Unique irredundant intersections of completely irreducible ideals

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Journal of Algebra 287 (2005) 432–448

www.elsevier.com/locate/jalgebr

Unique irredundant intersections of completelyirreducible ideals

William Heinzera,∗, Bruce Olberdingb

a Department of Mathematics, Purdue University, West Lafayette, IN 47907, USAb Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM 88003-8001,

Received 12 March 2004

Available online 5 April 2005

Communicated by Craig Huneke

Abstract

An ideal of a commutative ring is completely irreducible if it is not the intersection of any sproper overideals. It is known that every ideal is an intersection of completely irreducible ideacharacterize the rings for which every ideal can be represented uniquely as an irredundant inteof completely irreducible ideals as precisely the rings in which every proper ideal is an irreduintersection of powers of maximal ideals. We prove that every nonzero ideal of an integral domR

has a unique representation as an intersection of completely irreducible ideals if and only ifR is analmost Dedekind domain with the property that for each proper idealA the ringR/A has at leasone finitely generated maximal ideal. We characterize the rings for which every proper ideairredundant intersection of powers of prime ideals as precisely the ringsR for which (i) RM is aNoetherian valuation ring for each maximal idealM, and (ii) every ideal ofR is an irredundanintersection of irreducible ideals. 2005 Elsevier Inc. All rights reserved.

Keywords:Arithmetical ring; Valuation ring; Chain ring; Prüfer domain; Almost Dedekind domain; Irreduciideal; Irredundant intersection

* Corresponding author.
E-mail addresses:[email protected] (W. Heinzer), [email protected] (B. Olberding).
0021-8693/$ – see front matter 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2005.03.001

W. Heinzer, B. Olberding / Journal of Algebra 287 (2005) 432–448 433

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1. Introduction

Let R denote throughout a commutative ring with 1. An ideal ofR is calledirreducibleif it is not the intersection of two proper overideals; it is calledcompletely irreducibleifit is not the intersection of any set of proper overideals. In this paper we characterirings for which every ideal can be represented uniquely as an irredundant interseccompletely irreducible ideals. We prove in Theorem 2.2 that such rings are necearithmetical, and in Theorem 3.7 that the completely irreducible ideals of such a rinprecisely the powers of maximal ideals.

We recall that a ring is said to bearithmeticalif its localization at each maximal ideala valuation ring, where by avaluation ringwe mean a ring in which the ideals are lineaordered with respect to inclusion, i.e., the ideals form a chain. An arithmetical intdomain is aPrüfer domain. An integral domainR is said to bealmost Dedekindif RM is aNoetherian valuation domain for each maximal idealM . The completely irreducible ideaof an arithmetical ring are explicitly described in [4] (see Remark 3.1 below). Thuideals of a ring in which every ideal is uniquely represented as an irredundant intersof completely irreducible ideals can be decomposed into “simpler” ideals belongingwell-understood class.

Let A be the set of ideals of the ringR that are finite intersections of completely irrducible ideals. We prove in Theorem 2.2 that every idealA ∈ A is a unique irredundanintersection of completely irreducible ideals if and only ifR is arithmetical. Moreoverif R is arithmetical, then the components are unique in every irredundant intersectirreducible ideals, even in the case of an infinite intersection.

In Corollary 2.9 we prove that in a zero-dimensional ringR every ideal is a uniqueirredundant intersection of irreducible ideals if and only if Spec(R) is a scattered topological space. Combining this with Theorem 2.8, we obtain that for a zero-dimensionaR every ideal is a unique irredundant intersection of irreducible ideals if and only ifR isan arithmetical ring such that for every radical idealJ of R, R/J has a principal maximal ideal. As we record in Question 2.12, the classification of rings of positive dimefor which every ideal is a unique irredundant intersection of irreducible (not necescompletelyirreducible) ideals remains an open problem.

In Theorem 3.5 we prove that every proper irreducible ideal of a ringR is a powerof a maximal ideal if and only ifRM is a Noetherian valuation ring for every maximidealM of R. We then characterize in Theorem 3.7 the rings for which every ideal carepresented uniquely as an irredundant intersection of completely irreducible ideashow that this class of rings coincides with the class of rings for which every properis an irredundant intersection of powers of maximal ideals.

Theorem 3.7 motivates our consideration in Section 4 of the class of ringsR in whichevery proper ideal is an irredundant intersection of powers of prime ideals. We obserthis class of rings properly includes the ZPI rings of classical interest, i.e., those rinwhich every proper ideal is a product of prime ideals. We prove in Theorem 4.1 thfollowing two conditions are equivalent in a ringR:

(i) every ideal ofR can be represented as an irredundant intersection of powers of
ideals, and

434 W. Heinzer, B. Olberding / Journal of Algebra 287 (2005) 432–448

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(ii) RM is a Noetherian valuation ring for each maximal idealM of R and every ideal oR can be represented as an irredundant intersection of irreducible ideals.

Notation 1.1. For idealsI, J of the ringR, the residualI : J is defined as usual by

I : J = {x ∈ R: xJ ⊆ I }.

For an idealA and for a prime idealP of R, we use the notation

A(P) = {x ∈ R: sx ∈ A for somes ∈ R \ P } =⋃

s∈R\PA : s

to denote theisolatedP -component(isoliertes Komponentenideal) ofA in the sense oKrull [10, p. 16]. Notice thatx ∈ A(P) if and only if A : x � P . If R is a domain, thenA(P) = ARP ∩ R, whereRP denotes the localization ofR atP .

Two different notions of associated primes of a proper idealA of the ringR are usefulfor us. One of these was introduced by Krull [9, p. 742], and following [7] we call a pidealP of R aKrull associated primeof A if for everyx ∈ P , there existsy ∈ R such thatx ∈ A : y ⊆ P . The prime idealP is said to be aZariski–Samuelassociated prime ofA ifthere existsx ∈ R such that

√A : x = P . We denote by Ass(A) the set of Krull associate

primes ofA and byZ(A) the set of Zariski–Samuel associated primes ofA. It is true ingeneral thatZ(A) ⊆ Ass(A).

Remark 1.2. In [1] Fuchs defines aprimal idealof a ringR as an idealA having the prop-erty that the zero divisors inR/A form an ideal. This ideal is necessarily prime and heof the formP/A for some prime idealP of R. The idealP is called theadjoint primeof A.If A is aP -primal ideal (that is,A is a primal ideal with adjoint primeP ) thenA = A(P)

[2, Theorem 3.4]. Moreover a prime idealP of a ringR is a Krull associated prime of aidealA if and only if A(P) is a P -primal ideal ofR [2, Theorem 3.4]. Every irreduciblideal is primal [1]; hence every completely irreducible ideal is primal. In addition,adjoint prime of a completely irreducible ideal is a maximal ideal [4, Proposition 1.2]

2. Unique irredundant intersections of irreducible ideals

In this section we consider irredundant intersections of irreducible ideals. This aus to develop several technical characterizations needed in Section 3 and Sectiongive sufficient conditions for a ring to have the property that every ideal is a uniquredundant intersection of irreducible ideals, and we characterize this class of ringszero-dimensional case. We observe in Theorem 2.2 that the requirement of uniqplaces this problem in the setting of arithmetical rings, that is:If every ideal of a ringRcan be represented uniquely as an irredundant intersection of irreducible ideals, theR is
arithmetical.


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Lemma 2.1. Let (R,M) be a quasilocal ring. IfR is not a valuation ring, then therexists an ideal ofR that has two different representations as an irredundant intersectiotwo completely irreducible ideals. More precisely, there exist an idealA of R and distinctcompletely irreducible idealsC1,C2,C3 of R such thatA = C1 ∩ C2 = C1 ∩ C3 are twodifferent irredundant representations ofA.

Proof. AssumeR is not a valuation ring. Since every ideal inR is an intersection ocompletely irreducible ideals, there exist incomparable completely irreducible ideaC1and C2 of R. Let A = C1 ∩ C2. Let C∗

1 and C∗2 denote the unique minimal overidea

to the completely irreducible idealsC1 and C2, respectively. There exist elementsx ∈(C∗

1 ∩ C2) \ A andy ∈ (C1 ∩ C∗2) \ A. Sincex ∈ C2 andy ∈ C1, we have SocR/A =

(A + (x, y)R)/A is a 2-dimensional vector space overR/M andx + y /∈ C1 ∪ C2. Let C3be an ideal containingA + (x + y)R that is maximal with respect tox /∈ R. ThenC3 iscompletely irreducible, distinct fromC1 andC2, andA = C1 ∩ C3. �Theorem 2.2. LetA be the set of ideals of the ringR that are finite intersections of completely irreducible ideals. Every idealA ∈ A has a unique representation as an irredundaintersection of completely irreducible ideals if and only ifR is arithmetical. Moreover, ifR is arithmetical, then the components are unique in every irredundant intersectiirreducible ideals, even in the case of an infinite intersection.

Proof. By Corollary 5.6 of [2] in an arithmetical ring the components in any irredundintersection of irreducible ideals are unique. Assume, conversely, that for eachA ∈ A,the representation ofA as an irredundant intersection of completely irreducible ideaunique. This property then also holds inRM for each maximal idealM of R. For if M isa maximal ideal ofR andC is a completely irreducible ideal ofRM , then the preimagB of C in R under the mappingR → RM is completely irreducible andC = BRM (see[4, Theorem 1.3]). By Lemma 2.1,RM is a valuation ring. ThereforeR is arithmetical. �Lemma 2.3. LetR be a ring in which every proper ideal has a Zariski–Samuel associprime ideal. IfA is a proper ideal ofR, thenA = ⋂

P∈Z(A) A(P ).

Proof. To show thatA = ⋂P∈Z(A) A(P ), it suffices to verify the inclusion⊇, so sup-

posex ∈ R \ A. Then the proper idealA : x has a Zariski–Samuel associated prime idthat is, there exist a prime idealP andy ∈ R \ (A : x) such thatP = √

(A : x) : y. Since(A : x) : y = A : xy, we haveP ∈ Z(A). Moreover,A : xy ⊆ P implies thatxy /∈ A(P).Thusx /∈ A(P). It follows that

⋂P∈Z(A) A(P ) ⊆ A, and the proof is complete.�

Lemma 2.4.

(i) Let R be a ring in which every radical idealJ has a minimal prime divisorP suchthat P/J is the radical of a finitely generated ideal ofR/J . If A is a proper ideal ofR, thenA = ⋂

P∈Z(A) A(P ).(ii) Assume the ringR satisfies the ascending chain condition on prime ideals. Fo

proper ideal A of R, let Z∗(A) denote the maximal elements ofZ(A). If A =


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⋂P∈Z(A) A(P ), thenA = ⋂

P∈Z∗(A) A(P ) and this second intersection is irredundaIf also R is arithmetical, then this second intersection is the unique representatiA as an irredundant intersection of irreducible ideals.

Proof. (i) To prove the first claim it suffices by Lemma 2.3 to show that every properB of R has a Zariski–Samuel associated prime ideal. SetJ = √

B. By assumption therexists a minimal prime divisorP of J such thatP = √

J + C for some finitely generateideal C of R. SinceP is minimal overJ , we haveJRP = PRP . Thus J(P ) = P , soC ⊆ J(P ) and sinceC is finitely generated, there existsx ∈ R \ P such thatxC ⊆ J .HenceJ + C ⊆ J : x, and sincex /∈ P we haveJ : x ⊆ P . ThusP = √

J : x. It followsthatP = ⋃

n>0

√B : xn. Forx /∈ P andB ⊆ P implies

√B : xn ⊆ P for all n > 0, and if

a ∈ P = √J : x, then there existsk > 0 such thatakx ∈ J , soaknxn ∈ B for somen > 0;

hencea ∈ √B : xn. SinceC is finitely generated and contained inP , we haveC ⊆ √

B : xn

for somen > 0. Also, J = √B ⊆ √

B : xn, so we haveP = √C + J ⊆ √

B : xn ⊆ P ,proving thatP is a Zariski–Samuel associated prime ofB.

(ii) Now assume thatA is a proper ideal of a ringR that satisfies the ascendinchain condition on prime ideals. Then for eachQ ∈ Z(A) there existsP ∈ Z∗(A) suchthat Q ⊆ P . It follows that A(P) ⊆ A(Q). ThereforeA = ⋂

P∈Z(A) A(P ) implies thatA = ⋂

P∈Z∗(A) A(P ). Moreover, this second intersection is irredundant. For ifP ∈ Z∗(A)

andx ∈ R \ A is such thatP = √A : x, thenx ∈ ⋂

Q∈Z∗(A)\{P } A(Q) \ A(P). If we also as-sume thatR is arithmetical, then eachA(P) is irreducible (cf. Remark 1.6 of [2]), and as whave observed above in Theorem 2.2, in an arithmetical ring any representation of aas an irredundant intersection of irreducible ideals is unique, so the proof of Lemmacomplete. �Remark 2.5. The prime spectrum of a ringR is Noetherian as a topological space if aonly if every radical ideal ofR is the radical of a finitely generated ideal [12]. Thus ifR

is an arithmetical ring with Noetherian prime spectrum, then by Lemma 2.4 every idR can be represented uniquely as an irredundant intersection of irreducible idealsever, even for zero-dimensional arithmetical rings, there exist examples of rings wNoetherian prime spectrum for which every ideal is a unique irredundant intersectirreducible ideals. We postpone a discussion of such examples until Remark 2.11.

Lemma 2.6. LetA be a proper ideal of the ringR and assume thatJ = √A is an irredun-

dant intersectionJ = M ∩ B, whereM is a maximal ideal ofR andB is a proper idealnot contained inM . ThenM is a minimal prime divisor ofA andM/A is the radical of aprincipal ideal ofR/A. MoreoverM/J is a principal maximal ideal ofR/J .

Proof. SinceJ = M ∩ B is irredundant andM is maximal we haveR = M + B, soJ =MB. Thus there existx ∈ M \ B and y ∈ B \ M such that 1= x + y. It follows thatM = xR + J , for if w ∈ M , thenw = wx + wy ∈ xR + MB = xR + J . HenceM/J

is a principal maximal ideal ofR/J . To show thatM is minimal overA is equivalent toshowing thatM is minimal overJ , and for this it is enough to show thatRM/JRM is afield. Since the maximal ideal ofRM/JRM is generated byx/1+JRM , it suffices to show
x/1 is in the idealJRM . This is indeed the case sinceyx ∈ J andy /∈ M .


s

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of

Since the prime ideals containingA are precisely those that containJ , andM is theonly prime ideal that contains bothJ andx, we haveM = √

xR + A. It follows thatM/A

is the radical of a principal ideal ofR/A. �Lemma 2.7. Let A be an ideal of the ringR. If M and N are distinct maximal idealsof R with A ⊆ M ∩ N and if A(M) ⊆ N , then there is a prime idealP of R such thatA(M) ⊆ P � M ∩ N .

Proof. DefineS to be the multiplicatively closed set{xy: x ∈ R \ M, y ∈ R \ N}. Weobserve thatA(M) ⊆ N implies A(M) ∩ S = ∅. For suppose there exists an elementr ∈A(M) ∩ S. Then r = xy, with x ∈ R \ M and y ∈ R \ N . Also r ∈ A(M) implies thereexistsx′ ∈ R \ M such thatx′r = a ∈ A, anda = x′xy impliesy ∈ A(M). But y /∈ N . ThusA(M) ⊆ N implies A(M) ∩ S = ∅. Hence there is a prime idealP of R containingA(M)

such thatP ∩ S = ∅. It follows thatA(M) ⊆ P � M ∩ N . �A topological spaceX is scatteredif every nonempty subset ofX contains a point that i

isolated in the relative topology. IfR is a zero-dimensional ring, then Spec(R) is scatteredif and only if for each nonempty family{Mi : i ∈ I } of maximal ideals ofR, there existsi ∈ I such that

⋂j �=i Mj � Mi .

Theorem 2.8. LetR be a zero-dimensional ring. The following statements are equiva

(i) Spec(R) is a scattered space.(ii) For every radical idealJ of R, there is a maximal idealM containingJ such that

M/J is a principal ideal ofR/J .(iii) For every radical idealJ of R, there is a maximal idealM containingJ such that

M/J is the radical of a finitely generated ideal ofR/J .(iv) For every proper idealA of R, the setZ(A) of Zariski–Samuel associated primes

A is nonempty.(v) For every proper idealA of R, A = ⋂

P∈Z(A) A(P ).(vi) Every radical ideal ofR is an irredundant intersection of maximal ideals.

Moreover, ifR satisfies(i)–(vi) andA is a proper ideal ofR, thenA = ⋂P∈Z(A) A(P ) is

an irredundant intersection.

Proof. First observe that ifA is a proper ideal ofR, then each member ofZ(A) isa maximal ideal ofR sinceR is zero-dimensional. Therefore by Lemma 2.4, ifA =⋂

P∈Z(A) A(P ), then this intersection is irredundant.(i) ⇒ (ii) If J = ⋂

i∈I Mi is an intersection of maximal idealsMi of R, then by (i) thereexistsi ∈ I such that

⋂j �=i Mj � Mi . By Lemma 2.6,Mi/J is a principal ideal ofR/J .

(ii) ⇒ (iii) This is clear.(iii) ⇒ (iv) Apply Lemma 2.4.(iv) ⇒ (v) Apply Lemma 2.3.(v) ⇒ (vi) Let J be a radical ideal ofR. By (v) J = ⋂

P∈Z(J ) J(P ), and by the remark
at the beginning of the proof this intersection is irredundant, so to complete the proof it


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suffices to observe that eachJ(P ), P ∈Z(J ), is a maximal ideal ofR. Indeed, sinceJ is aradical ideal, so isJ : x for everyx ∈ R. It follows thatJ(P ), as a set union of radical idealis a radical ideal. SinceR is zero-dimensional,J(P ) is contained only inP (Lemma 2.7).ThusJ(P ) = P , and it follows thatJ is an irredundant intersection of maximal ideals ofR.

(vi) ⇒ (i) Let {Mi : i ∈ I1} be a collection of maximal ideals ofR, and letJ = ⋂i∈I1

Mi .By (vi) there is a collection{Ni : i ∈ I2} of maximal ideals such thatJ = ⋂

i∈I2Ni is an

irredundant intersection. Fixi2 ∈ I2, and letJ2 = ⋂i �=i2

Ni . The intersectionJ = Ni2 ∩ J2is irredundant, andJ ⊆ Mi implies eitherNi2 = Mi or J2 ⊆ Mi . If Ni2 �= Mi for all i ∈ I1,thenJ2 ⊆ ⋂

i∈I1Mi = J , a contradiction. HenceNi2 = Mi1 for somei1 ∈ I1. SetJ1 =⋂

i �=i1Mi . SinceNi2 ∩ J2 = J ⊆ J1 andNi2 � Mi for any i ∈ I1 with i �= i1, it follows

that J2 ⊆ J1. ThusJ = Mi1 ∩ J1 and J1 � Mi1 since otherwiseJ2 ⊆ J1 ⊆ Mi1 = Ni2,a contradiction. This proves Spec(R) is scattered. �Corollary 2.9. LetR be a zero-dimensional ring. The following statements are equiva

(i) Every ideal ofR can be represented uniquely as an irredundant intersection ofducible ideals.

(ii) R is an arithmetical ring with scattered prime spectrum.

Proof. (i) ⇒ (ii) By Theorem 2.2,R is arithmetical. If an idealA of an arithmetical ringR is an irredundant intersectionA = ⋂

i Ai of irreducible idealsAi , then eachAi is anisolated component ofA, that is,Ai = A(Pi) for some prime idealPi of R (cf. Lemma 5.5of [2]). If A is a radical ideal, thenA(Pi) is a prime ideal by Lemma 2.7, so it follows thA is an irredundant intersection of prime ideals. SinceR is zero-dimensional,R satisfiesthe equivalent conditions of Theorem 2.8.

(ii) ⇒ (i) Apply Lemma 2.4 and Theorem 2.8.�In Corollary 2.9 we have characterized the zero-dimensional rings for which every

can be represented uniquely as an irredundant intersection of irreducible ideals. Wethe following characterization for one-dimensional integral domains.

Corollary 2.10. LetR be a one-dimensional integral domain. The following statementequivalent.

(i) Every ideal ofR can be represented uniquely as an irredundant intersection ofducible ideals.

(ii) R is a Prüfer domain and for each nonzero proper idealA, the ring R/A has ascattered spectrum.

(iii) R is a Prüfer domain and for each proper idealA, the ringR/A contains at least onemaximal ideal that is the radical of a finitely generated ideal.

Remark 2.11. (i) An almost Dedekind domainR with at most finitely many maximaideals that are not finitely generated satisfies (i)–(iii) of Corollary 2.10. Thus ifA is anyproper nonzero ideal ofR, thenR/A is satisfies the equivalent conditions of Corollary 2
One reference for such an example that is non-Noetherian is [6, Example 2.2].


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(ii) In [13], the example of (i) is generalized to show that ifX is a compact totally disconnected topological space, then there is an almost Dedekind domainR such that Max(R)

is homeomorphic toX. Since a compact scattered space is totally disconnected, sspace can be realized as the maximal spectrum of an almost Dedekind domain.examples are discussed in [13]; we mention one here. Let(X,�) be a well-ordered seThenX is a compact scattered space with respect to the order topology onX, and theisolated points ofX are precisely the smallest element ofX and the immediate successoof elements inX [8, Example 17.3, p. 272]. Hence by suitable choices of the spacX,one obtains examples of scattered spaces having infinitely many non-isolated poinR

is an almost Dedekind domain with Max(R) homeomorphic toX, thenR has nonzero Jacobson radical and the isolated points ofX correspond to finitely generated maximal ideof R [13]. The ringR satisfies the equivalent conditions of Corollary 2.10 and every prhomomorphic image ofR satisfies the conditions of Corollary 2.9.

It would be interesting to discover how to extend the characterizations of this sectresolve the following:

Question 2.12. What ringsR have the property that every ideal ofR can be representeuniquely as an irredundant intersection of irreducible ideals?

Necessarily such rings are arithmetical. Moreover this question is equivalent to Qu5.17(1) of [2], which asks for a classification of the arithmetical ringsR having the propertythat every proper idealA can be written as an irredundant intersectionA = ⋂

P∈C A(P ) forsome setC of Krull associated primes ofA. (Apply Lemma 5.5 and Corollary 5.6 of [2].

3. Irredundant intersections of completely irreducible ideals

In this section we characterize the rings in which every ideal is a unique irreduintersection of completely irreducible ideals.

Remark 3.1. In [4, Theorem 4.3] the completely irreducible ideals of an arithmeticalare explicitly described:A ring R is arithmetical if and only if the proper completely ireducible ideals ofR are precisely the ideals of the formMB(M), whereM is a maximalideal andB is a principal ideal having the property thatBRM �= 0.

Lemma 3.2. If R is a valuation ring having a proper nonzero principal idealrR that iscompletely irreducible, then the maximal idealM of R is principal and everyM-primaryideal ofR is a power ofM .

Proof. By Remark 3.1,rR = sM for somes ∈ R. If M = M2, thenrM = sM2 = sM =rR, but this impliesr ∈ rM which contradicts the fact thatr �= 0. HenceM �= M2, andsinceM2 is irreducible, it follows that theR/M-vector spaceM/M2 has dimension oneConsequently,M is a principal ideal ofR. It follows that everyM-primary ideal ofR is a
power ofM . �


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We recall that a principal ideal ringR is called aspecialPIR if R is a local ring withmaximal idealM and every ideal ofR is a power ofM [14, p. 245].

Lemma 3.3. Every ideal of a ringR is completely irreducible if and only ifR is a specialPIR.

Proof. Suppose that every ideal ofR is completely irreducible. Then any two ideals ofR

are comparable, soR is a valuation ring and hence is quasilocal. In Remark 1.6 of [is noted that if every irreducible ideal of a ring is completely irreducible, then the rizero-dimensional. ThusR is zero-dimensional, and the claim follows from Lemma 3.2

Conversely, suppose thatR is local with maximal idealM and every ideal ofR is apower ofM . ThenR is a valuation ring and eitherR is a field orM �= M2, so as in theproof of the preceding lemma,M is a principal ideal ofR. The claim now follows fromRemark 3.1. �Proposition 3.4. Every primal ideal of a ringR is completely irreducible if and only ifRM

is a special PIR for each maximal idealM of R.

Proof. Assume every primal ideal ofR is completely irreducible. By Theorem 1.8 of [2R is arithmetical. LetM be a maximal ideal ofR. If B is an ideal ofRM andA is thepreimage ofB in R under the canonical mappingR → RM , thenARM = B andA = A(M)

is irreducible and hence primal and therefore completely irreducible. (The irreducof A = A(M) follows becauseR is arithmetical, soARM is irreducible; see Remark 1of [2].) It is shown in Theorem 1.3 of [4] that an idealC of (any ring)R is completelyirreducible if and only if for some maximal idealN of R, C = C(N) andCRN is a com-pletely irreducible ideal ofRN . Thus we conclude thatB is a completely irreducible ideaof RM . Consequently, every ideal ofRM is completely irreducible. By Lemma 3.3,RM isa special PIR.

Assume, conversely, thatRM is a special PIR for every maximal idealM of R. ThenR is arithmetical and zero-dimensional. IfA is a primal ideal ofR, thenA is M-primalfor some maximal idealM . By Lemma 3.3, every ideal ofRM is completely irreducibleIn particular,ARM is completely irreducible, and it follows (as above) from Theoremof [4] thatA = A(M) is completely irreducible. �Theorem 3.5. The following statements are equivalent for a ringR.

(i) Every proper irreducible ideal ofR is a power of a prime ideal.(ii) For every maximal idealM of R, RM is a Noetherian valuation ring.

Proof. (i) ⇒ (ii) We first establish the following three claims.Claim (1) If R is a quasilocal ring with maximal idealM �= M2 and every proper irre-

ducible ideal ofR is a power of a prime ideal, thenR is a Noetherian valuation ring.Weshow first thatM is a principal ideal. Letm ∈ M \ M2. Since every ideal is an intersectioof completely irreducible ideals, it follows from our assumption thatmR = ⋂

i∈I Pei

i for
some prime idealsPi of R such that for eachi, P
ei

i is a completely irreducible ideal ofR.


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Since the only completely irreducible prime ideal ofR is M , it follows that for eachi withPi �= M , it must be thatei > 1. In fact, if ei > 1, thenm ∈ P

ei

i ⊆ P 2i ⊆ M2, contrary to

assumption. This forces{Pi}i∈I = {M} andmR = M , proving thatM is a principal idealof R.

Under the assumptions of claim (1), we show thatR is a zero-dimensional ring orone-dimensional domain. IfP is a prime ideal ofR properly contained inM = mR, thenP � mR, so thatP = mA for some proper idealA of R. SinceP is a prime ideal ofR andm /∈ P , it follows thatA = P . Thus for all prime idealsP properly contained inM , P =mP andP ⊆ ⋂∞

k=1 Mk . Now suppose that there exists a nonmaximal prime idealP of R.We claim thatP = (0). Let y ∈ P . Since every ideal ofR is an intersection of irreduciblideals,yR = ⋂

i∈I Pei

i for some prime idealsPi of R. If {Pi}i∈I = {M}, then⋂∞

k=1 Mk ⊆yR ⊆ P . As already noted,P ⊆ ⋂∞

k=1 Mk , so we have in this case thatyR = P . ThusP = mP implies y = ymr for somer ∈ R and y(1 − mr) = 0. SinceR is quasilocal,1 − mr is a unit andy = 0. Assume{Pi}i∈I �= {M}. SincePi �= M impliesPi ⊆ Mk forall k > 0, we may assume thatPi �= M for eachi. In particular for eachi, Pi = mPi andwe haveyR = ⋂

i∈I Pei

i = m(⋂

i∈I Pei

i ) = ymR. From the fact thatR is quasilocal, weconcludey = 0. This shows that if there exists a nonmaximal prime idealP of R, thenP = 0. We conclude thatR is either a zero-dimensional ring or a one-dimensional domIn the case thatR is zero-dimensional, by assumption every irreducible ideal ofR is apower ofM ; in the case thatR is one-dimensional, every nonzero irreducible ideal oR

is a power ofM . Since every ideal of a ring is an intersection of irreducible idealsconclude thatR is a Noetherian valuation ring.

Claim (2) If R is a ring such that every proper irreducible ideal is a power of a priideal, then for each prime idealQ of R, every proper irreducible ideal ofRQ is a power ofa prime ideal.Indeed, ifA is a proper irreducible ideal ofRQ, let B denote the preimagof A under the mappingR → RQ. ThenB is an irreducible ideal ofR, and by assumptioB = P e for some prime idealP of R ande > 0. ThusA = BRQ = P eRQ, so that everyproper irreducible ideal ofRQ is a power of a prime ideal ofRQ.

Claim (3) If R is a quasilocal domain in which every proper irreducible ideal ispower of a prime ideal, thenR is either a field or a discrete rank-one valuation doma(DVR). Suppose thatR is not a field and letM be the maximal ideal ofR. Suppose thathere exists a prime idealP of R properly contained inM . We claim thatP = (0). Lety ∈ M \P , and letQ be a prime ideal ofR minimal overP +Ry. SinceQRQ is the radicalof PRQ + yRQ, it is also the radical ofA := PRQ + y2RQ. FurthermoreA �= QRQ, soby claim (2)A = QeRQ for somee > 1 sinceA is an intersection of irreducible ideaand the only prime ideal containingA is QRQ. In particularQRQ �= Q2RQ. ThusRQ

is a quasilocal ring such that every proper irreducible ideal ofRQ is a power of a primeideal andQRQ �= Q2RQ. This places us in the setting of claim (1), so we concludeRQ is a DVR. In particular,PRQ = 0. Thus (sinceR is a domain)P = 0 andR is aone-dimensional domain. Now ifB is a nonzero irreducible ideal ofR properly containedin M , thenB = Mk for somek > 1. HenceM �= M2 and by claim (1),R is a DVR.

We now prove that (i)⇒ (ii) in full generality. LetR be a ring satisfying (i) and letMbe a maximal ideal ofR. By claim (2), every proper irreducible ideal ofRM is a power ofa prime ideal. Thus to prove thatRM is a Noetherian valuation ring, it suffices by claim (
to show thatRM is a field orMRM �= M2RM . Suppose there exists a prime idealP of R


le

n

g

le

l

eals,

f

om-

for

properly contained inM . Then every proper irreducible ideal ofRM/PM is a power of aprime ideal, so by claim (3)RM/PM is a DVR. In particular,MRM �= M2RM , as claimed.Otherwise, ifRM is zero-dimensional and ifRM is not a field, then there is an irreducibidealB of RM properly contained inMRM . By assumption,B = MkRM for somek > 1.HenceMRM �= M2RM in this case also. Thus by claim (1)RM is a Noetherian valuatioring.

(ii) ⇒ (i) Let A be a proper irreducible ideal ofR with adjoint prime idealP , andlet M be a maximal ideal ofR containingP . ThenA = A(M) sinceA is P -primal. NowARM is the zero ideal ofRM or a power ofMRM . If ARM = MkRM for somek, thenthe preimagesA(M) and(Mk)(M) of ARM andMkRM , respectively, under the mappinR → RM are equal; that is,A = A(M) = (Mk)(M). SinceM is a maximal ideal ofR,then(Mk)(M) = Mk . For it is enough to verify this equality locally and ifN is a maximalideal ofR, then(Mk)(M)RN = MkRN . ThusA = Mk in caseARM is a power ofMRM .Otherwise,ARM is not a power ofMRM , so by (ii) ARM = (0)RM , where(0)RM isa prime ideal ofRM ; in particular,ARM = QRM for some prime idealQ of R. NowA = A(M) = Q(M) = Q, so thatA is a prime ideal ofR. This proves that every irreducibideal ofR is a power of a prime ideal ofR. �Corollary 3.6. If A is a proper ideal of an arithmetical ringR such that every primaideal ofR/A is completely irreducible, thenA is an intersectionA = ⋂

i Mei

i of powers ofmaximal idealsMi of R.

Proof. By Proposition 3.4RM/AM is a special PIR for all maximal idealsM of R con-taining A. Since every ideal of a ring is an intersection of completely irreducible idwe may writeA = ⋂

i Ai , where eachAi is completely irreducible. SinceAi is completelyirreducible,Ai is primal; henceAi = (Ai)(Mi) for some maximal idealMi of R. SinceR/Ai is a zero-dimensional ring,Mi is by Lemma 2.7 the unique maximal ideal ofR con-tainingAi . Moreover by Theorem 3.5 for eachi, Ai = M

ei

i + A for someei > 0. HenceAi = (M

ei

i )(Mi) + A(Mi). SinceRMiis a valuation ring, the idealsA(Mi) and(M

ei

i )(Mi) arecomparable, and sinceA(Mi) ⊆ Ai , we concludeAi = (M

ei

i )(Mi). As noted in the proof o(ii) ⇒ (i) of Theorem 3.5,Mei

i = (Mei

i )(Mi). ThusAi = Mei

i , proving thatA is an intersec-tion of powers of maximal ideals ofR. �Theorem 3.7. The following statements are equivalent for a ringR.

(i) Every ideal ofR can be represented uniquely as an irredundant intersection of cpletely irreducible ideals.

(ii) R has a scattered prime spectrum andRM is a special PIR for each maximal idealM

of R.(iii) For every proper idealA of R, R/A has a finitely generated maximal ideal, and

every maximal idealM of R, RM is a special PIR.(iv) Every proper idealA of R is an irredundant intersectionA = ⋂

i Mei

i of powers of
maximal idealsMi .


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enceery

d

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wer of

.7 isatisfy-g is a

s de-cular,s, then

.7.

eachweveritely

Proof. (i) ⇒ (ii) Since an irreducible ideal cannot be expressed as an irredundatersection of two distinct proper overideals, (i) implies that every irreducible idealR

is completely irreducible, so by Proposition 3.4RM is a special PIR for every maximaideal M of R. ThusR is zero-dimensional and by Corollary 2.9 Spec(R) is a scatteredspace.

(ii) ⇒ (iii) Let A be a proper ideal ofR. By Theorem 2.8 there existsy ∈ R and amaximal idealM of R such thatM = yR + √

A. SinceRM is a special PIR, there exists z ∈ M such thatMRM = zRM . DefineB = (y, z)R + A. We claimM = B. ClearlyMRM = BRM and by constructionB has radicalM . ThereforeMRN = BRN for eachmaximal idealN of R, soB = M . HenceM/A is a finitely generated ideal ofR/A.

(iii) ⇒ (iv) By (iii) R is an arithmetical ring since each localization at a maximal idis a valuation ring. LetA be a proper ideal ofR. By Theorem 2.8 and Corollary 2.9A is anirredundant intersection of irreducible ideals, and by Proposition 3.4 every primal (hirreducible) ideal is completely irreducible. Also, it follows from Corollary 3.6 that evcompletely irreducible proper ideal ofR is a power of a maximal ideal ofR. HenceA isan irredundant intersection of powers of maximal ideals ofR.

(iv) ⇒ (i) From (iv) it follows that every proper irreducible ideal ofR is a power ofa maximal ideal ofR, soR is zero-dimensional. Thus, by Theorem 3.5,RM is a specialPIR for each maximal idealM of R. Now to show that (i) holds, it suffices by (iv) anTheorem 2.2 to show thatMe is completely irreducible for each maximal idealM of R

and integere > 0. As noted in the proof of Theorem 3.5 (ii)⇒ (i), Me = (Me)(M). SinceRM is a valuation ring, the idealMeRM of RM is irreducible, soMe, as the preimage oMeRM under the mappingR → RM , is an irreducible ideal ofR. Thus every power of amaximal ideal ofR is irreducible. Since for each maximal idealM of R the ringRM is asspecial PIR, we have by Proposition 3.4 that every irreducible ideal, hence every poa maximal ideal, is completely irreducible.�Remark 3.8. (i) A Noetherian ring satisfying the equivalent conditions of Theorem 3necessarily semilocal since it is zero-dimensional. Conversely, any semilocal ring sing the equivalent conditions of Theorem 3.7 is Noetherian. It follows that such a rinfinite product of special PIRs (see for example Corollary 4.5).

(ii) Rings that are proper homomorphic images of the almost Dedekind domainscribed in Remark 2.11(ii) satisfy the equivalent conditions of Theorem 3.7. In partiif X is a compact scattered space having both isolated and non-isolated pointthere is an almost Dedekind domainR such that Max(R) is homeomorphic toX. SinceMax(R) has an isolated point, the Jacobson radicalJ of R is nonzero. The ringR/J hasa scattered prime spectrum andR/J satisfies the equivalent conditions of Theorem 3Since Spec(R/J ) has a non-isolated point,R/J is not semilocal; hence by (i)R/J is notNoetherian.

A ring satisfying the equivalent conditions of Theorem 3.7 is locally Noetherian atmaximal ideal, but, as noted in Remark 3.8, the ring itself need not be Noetherian. Hofrom Theorem 3.7(iii) it follows that every non-unit in such a ring is contained in a fin
generated maximal ideal.


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Corollary 3.9. The following statements are equivalent for a domainR.

(i) Every nonzero proper ideal ofR has a unique representation as an irredundant intsection of powers of maximal ideals.

(ii) R is almost Dedekind and for every proper idealA, the ringR/A has at least onefinitely generated maximal ideal.

(iii) R is an almost Dedekind domain and for each proper nonzero idealA of R,Spec(R/A) is a scattered space.

(iv) Every nonzero proper ideal ofR can be represented uniquely as an irredundant insection of completely irreducible ideals.

Proof. (i) ⇒ (ii) Statement (i) implies thatR/A is zero-dimensional for every nonzeproper idealA of R. Therefore dimR � 1. If A andB are nonzero ideals contained in tmaximal idealM , then by Theorem 3.7,RM/(A ∩ B)RM is a special PIR. ThusARM andBRM are comparable ideals ofRM . It follows thatRM is a valuation domain. Indeed, sinRM/ARM is a special PIR for any nonzero idealA ⊆ M , it follows thatRM is a Noetherianvaluation domain. This proves thatR is almost Dedekind. Now letA be a proper ideal ofR.If A is nonzero, then by (i) every proper ideal ofR/A can be represented as an irredundintersection of powers of maximal ideals ofR/A. Hence by Theorem 3.7R/A has a finitelygenerated maximal ideal. On the other hand, if dimR = 1 andA = 0, let a be a nonzerononunit ofR. ThenR/aR has a finitely generated maximal ideal, sayM/aR, whereM isa maximal ideal ofR. ThusM is a finitely generated ideal ofR.

(ii) ⇒ (iii) Apply Theorem 2.8.(iii) ⇒ (iv) If A is a nonzero proper ideal ofR, then by (iii),R/A satisfies the equiv

alent conditions of Theorem 3.7. HenceA is an irredundant intersection of completeirreducible ideals.

(iv) ⇒ (i) SupposeR is not a field. For any nonzero idealB of R, R/B is by Theo-rem 3.7 a zero-dimensional ring. HenceR is a one-dimensional domain. LetA be a propernonzero ideal ofR. ThenA is an irredundant intersectionA = ⋂

i∈I Ai of completelyirreducible idealsAi of R. Let i ∈ I . For each maximal idealM of R containingAi ,RM/(Ai)M is a special PIR, so by Proposition 3.4 every primal ideal ofR/Ai is com-pletely irreducible. SinceAi is completely irreducible, we have by Corollary 3.6 thatAi isa power of a maximal ideal ofR. Thus (i) follows. �Remark 3.10. As mentioned in Remark 2.11, it is shown in [13] that every compact stered space can be realized as Max(R) for an almost Dedekind domainR. If R is such adomain, then for every nonzero idealA, Spec(R/A) is a scattered space andR satisfies theequivalent conditions of Corollary 3.9.

Corollary 3.11. The following statements are equivalent for a ringR.

(i) Every proper ideal ofR can be represented uniquely as a finite irredundant interstion of completely irreducible ideals.

(ii) Every proper ideal ofR is a finite product of maximal ideals.
(iii) R is isomorphic to a finite direct product of special PIRs.


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is aekindy if itter IX,tion ation off rings

tion

ible

Proof. (i) ⇒ (ii) By Theorem 3.7R is a zero-dimensional ring. Now(0) = A1 ∩ · · · ∩ An

for some completely irreducible idealsAi of R. For eachi, Ai is primal, soAi = (Ai)(Mi)

for some maximal idealMi of R. By Lemma 2.7Mi is the unique maximal ideal ofRcontainingAi . It follows that the only maximal ideals ofR are M1,M2, . . . ,Mn. Nowapply Theorem 3.7 to obtain (ii).

(ii) ⇒ (iii) By (ii) the zero ideal(0) can be represented uniquely as a finite intersecof powersMei

i , i = 1, . . . , n, of distinct maximal idealsMi of R. ThusR ∼= ∏ni=1 R/M

ei

i .By Theorem 3.7, eachR/M

ei

i is a special PIR.(iii) ⇒ (i) SinceR is Noetherian, every ideal ofR is an intersection of finitely man

irreducible ideals. Also, by Proposition 3.4 every irreducible ideal ofR is completely irre-ducible, so (i) follows from Theorem 2.2.�

In view of the fact that every ideal of a ring is an intersection of irreducible ideaa ring has the property that every irreducible ideal is a power of a prime ideal, thenproper ideal is an intersection of powers of prime ideals. This motivates us to ask:

Question 3.12. If every proper idealA of a ringR is an intersectionA = ⋂i P

ei

i of powersof prime ideals, doesR satisfy the equivalent conditions of Theorem3.5?

Since every ideal of a ring is an intersection of completely irreducible ideals, thistion is equivalent to:If every completely irreducible proper ideal ofR is a power of a primeideal, is every irreducible proper ideal ofR a power of a prime ideal?

4. A generalization of ZPI rings

A ring is a ZPI (Zerlegung in Primideale) ring if every proper ideal in the ringproduct of prime ideals. It is well known that the ZPI domains are precisely the Deddomains. The general case is also well-understood: a ring is a ZPI ring if and onlis isomorphic to a finite product of Dedekind domains and special PIRs (see ChapSection 2, of [11]). As an application of our previous results, we consider in this secrelated class of rings, those for which every proper ideal is an irredundant intersecpowers of prime ideals. We observe in Remark 4.4 and Corollary 4.5 that this class oproperly contains the class of ZPI rings.

Theorem 4.1. The following statements are equivalent for a ringR.

(i) Every proper idealA of R can be represented uniquely as an irredundant intersecA = ⋂

i Pei

i of powers of prime idealsPi .(ii) Every proper idealA of R is an irredundant intersectionA = ⋂

i Pei

i of powers ofprime idealsPi .

(iii) Every ideal ofR can be represented as an irredundant intersection of irreduc
ideals andRM is a Noetherian valuation ring for each maximal idealM of R.


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Proof. Before proving Theorem 4.1, we establish (�): If R is a ring in which every propeirreducible ideal is a power of a prime ideal, then every power of a prime ideal ofR is anirreducible ideal.By Theorem 3.5,RM is a Noetherian valuation ring for each maximideal M of R. Let P be a prime ideal ofR and letk > 0. Let M be a maximal ideal oR containingP . It suffices to prove that(P k)(M) = P k , since this implies thatP k , as apreimage of an irreducible ideal under the mappingR → RM , is itself irreducible. IfPis a maximal ideal ofR (i.e. P = M), then as in the proof of (ii)⇒ (i) of Theorem 3.5,(P k)(M) = P k . Otherwise ifP is not a maximal ideal ofR, thenP is a minimal primeideal andPRM = (0)RM . HenceP kRM = PRM , so that(P k)(M) = P(M) = P . Thus forany maximal idealM containingP , (P k)(M) = P . Now P k = ⋂

N⊇P (P k)(N), whereN

ranges over the set of maximal ideals ofR containingP . Therefore we concludeP k = P ,so thatP k is clearly an irreducible ideal ofR. This shows that every power of a prime ideis an irreducible ideal.

We now verify the theorem.(i) ⇒ (ii) This is clear.(ii) ⇒ (iii) Since an irreducible ideal cannot be expressed as an irredundant inters

of distinct overideals, (ii) implies that every proper irreducible ideal is a power of a pideal. By Theorem 3.5, for each maximal idealM , RM is a Noetherian valuation ring. Bour above observation (�), every power of a prime ideal is irreducible so the assertionevery proper ideal ofR can be represented as an irredundant intersection of irreduideals is a consequence of (ii).

(iii) ⇒ (i) By Theorem 3.5, every proper irreducible ideal ofR is a power of a primeideal of R. If A is a proper ideal ofR, then by (iii) there is an irredundant intersectiA = ⋂

i Pei

i of powers of prime idealsPi of R. By (�), the set of proper irreducible ideaof R is precisely the set of powers of prime ideals ofR. Thus from Theorem 2.2 it followthat the representationA = ⋂

i Pei

i is unique among irredundant intersections of powerprime ideals. �Corollary 4.2. A domainR satisfies the equivalent conditions of Theorem4.1 if and onlyif R is an almost Dedekind domain such that for each proper idealA of R, the ringR/A

has at least one finitely generated maximal ideal.

Proof. Apply Theorem 4.1 and Corollary 3.9.�Remark 4.3. If P is a nonmaximal prime ideal of a ringR satisfying (i)–(iii) of Theo-rem 4.1, then as noted in the proof of (�) of this theorem,P k = P for all k > 0. Thus by(i) every proper idealA of R is an irredundant intersection:A = (

⋂i∈I M

ei

i )∩ (⋂

j∈J Pj ),where eachMi is a maximal ideal ofR and eachPj is a one-dimensional prime ideal ofR.(We admit here the possibility thatI or J is empty.) Moreover, sinceR is arithmetical, theset{Mi} ∪ {Pj } consists of pairwise comaximal prime ideals.

Remark 4.4. (i) By Theorem 3.7, ifR is an almost Dedekind domain with nonzero Jacson radical andR has a scattered prime spectrum and at least one non-isolated poinR satisfies the conditions of Corollary 4.2 but is not a ZPI ring. See Remark 2.11 for
an example.


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(ii) If R = R1 × R2 × · · · × Rn is a finite product of rings satisfying the equivalent coditions of Theorem 4.1, thenR also satisfies the conditions of the theorem. To provewe may assumen = 2. Let e1 ande2 be idempotent elements ofR such thatRei = Ri ,i = 1,2. A maximal idealM of R contains eithere1 or e2, but not both. Ife1 ∈ M , thenM = R1×M2, whereM2 is a maximal ideal ofR2 andRM

∼= (R2)M2. Similarly, if e2 ∈ M ,thenM = M1 × R2, whereM1 is a maximal ideal ofR1 andRM

∼= (R1)M1. Thus each lo-calization ofR at a maximal ideal is a Noetherian valuation ring. IfB = B1 × B2 is anarbitrary ideal ofR, we obtain an irredundant representation ofB as an intersection oirreducible ideals as follows: there exist irreducible idealsA1i of R1 that intersect irre-dundantly inB1 and irreducible idealsA2i of R2 that intersect irredundantly inB2. Eachof the idealsA1i × R2 andR1 × A2i is irreducible inR andB = B1 × B2 is representedirredundantly as the intersection of this collection of ideals. ThereforeR = R1 × R2 alsosatisfies the conditions of Theorem 4.1.

(iii) If R is a finite product of domains satisfying the conditions of Corollary 4.2rings satisfying the conditions of Theorem 3.7, then by (ii)R is a one-dimensional ring witzero-divisors having the property that every proper ideal is an irredundant intersecpowers of prime ideals.

Corollary 4.5. The following statements are equivalent for a ringR.

(i) R is a ZPI ring.(ii) There is an isomorphism of rings,R ∼= R1 × R2 × · · · × Rn, where eachRi is a

Dedekind domain or a special PIR.(iii) Every proper idealA of R is a finite intersectionA = ⋂n

i=1 Pei

i of powers of primeidealsPi of R.

(iv) Every proper idealA of R is a finite irredundant intersectionA = ⋂ni=1 P

ei

i of powersof prime idealsPi of R, and this representation ofA is unique among irredundanrepresentations ofA as an intersection of powers of prime ideals.

Proof. (i) ⇔ (ii) This equivalence can be found in Theorem IX.9.10 of [11].(ii) ⇒ (iii) Write R = Re1 ⊕ · · · ⊕ Ren, where theei are orthogonal idempotents an

for eachi, Ri∼= Rei . For eachi defineBi = ∑

j �=i Rej . If A is an ideal ofR, thenA =⋂ni=1(A+Bi), and since eachRi is a Dedekind domain or a special PIR, each idealA+Bi

is a finite intersection of powers of prime ideals. Hence (iii) follows.(iii) ⇒ (iv) By (iii) R satisfies the equivalent conditions of Theorem 4.1, so tha

property (�) in the proof of this theorem, every power of a prime ideal ofR is irreducible.In an arithmetical ring, an irredundant intersection of irreducible ideals ofR is unique(Theorem 2.2). Thus (iv) follows.

(iv) ⇒ (i) Let A be a proper ideal ofR. By Theorem 4.1 and Remark 4.3,A is anirredundant intersection of comaximal powers of prime ideals. By the uniqueness asin (iv) this intersection must be finite, soA is a product of these same powers of pri
ideals. �


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Acknowledgment

Our work in this paper is an outgrowth of our collaboration with Laszlo Fuchs in thticles [2–5]. We thank Laszlo for generously sharing with us his keen mathematical inas we have collaborated with him on these topics.

References

[1] L. Fuchs, On primal ideals, Proc. Amer. Math. Soc. 1 (1950) 1–6.[2] L. Fuchs, W. Heinzer, B. Olberding, Commutative ideal theory without finiteness conditions: primal i

Trans. Amer. Math. Soc., in press.[3] L. Fuchs, W. Heinzer, B. Olberding, no date, Maximal prime divisors in arithmetical rings, in: Rings,

ules, Algebras and Abelian Groups, Proceedings of the Algebra Conference—Venezia, vol. 236,Dekker, pp. 189–203.

[4] L. Fuchs, W. Heinzer, B. Olberding, Commutative ideal theory without finiteness conditions: Compirreducible ideals, Trans. Amer. Math. Soc., in press.

[5] L. Fuchs, W. Heinzer, B. Olberding, Commutative ideal theory without finiteness conditions: Irreducin the quotient field, in preparation.

[6] W. Heinzer, J. Ohm, Locally Noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971) 273[7] J. Iroz, D. Rush, Associated prime ideals in non-Noetherian rings, Canad. J. Math. 36 (1984) 344–3[8] S. Koppelberg, Handbook of Boolean Algebras, vol. 1, North-Holland, 1989.[9] W. Krull, Idealtheorie in Ringen ohne Endlichkeitsbedingung, Math. Ann. 101 (1929) 729–744.

[10] W. Krull, Idealtheorie, Ergebnisse d. Math., Berlin, 1935.[11] M. Larsen, P. McCarthy, Multiplicative Theory of Ideals, Academic Press, 1971.[12] J. Ohm, P. Pendleton, Rings with Noetherian spectrum, Duke Math. J. 35 (1968) 631–640.[13] B. Olberding, Factorization into radical ideals, in: Arithmetical Properties of Commutative Rings

Monoids, Taylor and Francis, Boca Raton, FL, in press.
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Unique irredundant intersections of completely irreducible ideals

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