa=0. So, a∈SI∩J.(2)
LetI be a left ideal and a∈SSI,s∈S. Then, we write a∈S and aIa=0. For arbitrary x∈I, since I is a left ideal, we have sx∈I. So, the equation(21)asxas=asxas=0s=0is satisfied. This means thatas∈SSI, and so, SSI is right ideal.(4)
Letab∈SSI.SSJ. Then, we get aIa=0 and bJb=0. Since I is an ideal, we obtain bIJ⊆I. So, we get(22)abIJab=abIJab⊆aIab=0.This result means thatab∈SSIJ. So, SSI.SSJ⊆SSIJ is provided.Theorem 2.
LetS be a commutative semigroup. S is a SS−regular semigroup if and only if I∩J−SI∩J=IJ−SIJ for every ideals I and J.Proof.
LetSbe a SS−regular semigroup and x∈I∩J−SI∩J. Thus, x∈I∩J⊆S and x∉SI∩J. Assume that x∈SS. From Lemma 1 and Proposition 2, we get x=0. This result contradicts with x∉SI∩J. Then, x∈S and x∉SS. This gives us x as a regular element. Therefore, there exists y∈S, such that xyx=x. In this equation, since I and J are ideals, we write x=xyx∈IJ. Also, since x is a regular element and using Proposition 1, x∉SIJ is provided. Then, we get x∈IJ−SIJ. From this expression, we obtain(23)I∩J−SI∩J⊆IJ−SIJ.
On the other hand, ifx∈IJ−SIJ, then x is a regular element because x∈IJ⊆S and x∉SIJ⊆SS. Also, since IJ⊆I∩J, we have x∈I∩J. In this expression, using x∉SI∩J for regular element x, we obtain x∈I∩J−SI∩J. Therefore,(24)IJ−SIJ⊆I∩J−SI∩Jis provided. So, we get I∩J−SI∩J=IJ−SIJ.
To prove the converse, letI∩J−SI∩J=IJ−SIJ for right ideal I and left ideal J, and let a∈S−SS. Let us consider the ideal a=ax∣x∈S∪a by generated a. From the hypothesis, we get(25)a−Sa=a∩S−Sa∩S=aS−SaS=aS−SaS.
Usinga∉SS and Sa⊆SS, we obtain a∈a−Sa. Therefore, we have a∈aS−SaS. Also, we know that aS=Sa for commutative semigroup S. From the hypothesis, we get(26)aS−SaS=aS∩Sa−SaS∩Sa=aSSa−SaSSa=aS2a−SaS2a.
This equation gives usa∈aS2a−SaS2a. This means that a is a regular element such that a=aba for some b∈S2⊆S. So, S is a SS−regular semigroup.Theorem 3.
LetS be a commutative semigroup. S is a SS−regular semigroup if and only if I2−SI2=I−SI is provided for each ideal I of S.Proof.
LetS be a SS−regular semigroup. For ideal I, using Theorem 2, we get(27)I2−SI2=II−SII=I∩I−SI∩I=I−SI.
Conversely, letI2−SI2=I−SI for any ideal I. Then, since I∩J is an ideal, we get(28)I∩J−SI∩J=I∩J2−SI∩J2⊆I∩JI∩J−SI∩J2.
In above equation, usingSI∩J⋅SI∩J⊆SI∩J2 from Lemma 4 (5) and IJ⊆I∩J, we obtain(29)I∩J−SI∩J=I∩J2−SI∩J2⊆IJ−SI∩J⋅SI∩J.
So, usingSIJ⋅SIJ⊆SIJ, we get(30)I∩J−SI∩J⊆IJ−SIJ.
Also, sinceIJ⊆I∩J,(31)IJ−SIJ⊆I∩J−SI∩Jis provided. Then, I∩J−SI∩J=IJ−SIJ. From Theorem 2, S is a SS−regular semigroup.Now, let us give the relation betweenSS−regular and SS−reduced semigroups. The following example shows that the commutative property in this theorem is necessary.Theorem 4.
IfSis a commutative SS−regular semigroup, then S is a SS−reduced semigroup.Proof.
LetS be a commutative SS−regular semigroup, and let a∈S−SS be a nilpotent element. In this case, there exists n∈ℤ+, such that an=0 and an−1≠0. a is also a regular element. Since S is commutative, there exists b∈S, such that a2b=a. Thus, we get(32)0=an=an−2a2=an−2a2b=an−2a=an−1.
However, this result contradictsan−1≠0. Therefore, there is no nilpotent element in S−SS. So, S is a SS−reduced semigroup.Example 9.
Let the operation table of the semigroupS be given as follows.(33)⋅0abcde0000000a0a0c00b00b0d0c00c0a0d0d0b00e000000.
Using the table, we get(34)SS=0,e,S−SS=a,b,c,d.
Sincec∈S−SS is not an idempotent element, S is not a SS−idempotent semigroup. On the other hand, since aaa=a, ccc=c, and cdc=c, every element in S−SS is regular. So, S is a SS−regular semigroup. However, S is not a SS−reduced semigroup because c and d are the nilpotent elements. Therefore, S is not a SS−nonzero divisor either. As shown in this example, in a noncommutative semigroup, the above result is not valid.Finally, we will adapt one of the well-known results in semigroup theory toSS−regular semigroups. An element xof a semigroup S is called inverse element if there exist at least one y∈S, such that xyx=x and yxy=y. If there is an inverse element with uniqueness for each element of the S semigroup, the S semigroup is called the inverse semigroup. As can be seen in [8], Proposition 2.6, S is an inverse semigroup if and only if S is a regular and the idempotents of S commute with each other. We reach similar results for SS−regular semigroups, and a relation establishes between SS−regular semigroups and inverse semigroups.Lemma 5.
Every elements ofS−SS are inverse if and only if S is a SS−regular semigroup.Proof.
If every elements ofS−SS are inverse, then these elements are also regular elements. It is clear that Sis a SS−regular semigroup.
Conversely, letS be a SS−regular semigroup. Then, each element a∈S−SS is regular, and there exists b∈S, such that aba=a.(35)bababab=bababab=babab=babab=babandababa=ababa=aba=a.
So,bab and a are inverse for each other. So, every elements in S−SS are inverse elements.Theorem 5.
Every elements ofS−SS are inverse elements with uniqueness if and only if Sis a SS−regular semigroup and the idempotents of S commute with each other.Proof.
From Lemma5, every elements of S−SS are inverse if and only if S is a SS−regular semigroup. As can be seen in [8], Proposition 2.6, S is an inverse semigroup if and only if S is a regular and the idempotents of S commute with each other. From this property, all idempotent elements of S commute with each other.


## 4. Conclusions

We have shown that some properties of the source of the semiprimeness defined asSS=a∈S∣aSa=0 for a semigroup S are given. Moreover, the relations of the source of the semiprimeness with idempotent, regular, nilpotent, and zero divisor elements, which are the basis of the new semigroup structures, are investigated. Additionally, we define the SS−idempotent semigroup, SS−regular semigroup, SS− reduced semigroup, and SS−nonzero divisor semigroup structures. Thus, the mentioned semigroups are generalized. Also, we give examples for each semigroup. In particular, the monoid ℤn,⋅ is investigated and generalizations are obtained. Furthermore, we adapt some well-known results in semigroup theory to new semigroup structures. The source of primeness can be investigated in the sense of this article in future works.



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