Vol. 29, No. 01 (2023), pp. 93–98.
S-PRIME IDEALS IN PRINCIPAL DOMAIN
Mohamed Aqalmoun
11Department of Mathematics, Higher Normal School, Sidi Mohamed Ben Abdellah University, Fez, Morocco, [email protected]
Abstract.LetRbe a commutative ring andSbe a multiplicative subset ofR. The S-prime ideal is a generalization of the concept of prime ideal. In this paper, we completely determine allS-prime andS-maximal ideals of a principal domain. It is shown that the intersection of any descending chain ofS-prime ideals in a principal domain is anS-prime ideal, also theS-radical is investigated.
Key words and Phrases: Principal domain, S-prime ideal, S-maximal ideal, S- radical.
1. Introduction
Throughout this paper all rings are commutative with identity 6= 0. LetR be a commutative ring andS be a multiplicative subset ofR. Recently, Sevim et al. [11], studied the concept of S-prime ideal which is a generalization of prime ideal and used it to characterize integral domains, certain prime ideals, fields and S-Noetherian rings. An idealP withP∩S=∅is said to beS-prime ideal if there exists an element s ∈ S such that, whenever a, b ∈ R, if ab ∈ P then sa ∈ P or sb∈P. Note that ifS consist of units ofR, then the notions ofS-prime ideal and prime ideal coincide. Recall from [4] that an ideal P ofR is said to beS-maximal ideal ifP∩S=∅and there existss∈S such that wheneverP ⊆Qfor some ideal Qof R, then eithersQ⊆P orQ∩S 6=∅. TheS-radical of an ideal I is defined by √S
I={a∈R /san∈Ifor some s∈S andn∈N}. In this paper we study the concept of S-prime ideal in a principal ideal domain, for instance, we completely determine allS-prime ideals of a principal ideal domain. In [4], the author showed that any S-maximal ideal is S-prime. If R is a principal ideal domain, we show that every non-zeroS-prime ideal is S-maximal. Also the S-radical of an ideal is given.
Recall from [5] that a multiplicative subset S of R is said to be strongly multiplicative if for each family (sα)α∈Λ we have ∩α∈Λ(sαR)∩S 6=∅. In [5], the
2020 Mathematics Subject Classification: 13F10, 13A15, 13E15.
Received: 18-04-2022, accepted: 22-02-2023.
93
author showed that if S is a strongly multiplicative subset, then the intersection of any chain of S-prime ideals is an S-prime ideal, and in particular, any ideal disjoint withSis contained in a minimalS-prime ideal. Then the author asked the following question;
Question: Is the assumption “S strongly multiplicative subset” necessary for the theorem?
As part of our study, we give a negative answer to this question.
Her, we fix some notations that will be used throughout this paper. If R is a principal ideal domain. The set of all irreducible ( prime ) elements of R is denoted by P. For a multiplicative subset S the set PS is defined by PS = {p∈ P /(p)∩S 6=∅}, that is, PS is the set of all irreducible elements ofR that belong to some element of S. An irreducible elementpis inPS if and only if there exists s∈S andb∈R such thats=bp. Note that ifS=R\ {0}, then PS =P.
2. S-prime ideal in principal domain
We start this section by recalling the concept of S-prime ideals of a com- mutative ring R in order to give the form of allS-prime ideals in principal ideal domain.
Definition 2.1. Let R be a commutative ring, S be a multiplicative subset of R andP be an ideal ofRdisjoint withS. ThenP is said to beS-prime ideal if there exists ans∈S such that for all a, b∈R withab∈P, we havesa∈P or sb∈P.
The following result will be frequently used and can be found in [5].
Proposition 2.2. LetR be a commutative ring andS be a multiplicative subset of R. LetP be an ideal ofR. The following statements are equivalent
(1) P is an S-prime ideal ofR.
(2) There existss∈S such that(P :s)is a prime ideal of R.
TheS-prime ideals of a principal ideal domain are completely determined in the following result.
Theorem 2.3. Let R be a principal ideal domain andS be a multiplicative subset of R and letI be an ideal ofR. The following statements are equivalent:
(1) I is an S-prime ideal ofR,
(2) I= (0)or I= (vp)for some p∈P−PS andv∈R such that(v)∩S6=∅.
Proof. (2)⇒ (1). IfI = (0), then I is an S-prime ideal since it is a prime ideal.
Now, let I = (vp) where p ∈ P−PS and (v)∩S 6= ∅. There exists an s0 ∈ S and v0 ∈R such that s0 =vv0. Let x∈ (I :s0), then xs0 ∈I so xs0 =αvp for some α ∈ R, therefore s0x ∈ (p), hence x ∈ (p) since s0 6∈ (p). It follows that (I:s0)⊆(p). On the other hand, we haveps0 =v0vp∈I, that isp∈(I:s0), so that (I:s0) = (p) is a prime ideal of R. ThusI is anS-prime ideal ofR.
(1) ⇒ (2). LetI = (a) be a non-zero S-prime ideal of R. Let s0 ∈S such that (I :s0) is a prime ideal of R. Since (0) 6=I⊆(I :s0) there exists an irreducible
element pof R such that (I :s0) = (p). As ps0 ∈I, we have ps0 =a0afor some a0 ∈R, in particular a0a∈ (p) so a0 ∈(p) ora ∈(p). If a0 ∈(p), then a0 =a00p wherea00∈R, that isps0=a0a=aa00p, so thats0=a00a∈(a)∩S, a contradiction.
Thusa∈(p), hencea=vpwhere v∈R. Nowps0=a0a=a0vp, sos0=a0v, that is (v)∩S 6=∅. It follows thatI = (vp) and (v)∩S 6=∅ and p∈P−PS; in fact if p∈ PS then (p)∩S 6=∅, so there is an elements∈ S such thats=cp where c ∈ R. Then clearlyss0 =cs0p∈ I, which is not compatible with the fact that
I∩S=∅.
Remark 2.4. (1) IfS is a multiplicative subset of a commutative ringR, then there exists a saturated multiplicative subset S0 of R such that SpecSR = SpecS0R (see the appendix).
(2) IfSis a saturated multiplicative subset of a principal ideal domainR. Then an ideal P isS-prime if and only if P is the zero ideal or P = (sp) where s∈S andp∈P−PS.
Example 2.5. Let R=Zand S={2k /k∈N}. Note that PS ={2}. LetI be a non-zero S-prime ideal ofZ. Then I= (vp)where pis a prime integer andv∈Z such thatp6= 2 and(v)∩S6=∅ that ismv= 2k for somem∈Zandk∈N. Thus v =±2l for some l ∈N. It follows that the S-prime ideals of Z are the zero ideal and the ideals of the form (2lp)wherep6= 2 is a prime integer andl∈N.
Lemma 2.6. LetRbe a principal ideal domain. Let(In)n∈Nbe a descending chain of ideals of R. ThenIn stabilize or ∩nIn = (0).
Proof. LetI= (a) =∩nInand assume thata6= 0. Ifais invertible, then the chain stabilize. Ifa is not invertible, consider the commutative ring R0 =R/(a). Then R0 is Noetherian and dimR0 = 0, so R0 is an Artinian ring. ThusIn stabilize (in R0). There existsN such that for alln≥N,In =IN, so In =IN. Proposition 2.7. Let R be a principal ideal domain. If (Qn)n is a descending chain ofS-prime ideals ofR, then∩nQn is an S-prime ideal ofR.
Proof. This follows from the previous lemma.
Theorem 2.8. Let R be a principal ideal domain. Then every ideal which is disjoint withS is contained in a minimalS-prime ideal.
Proof. LetIbe an ideal ofR withI∩S=∅. Let
Γ ={Q /Qis anS-prime ideal andI⊆Q}
Note that Γ is not empty sinceI⊆P for some prime idealP ofRwithP∩S6=∅, which is anS-prime ideal ofR. If (Qn)n is a descending chain ofS-prime ideals of R containingI, then by the previous Proposition, Q=∩nQn is an S-prime ideal containingI. By applying the Zorn’s lemma, we get the desired results.
Proposition 2.9. Let R be a principal ideal domain and S be a multiplicative subset ofR. The following statements are equivalent.
(1) S is a strongly multiplicative subset ofR.
(2) S⊆U(R), whereU(R)is the set of invertible elements of R.
Proof. If S ⊆ U(R), then S is clearly a strongly multiplicative subset since for any s ∈ S we have sR = R. Now, assume that S 6⊆U(R). Then there exists a nonzero element s ∈ S which is not invertible. Let p ∈ P such that (s) ⊂ (p), then for any n∈ N, (sn)⊆(pn), thus∩n∈N(sn)⊆ ∩n∈N(pn) = (0), in particular
∩n∈N(sn)∩S=∅. ThusS is not a strongly multiplicative subset ofR.
Example 2.10. Let p be an irreducible element of a principal ideal domain R and S = {pn / n ∈ N}. Then S is not a strongly multiplicative subset since
∩n∈N(pnR)∩S =∅. But the intersection of a chain of S-prime ideals pf R is an S-prime ideal ofR.
3. S-maximal ideal in principal domain
Definition 3.1. LetR be a commutative ring andSbe a multiplicative subset. Let P be an ideal of R withP∩S =∅. ThenP is said to be anS-maximal ideal of R if there exists s∈S such that wheneverP ⊆Qfor some ideal Q of R then either sQ⊆P orQ∩S6=∅.
Remark 3.2. EveryS-maximal ideal of R is anS-prime ideal ofR (see [4]).
Lemma 3.3. Let Rbe a principal ideal domain andS be a multiplicative subset of R. Then(0)is an S-maximal ideal ofR if and only if PS =P
Proof. If (0) is anS-maximal ideal ofRandp∈Pthen (p)∩S6=∅sinces(p)6⊆(0), so p∈ PS. Conversely, assume that PS = P. Let Q = (a) be an ideal of R. If a= 0 then 1(Q) = (0)⊆(0). Ifa6= 0. Then either Q=R, in this caseQ∩S6=∅, or Q 6=R, in this case a=pn11· · ·pnmm where p1,· · · , pm ∈P and n1,· · · , nm are positive integers. Since pi ∈ P = PS there exists αi ∈ R such that αipi ∈ S, so Qm
i=1(αipi)ni ∈(a)∩S. ThusQ∩S 6=∅.
Classically, in a principal ideal domain every non-zero prime ideal is a maxi- mal ideal, it’sS-version is the following result.
Theorem 3.4. Let R be a principal ideal domain. Then every non-zeroS-prime ideal is an S-maximal ideal.
Proof. LetP be a non-zeroS-prime ideal ofR. ThenP = (vp) for somep∈P−PS
and v ∈ R with (v)∩S 6= ∅. Let Q= (a) be an ideal ofR with P ⊆ Q. Since vp∈(a),vp=bafor someb∈R. In particular ab∈(p), soa∈(p) orb∈(p).
First case, if a ∈ (p), then a = a0p for some a0 ∈ R, so that vp = ba0p, thus v=ba0. As (v)∩S 6=∅, there existst∈R such thats=tv=tba0∈S. Therefore sa=tva0p∈(vp). It follows thatsQ⊆P.
Second case, ifa6∈(p), thenb∈(p). So b=b0pfor someb0 ∈R. Hencev =b0a sincevp=ba=b0ap. Thus∅ 6= (v)∩S⊆(a)∩S. It follows that Q∩S6=∅.
4. S-radical in principal domain
Definition 4.1. Let R be a commutative ring andS be a multiplicative subset of R. TheS-radical of an idealI is defined by
√S
I={a∈R /san ∈I for somes∈S andn∈N}
Theorem 4.2. Let R be a principal ideal domain andS be a multiplicative subset ofR. LetI= (a)be a proper ideal ofRwritea=Qm
j=1qmj jQd
i=1pnii whereqj ∈PS
andpi∈P−PS. Then √S
I= (Qd i=1pi).
Proof. Since qj ∈PS, there existsαj ∈R such thatαjqj ∈S. Let n=max(ni), then Qm
j=1(αjq)mi(Qd
i=1pi)n ∈ I, thus Qd
i=1pi ∈ √S
I, that is (Qd
i=1pi) ⊆ √S I.
Conversely, let x ∈ √S
I, then sxn ∈ I for some s ∈ S and n ∈ N. Let b ∈ R such that sxn =bQm
j=1qmiQd
i=1pnii. Then for each 1≤i≤d, sxn ∈(pi), since (pi)∩S =∅ and (pi) is a prime ideal of R, we have xn ∈(pi), sox∈(pi). Thus x∈ ∩di=1(pi) = (Qd
i=1pi). It follows that √S
I= (Qd
i=1pi).
5. Appendix
Here we show, to studying the concept ofS-prime ideal, we can always assume that the multiplicative subsetS is saturated. So, for a multiplicative subsetS of a commutative ringR, denoteS0 the set defined byS0 ={a∈R /(a)∩S6=∅}.
Proposition 5.1. With the previous notations, we have (1) S⊆S0 andS0 is a saturated multiplicative subset.
(2) IfI is an ideal ofR, then I∩S=∅ if and only ifI∩S0=∅.
(3) IfP is an ideal ofR, then P isS-prime if and only ifP is S0-prime.
(4) IfP is an ideal of R, thenP isS-maximal if and only if P isS0-maximal.
(5) IfI is an ideal ofR, then √S I= S√0
I.
Proof. (1) Clearly S ⊆S0, 0 6∈S0 and 1 ∈S0. If a, b∈S0, then aa0 ∈ S and bb0 ∈ S for some a0, b0 ∈R, so (a0b0)(ab)∈S, that is ab∈ S0. Ifab ∈S0, thenabt∈S for somet∈R, soa, b∈S0.
(2) Clearly, ifI∩S6=∅, thenI∩S06=∅. IfI∩S06=∅, then there existsi∈I such that (i)∩S6=∅, soia∈S for somea∈R. Thusia∈I∩S.
(3) IfP is anS-prime ideal of R, then it is easy to see that P is also an S0- prime ideal of R. Conversely, assume that P is an S0-prime ideal of R.
Then (P :s0) is a prime ideal for somes0 ∈S0. We havets0 ∈S for some t ∈ R. Now; we show that (P : ts0) = (P : s0). If x ∈ (P : ts0), then xts0 ∈P, so xt∈(P :s0). Sincet 6∈(P :s0), we havex∈(P :s0), hence (P : ts0)⊆ (P : s0). If x∈ (P : s0), then xs0 ∈ P, hence xts0 ∈ P, thus x∈(P :ts0). It follows that (P :ts0) is a prime ideal ofR, thereforeP is anS-prime ideal ofR.
(4) IfP is an S-maximal ideal. We fix an elements∈S as in the definition, in particulars∈S0. IfP ⊆QandQ∩S0=∅, thenQ∩S =∅, sosQ⊆P.
It follows that P is an S0-maximal ideal ofR. Now, assume that Qis an
S0-maximal ideal and fix s0 ∈ S0 as in the definition. There exitst ∈ R such thatts0∈S. IfP ⊆QwithQ∩S=∅, thenQ∩S0 =∅, sos0Q⊆P, thusst0Q⊆tP ⊆P.
(5) From the definition we have √S I ⊆ S√0
I. Letx∈ S√0
I, thens0xn ∈I for somes0∈S0andn∈N. There existst∈Rsuch thatts0∈S, sots0xn∈I, thusx∈ √S
I.
Acknowledgement. The author would like to thank the referee for his/her great efforts in proofreading the manuscript.
REFERENCES
[1] Anderson, D. D.,A note on minimal prime ideals, Proceedings of the AMS. Vol. 122 (1), 1994.
[2] Anderson, D.D., Dumitrescu, T.:S-Noetherian rings. Commun. Algebr. 30, 44074416 (2002).
[3] Atiyah, M., McDonald, I. G. (2018).Introduction to Commutative Algebra. Oxford: Addison- Wesley Publishing Company.
[4] Eda Yldz, Bayram Ali Ersoy, nsal Tekir and Suat Ko (2020): On S-Zariski topology, Com- munications in Algebra
[5] Hamed, A., and Malek, A. (2019).S-prime ideals of a commutative ring. Beitrge Zur Algebra Und Geometrie.
[6] Hochster, M. (1971).The minimal prime spectrum of a commutative ring. Can. J. Math.
23(5):749758. DOI:10.4153/CJM-1971-083-8.
[7] Lim, J.W.:A note on S-Noetherian domains. Kyungpook Math. J.55, 507514 (2015) [8] Liu, Z.,On S-Noetherian rings. Arch. Math. (Brno) 43, 5560 (2007)
[9] McCoy, N.H.,Rings and ideals. Carus Math. Monogr. 8, 96107 (1948)
[10] Ohm, J., Pendleton, R.,Rings with Noetherian spectrum. Duke Math. J. 35:631639, (1968).
[11] Sevim, E. S., Arabaci, T., Tekir, ˙U., Ko, S. (2019).On S-prime submodules. Turk. J. Math.
43(2):10361046.
[12] Thomas W. Hungerford.On the structure of principal ideal rings. Pacific J. of math., Vol.
25, No. 3, 1968
[13] Wiegand, R., Wiegand, S.Prime ideals in Noetherian rings. Surv. Trend. Math. 13, 175193 (2010)
[14] Wiegand, S.,Intersections of prime ideals in Noetherian rings. Comm. Algebr. 11, 18531873 (1983)
[15] O. Zariski and P. Samuel,Commutative Algebra, volume I, Van Nostrand, Princeton, 1960.
