# Zariski-Like Topology on S-Quasi-Primary Ideals of a Commutative Ring

Authors: Bana Al Subaiei; Noômen Jarboui
Journal: Journal of Mathematics
 (2022)

---

## Abstract

LetR be a commutative ring with nonzero identity and, S⊆R be a multiplicatively closed subset. An ideal P of R is called an S-quasi-primary ideal if P∩S=∅ and there exists an (fixed) s∈S and whenever ab∈P for a,b∈R then either sa∈P or sb∈P. In this paper, we construct a topology on the set QPrimSR of all S-quasi-primary ideals of R which is a generalization of the S-prime spectrum of R. Also, we investigate the relations between algebraic properties of R and topological properties of QPrimSR like compactness, connectedness and irreducibility.


---

## Body

## 1. Introduction

All rings considered below are commutative with nonzero identity. The concept of prime ideals and their generalizations play an important role in Commutative Algebra and Algebraic Geometry since they are used to characterize some classes of rings and they have some applications to other areas of mathematics such as General Topology and Graph Theory (see for instance, [1–9]). A multiplicative closed subset (briefly, m.c.s) S of a commutative ring R is a subset of R such that 0∉S, 1∈S and SS⊆S. Let R be a commutative ring and S be a m.c.s of R. Recall [10, 11] that an ideal P of R is said to be S-prime if P∩S=∅ and there exists a fixed s∈S such that whenever ab∈P for some a,b∈R, then either sa∈P or sb∈P. The set of S-prime ideals of R is denoted by SpecSR. The concept of S-prime ideals is a generalization of prime ideals. Recall from [12] that an ideal is said to be quasi-primary if its radical is prime. The notion of quasi-primary ideals generalizes prime and primary ideals. By analogy, our aim in this paper is to generalize naturally the notion of S-prime ideals. In fact, if R is a commutative ring and S is a m.c.s of R, we say that an ideal P of R is S-quasi primary if P∩S=∅ and there exists a fixed s∈S such that whenever ab∈P for some a,b∈R, then either sa∈P or sb∈P. We let QPrimSR denote the set of all S-quasi primary ideals of R. Then we equip QPrimSR with a natural topology whose basis opens are of the form UaS=P∈QPrimSR|sa∉P,for alls∈S, where a∈R. The space QPrimSR is called the S-quasi-primary spectrum of R. This topology is a natural generalization of the S-Zariski topology introduced in [13]. More precisely, the S-prime spectrum SpecSR is a dense subspace of QPrimSR, see Corollary 2. The basic properties of the S-quasi-primary spectrum are studied. It is shown that the S-quasi-primary spectrum satisfies all the conditions of being a spectral space except the uniqueness of generic point, see Proposition 6, Lemma 2 and Theorem 4. Any unexplained terminology is standard as in [14, 15].


## 2. S-Quasi Primary Ideals

Throughout this paper,S will denote a m.c.s of a commutative ring R. Recall from [13] that the set of all S-prime ideals of R is denoted by SpecSR. In [12], L. Fuchs introduced the concept of quasi-primary ideals. More precisely, an ideal P of R is said to be quasi-primary if P is a prime ideal. By analogy, we introduce the following definition.Definition 1.
LetR be a commutative ring and S a m.c.s of R. An ideal P of a ring R is said to be S-quasi primary if P∈SpecSR. In other words, P is S-quasi-primary, if P∩S=∅ and there exists an (fixed) s∈S such that whenever ab∈P for some a,b∈R, then either sa∈P or sb∈P.
We letQPrimSR denote the set of all S-quasi primary ideals of R. It is worth noticing that QPrimSR is always nonempty because 0∩S=∅ and so there exists a prime ideal Q of R disjoint from S according to ([16], Theorem 3.44). Therefore, Q∈QPrimSR. The following result follows immediately from ([17], Lemma 2.3).Proposition 1.
LetRbe a commutative ring andSa m.c.s ofR. ThenSpecSR⊆QPrimSR.LetI be an ideal of R and J be a nonempty subset of R. The residual of I by J is defined as I:J=x∈R|xJ⊆I and also for the singleton J=a⊆R, we prefer I:a instead of I:a. Recall, from ([11], Lemma 2.16 and Theorem 2.18) (see also ([13], Proposition 2)), that if P is an S-prime ideal of R for a m.c.s S of R, then there exists an element sP∈S such that P:s′⊆P:sP for all s′∈S. Moreover, P:sP is a prime ideal of R. Using this fact we get immediately the following result which will be frequently used in the sequel.Proposition 2.
LetRbe a commutative ring andSa m.c.s ofR.IfP∈QPrimSR, then there existssP∈S:such that(1)
P:s′⊆P:sPfor alls′∈S.(2)
P:sPis a prime ideal ofR.


## 3.S Quasi Zariski Topology

LetR be a commutative ring and S a m.c.s of R. In [13], the authors have constructed a topology on SpecSR called the S-Zariski topology. The closed subsets of SpecSR are the S-varieties(1)VSE≔P∈SpecSR|sE⊆Pfor  somes∈S,where E is any subset of R. Thus, any open subset has the form DES=SpecSR\VSE for any subset E⊆R. The collection DaS|a∈R, where DaS=DaS:=P∈SpecSR|sa∉Pfor alls∈S, constitutes a basis for the S-Zariski topology. Our aim here is to built a natural topology on QPrimSR the set of all S-quasi primary ideals. To this end, we consider the following mapping:(2)θ:QPrimSR⟶SpecSR,P↦P,and we equip QPrimSR with the topology θ−1DES|E⊆R, which is, in fact, the initial topology with respect to the mapping θ. We call this topology the S-quasi Zariski topology. For any subset E of R, we let UES:=θ−1DES:=P∈QPrimSR|sE⊈P,for alls∈S. It follows that any closed subset for the S-quasi Zariski topology has the form ϑSE:=θ−1VSE:=P∈QPrimSR|sE⊆Pfor somes∈S, where E is any subset of R. If E=a, then we will let VSa, DaS, ϑSa and UaS denote respectively, VSa, DaS, ϑSa and UaS.We start our investigation with the following result.Theorem 1.
LetRbe a commutative ring andSa m.c.s ofR:Then the following hold true(1)
ϑSE=ϑSE for any subset E of R, where Edenotes the ideal generated by the setE⊆R.(2)
ϑSR=∅ and ϑS0=QPrimSR.(3)
∩i∈ΔϑSJi=ϑS∪i∈ΔJi=ϑS∑i∈ΔJi for any family of ideals Ji|i∈Δ of R.(4)
ϑSI∪ϑSJ=ϑSI∩J=ϑSIJ for any two ideals I and J of R.(5)
IfI⊆J, then ϑSJ⊆ϑSI for any two ideals IandJofR.Proof.
Statements (1)-(4) follow from ([13], Theorem 1) by taking the inverse images of the VS’s by the mapping θ. Statement (5) comes directly from ([13], Remark 2).
We derive from ([13], Theorem 2) the following result.Proposition 3.
LetR be a commutative ring and S a m.c.s of R. A basis for QPrimSR is the collection of all UaS=P∈QPrimSR|sa∉P,for alls∈S, where a∈R.Proposition 4.
LetR be a commutative ring and S a m.c.s of R. If I is an ideal of R, then ϑSI∩SpecSR=VSI.Proof.
LetP∈VSI. Then there exists s∈S such that sI⊆P. Thus, sI⊆P. Hence, VSI⊆ϑSI∩SpecSR. Conversely, let P∈ϑSI∩SpecSR. Then there exists s∈S such that sI⊆P. Let a∈I. There exists an integer n>0 such that snan∈P. Hence, an∈P:sn⊆P:sP. As P∈SpecSR, then P:sP is a prime ideal of R. Therefore, a∈P:sP. It follows that sPI⊆P. Hence, P∈VSI. This completes the proof.Corollary 1.
LetR be a commutative ring and S a m.c.s of R. Then UaS∩SpecSR=DaS for any a∈R.Proof.
Follows from Proposition 4 sinceUaS=QPrimSR\ϑSaand DaS=SpecSR\VSa.Corollary 2.
SpecSR is a dense subspace of QPrimSR.The following proposition determines the closure of any subsetY of QPrimSR.Proposition 5.
LetR be a commutative ring and S a m.c.s of R. Then for any Y⊆QPrimSR, we have:(3)Y¯=ϑS∩P∈YP:sP.Proof.
LetQ∈Y. As∩P∈YP:sP⊆Q:sQ, then sQ∩P∈YP:sP⊆Q. Thus, Q∈ϑS∩P∈YP:sP. It follows that Y⊆ϑS∩P∈YP:sP. Hence, Y¯⊆ϑS∩P∈YP:sP. Conversely, suppose Y⊆ϑSA for some subset A of R. Thus, for any P∈Y, there exists an element s∈S such that sA⊆P. This yields that A⊆P:s. It follows from Proposition 2 that A⊆P:sP. This shows that A⊆∩P∈YP:sP. Now, Theorem 1 ensures that ϑS∩P∈YP:sP⊆ϑSA. Since Y¯ is the smallest closed set containing Y, then we get immediately ϑS∩P∈YP:sP⊆Y¯. This completes the proof.
In Proposition6, we will characterize the closure P¯ of the singleton P for all P∈QPrimSR. But first we need the following useful lemma.Lemma 1.
LetR be a commutative ring and S a m.c.s of R. Then the following hold true:(1)
sI⊆J⇒sI⊆J for any ideals I, J of R and for any s∈S.(2)
ϑSI=ϑSI for any ideal I of R.(3)
P∈ϑSI⇒P∈ϑSI.Proof.
(1) Leta∈I and s∈S. Then there exists an integer n>0 such that an∈I. As sI⊆J, then san∈J. Hence, sanm∈J for some integer m>0. This implies that sanm∈J. Hence sa∈J.
(2) Follows by combining Theorem1 (5) and assertion (1).
(3) This is clear sinceP=P for any ideal P of R.Corollary 3.
LetE be a closed subset of QPrimSR. If P∈E, then P∈E.
The next result shows among other facts that generic points inQPrimSR are not unique.Proposition 6.
LetR be a commutative ring, S a m.c.s of R and P∈QPrimSR. Then the following hold true:(1)
P¯=ϑSP=ϑSP:sP=ϑSP=P¯.(2)
Suppose thatP∗∈QPrimSR, then(4)P¯=P∗¯⇔P:s=P∗:sfor somes∈S.Proof.
(1) It follows from Proposition5 that P¯=ϑSP:sP. Now we will show that ϑSP=ϑSP:sP. Since P⊆P:sP, then ϑSP:sP⊆ϑSP by Theorem 1. Conversely, let Q∈ϑSP. Then there exists s∈S such that sP⊆Q. This implies ssPP:sP⊆sP⊆Q, namely Q∈ϑSP:sP. The equality ϑSP=ϑSP follows from Lemma 1 Finally, P¯=P¯ by the fact that P=P.
(2) SinceP,P∗∈QPrimSR, then P,P∗∈SpecSR. The result follows from assertion (1) and the proof of ([13], Proposition 4 (ii)).
Notice that iff:R1⟶R2 is a (unital) ring homomorphism and S is a m.c.s of R1, then fS is a m.c.s of R2 if and only if 0∉fS.Proposition 7.
Letf:R1⟶R2 be a (unital) ring homomorphism and S a m.c.s of R1 with 0∉fS. If Q∈QPrimfSR2, then f−1Q∈QPrimSR1.Proof.
AsQ∈QPrimfSR2, then Q∈SpecfSR2. Thus, f−1Q∈SpecSR1 by ([13], p. 1216). As f is a ring homomorphism, then it is clear that f−1I=f−1I for any ideal I of the ring R2. It follows that f−1Q∈SpecSR1 and so f−1Q∈QPrimSR1. This completes the proof.Theorem 2.
Letf:R1⟶R2 be a (unital) ring homomorphism and S a m.c.s of R1 with 0∉fS. The map φ:QPrimfSR2⟶QPrimSR1, defined by φQ=f−1Q, is continuous.Proof.
LetE be a subset of R1. We have:(5)φ−1ϑSE=Q∈QPrimfSR2|f−1Q∈ϑSE,=Q∈QPrimfSR2|sE⊆f−1Qfor somes∈S=Q∈QPrimfSR2|sE⊆f−1Qfor somes∈S=Q∈QPrimfSR2|fsfE⊆Qfor somes∈S=ϑfSfE.
This shows thatφ is continuous.


## 4. Some Results on theS-Radical

Recall from ([13], Definition 1) that if R is a commutative ring, S is a m.c.s of R and I is an ideal of R, then the S-radical of I is defined as follows:(6)IS≔a∈R|san∈Ifor.somes∈Sandn∈ℕ.The main purpose of this section is to provide some new results concerning theS-radical. We start with the following result.Proposition 8.
LetR be a commutative ring and S a m.c.s of R. Let Y⊆QPrimSR such that θY⊆Y. Then(7)∩P∈YP:sP=∩P∈Y∩SpecSRP:sP.Proof.
Leta∈∩P∈YP:sP and let P∈Y∩SpecSR. Then sPa∈P. Thus, there exists an integer n>0 such that sPnan∈P. Hence, an∈P:sPn⊆P:sP. But as P:sP is a prime ideal of R, then we get a∈P:sP. It follows that a∈∩P∈Y∩SpecSRP:sP. This shows that∩P∈YP:sP⊆∩P∈Y∩SpecSRP:sP. Conversely, let a∈∩P∈Y∩SpecSRP:sP and let P∈Y. Since θY⊆Y, then P∈Y. Hence, P∈Y∩SpecSR and so a∈P:sP. This yields thata∈∩P∈YP:sP. Therefore,∩P∈Y∩SpecSRP:sP⊆∩P∈YP:sP. This completes the proof.
LetR be a commutative ring, S a m.c.s of R and I an ideal of R. It was proved in ([13], Proposition 5) thatIS=∩P∈VSIP:sP. The next result shows in fact that IS=∩P∈ϑSIP:sP.Corollary 4.
LetR be a commutative ring and S a m.c.s of R. If I is an ideal of R, then(8)IS=∩P∈ϑSIP:sP.Proof.
According to ([13], Proposition 5), IS=∩P∈VSIP:sP. AsϑSI∩SpecSR=VSI (Proposition 4) and θϑSI⊆ϑSI by Corollary 3, we derive from Proposition 8thatIS=∩P∈VSIP:sP=∩P∈ϑSIP:sP.Proposition 9.
LetR be a commutative ring and S a m.c.s of R. If I is an ideal of R, then ϑSI=ϑSIS.Proof.
AsI⊆IS, then ϑSIS⊆ϑSI by virtue of Theorem 1. Conversely, let P∈ϑSI, then I⊆P:s for some s∈S. Let a∈IS. Then s′an∈I for some s′∈S and n∈ℕ. Hence, an∈I:s′⊆P:ss′⊆P:sP. But as P:sP is a prime ideal, then a∈P:sP. So IS⊆P:sP, which shows that P∈ϑSIS. The proof is complete.Corollary 5.
LetR be a commutative ring and S a m.c.s of R. If I is an ideal of R, then VSI¯=ϑSI.Proof.
It follows from Proposition1 that θVSI⊆VSI. Combining Proposition 5 and Proposition 8 we get:(9)VSI¯=ϑS∩P∈VSIP:sP=ϑS∩P∈VSI∩SpecSRP:sP=ϑS∩P∈VSIP:sP=ϑSSI.
Hence,VSI¯=ϑSI by Proposition 9.


## 5. Irreducibility ofQPrimSR

Recall that a topological spaceX is called irreducible if X is nonempty and cannot be expressed as the union of two proper closed subsets, or equivalently, any two nonempty open subsets of X intersect (cf. [15, 18]).Proposition 10.
LetR be a commutative ring and S a m.c.s of R. If P∈QPrimSR, then ϑSP is an irreducible set in QPrimSR.Proof.
LetP∈QPrimSR. Since P∈SpecSR, then VSP is irreducible by virtue of ([13], Proposition 8). Hence ϑSP is irreducible according to Corollary 5. But ϑSP=ϑSP by Proposition 6. Therefore, ϑSP is also irreducible.Proposition 11.
LetR be a commutative ring and S a m.c.s of R. If I is an ideal of R such that ϑSI is irreducible, then IS∈SpecR.Proof.
AsϑSI is irreducible, then so is VSI by Corollary 5. Thus, ([13], Proposition 9) guarantees that IS∈SpecR.
Recall from ([13], Remark 3) that 0S is called the S-nilradical of R and denoted by NSR.Lemma 2.
LetR be a commutative ring and S a m.c.s of R. For any a,b∈R, we have:(1)
UaS∩UbS=UabS.(2)
UaS=∅⇔a∈NSR.Proof.
(1) Follows from ([13], Proposition 6 (i)) and from the fact that UxS=θ−1DxS for any x∈R.
(2) IfUaS=∅, then DaS=∅ by Corollary 1. Thus, a∈NSR by virtue of ([13], Proposition 6 (ii)). Conversely, if a∈NSR, then ([13], Proposition 6 (ii)) ensures that DaS=∅. Hence, UaS=θ−1DaS=∅.
For any ringR, we let NR denote the nilradical of R; that is, the set of all nilpotent elements.Theorem 3.
LetR be a commutative ring and S a m.c.s of R. Consider the following conditions:(1)
QPrimSR is an irreducible space;(2)
NR∈QPrimSR;(3)
NR∈SpecSR;(4)
NSR∈SpecR.
Then (2)⇔3⇒4⇔ (1). If moreover, S⊆RegR, then (4) ⇒ (3).Proof.
(2)⇔ (3) This clear, since NR=NR.
(3)⇒ (4) Let a,b∈R such that ab∈NSR. It follows from ([17], Lemma 2.1) that there exists s∈S such that sab=sab∈NR. As NR∈SpecSR, there exists an (fixed) s∗∈S such that either s∗sa=s∗sa∈NR or s∗b∈NR. Using again ([17], Lemma 2.1), we infer that either a∈NSR or b∈NSR. Therefore, NSR∈SpecR.
(4)⇒ (1) Let I=0. By assumption, IS=NSR is a prime ideal of R. Thus, IS∈QPrimSR. By Propositions 9 and 10, we deduce that ϑSI=ϑSIS=QPrimSR is an irreducible space.
(1)⇒ (4) Assume that QPrimSR is an irreducible space. Set I=0. Then ϑSI=QPrimSR is irreducible. By Proposition 11, IS=NSR∈SpecR.
To prove the “moreover” statement, it is enough to show that ifS⊆RegR, then NR=NSR. To this end, let a∈NSR. By using, ([17], Lemma 2.1), there exists an element s∈S such that sa∈NR. So there exists an integer n≥1 such that snan=san=0. As sn is a regular element in R, we get an=0. Hence, a∈NR. This proves that NSR⊆NR. As the reverse inclusion always holds, we get the equalityNR=NSR.
Recall from [11] that a ring R is said to be an S-integral domain if there exists a fixed s∈S and whenever ab=0 for some a,b∈R, then either sa=0 or sb=0. Note that every integral domain is an S-integral domain. Also a ring R is an S-integral domain if and only if the zero ideal is an S-prime ideal.Corollary 6.
LetR be a commutative ring and S a m.c.s of R. If R is an S-integral domain, then QPrimSR is an irreducible space.Proof.
AsR is an S-integral domain, then by using [13, Corollary 1], SpecSR is irreducible. But, Corollary 2 guarantees that SpecSR is a dense subspace of QPrimSR. Thus, QPrimSR is an irreducible space.


## 6. Compactness and Connectedness ofQPrimSR

Recall from [15] that a topological space X is called compact if any open cover of X has a finite subcover.Lemma 3.
LetR be a commutative ring and S a m.c.s of R. For any a,b∈R, we have:(10)UaS=UbS⇔aS=bS.Proof.
Note thatUaS=UbS⇔DaS=DbS. Thus, the result follows immediately from ([13], Proposition 6 (iii)).Theorem 4.
LetR be a commutative ring and S a m.c.s of R. Then UaS is compact for any a∈R. In particular, QPrimSR=U1S is compact.Proof.
Assume thatUaS⊆∪i∈ΔUaiS, where ai∈R for any i∈Δ. Then, by using Theorem 1, we obtainϑSa⊇∩i∈ΔϑSai=ϑSI, whereI=ai,i∈Δ is the ideal of R generated by the ai’s. It follows from Proposition 4 that VSa⊇VSI. Therefore, ([17], Lemma 3.1) ensures that aS⊆IS. Hence, a∈IS. Thus, there exist s∈S and n∈ℕ such that san∈I. Therefore, there exists a finite subset δ of Δ such that san∈ai,i∈δ. This shows that a∈ai,i∈δS. Thus, ϑSai,i∈δS⊆ϑSa by virtue of Theorem 1. Hence, ϑSai,i∈δ⊆ϑSa (see Proposition 9); or equivalently, ∩i∈δϑSai⊆ϑSa. It follows that UaS⊆Ui∈δUaiSas desired.
As a consequence, we derive the following corollary.Corollary 7.
LetR be a commutative ring and S a m.c.s of R. Assume that P∈SpecR|P∩S=∅⊆X⊆QPrimSR. Then X is compact.Proof.
Using Theorem 4, it is enough to show that ifO is an open subset of QPrimSR containing X, then O=QPrimSR. To this end, let P∈QPrimSR. As P∩S=∅, then it follows from ([16], Theorem 3.44) that there exists Q∈SpecR such that P⊆Q and Q∩S=∅. Since Q∈X⊆O, then there exists a∈R such that Q∈UaS⊆O. Thus, for any s∈S, sa∉Q. In particular, for any s∈S, sa∉P. This proves that P∈UaS⊆O. Therefore, we have proved that QPrimSR⊆O. As the reverse inclusion is obvious, we get O=QPrimSR. The proof is complete.
We close the paper with the following result. But first recall [15] that a topological space X is said to be connected if it cannot be written as a union of two disjoint proper open sets.Theorem 5.
LetS be a m.c.s of R. Then the following statements are equivalent:(1)
QPrimSR is a connected space.(2)
For any idempotente/s in S−1R, either e∩S≠∅ or s−e∩S≠∅ for some s∈SProof.
(1)⇒ (2) It is enough to show that SpecSR is connected and then apply ([13], Theorem 8). To this end, assume the contrary. Write SpecSR=O1∪O2 where O1 and O2 are disjoint proper open subsets of SpecSR. Then, we get QPrimSR=θ−1SpecSR=θ−1O1∪θ−1O2. This is a partition of QPrimSR into two proper open sets according to Corollary 1, which is a contradiction.
(2)⇒ (1) The hypotheses of assertion (2) imply that SpecSR is connected. As QPrimSR=SpecSR¯, then clearly QPrimSR is connected. This completes the proof.


## 7. Conclusion

In this work, we have generalized the concept of quasi-primary ideals already studied by Fuchs in [12] to that of S-quasi-primary ideals, where S is a m.c.s of a commutative unital ring R. We have equipped the set QPrimSR of S-quasi-primary ideals of R with a topology called the S-quasi-Zariski topology, which is a natural generalization of the S-Zariski topology, already investigated by Yildiz et al. in [13]. Several satisfactory results are obtained. We have proved that this topology satisfies all the conditions of being a spectral space except the uniqueness of generic point. Concerning the future plans for our work, we will study topological spaces which are S-spectral or S-Zariski spectral, in the sense that they are homeomorphic to the S-spectrum or the S-quasi-spectrum of a commutative ring R. Another direction for research, is to provide conditions on R under which QPrimSR is a normal, totally disconnected, extremely disconnected topological space.



---

*Source: 9177320-2022-11-04.xml*