MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS

LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

Abstract. We investigate for an ideal A of an arithmetical ring R the relationship between the set Max(A) of maximal prime divisors of A and the set X_{A} of maximal members of the set of Krull associated primes of A. We show that the arithmetical rings R such that X_{A} ⊆ Max(A) for every regular ideal A are precisely those satisfying the “strong” separation property. For a Prüfer domain R, we prove that every branched prime ideal of height greater than one is the radical of a finitely generated ideal if and only if End(A)_{M} = End(A_{M}) for every nonzero ideal A and maximal ideal M of R. We use this to prove that if in addition R is a QR-domain, then every maximal prime divisor of an ideal A of R is a Krull associated prime of A (i.e. X_{A} = Max(A)) if and only if each branched prime ideal of R of height greater than one is the radical of a finitely generated ideal.

Introduction

Let R be a commutative ring with identity. Associated to a proper ideal A of R is the set S(A) = {x ∈ R : xy ∈ A for some y ∈ R \ A} of elements of R that are non-prime to A. The complement R \ S(A) of elements prime to A is closed under multiplication. Krull proves [14, page 732] that ideals maximal with respect to not meeting a multiplicatively closed set are prime ideals and defines the maximal prime divisors of A as the prime ideals P of R that contain A and are maximal with respect to the property P ⊆ S(A).

Let Max(A) denote the set of maximal prime divisors of A. For P ∈ Max(A), Krull defines the ideal A_{(P)} = {x ∈ R : xy ∈ A for some y ∈ R \ P } to be the principal component of A with respect to P and establishes the decomposition of every proper ideal A of R as the intersection of its principal components A = ⋂_{P ∈Max(A)} A_{(P)} [14, Satz 2]. However, as we discuss in [4], a drawback to this decomposition is that we do not know what kind of ideals the A_{(P)} are. For instance, there sometimes exist elements of P that are prime to the principal component A_{(P)}; indeed, the ideal A_{(P)} is not in general a primal ideal, where an ideal B is said to be primal if the set S(B) of elements non-prime to B is an ideal. If B is primal, then S(B) is a prime ideal called the adjoint prime of B.

1991 Mathematics Subject Classification. Primary 13A15; Secondary 13F05.

Key words and phrases. Primal ideal, associated prime, arithmetical ring, Prüfer domain.

2 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

To obtain in a commutative ring without finiteness conditions a decomposition of the ideal A that involves components closely tied to A with structure we know, in [4] we define the set X_{A} of maximal Krull associated primes of A and obtain using the set X_{A} a canonical primal decomposition of A.

If R is a Noetherian ring and Ass(A) = {P_{1}, P_{2}, …, P_{n}} is the set of associated primes of a proper ideal A of R, then S(A) = ⋃_{i=1}^{n} P_{i}. It follows that the maximal prime divisors of A are exactly the prime ideals of R that are maximal members of Ass(A). In this sense, the maximal prime divisors of an ideal of a Noetherian ring are well-understood.

However if R is not Noetherian, the set Max(A) of maximal prime divisors of A is generally less transparent, and the primes in Max(A) need not be “associated.” There are several inequivalent notions of an associated prime of an ideal of a general commutative ring, but from our point of view, it is the Krull associated primes that are most natural. We review this and related notions in Section 1. Motivated by the Noetherian case, we are thus interested in the question of when the set of maximal prime divisors of an ideal A coincides with the set of maximal members of the set of Krull associated primes of A. We shall examine this question for the class of arithmetical rings, those rings R such that for every maximal ideal M of R, R_{M} is a valuation ring, that is, a ring for which the set of ideals is linearly ordered by inclusion. Of particular interest is the class of Prüfer domains, namely the arithmetical integral domains.

Our purpose in the present paper is to investigate for ideals A of an arithmetical ring or Prüfer domain the interrelationship between the sets Max(A) and X_{A}. The connection between these two sets of prime ideals is of special interest in the context of arithmetical rings. Indeed, if A is a proper ideal of an arithmetical ring R, then Max(A) = X_{A} if and only if every prime ideal containing A but not prime to A is a Krull associated prime (Corollary 1.3). Also, Krull mentions [15, p. 16] that he does not know whether the principal components A_{(P)} of A, P ∈ Max(A), are always P -primal ideals. (A primal ideal B is Q-primal if Q = S(B).) Examples show that the answer is in the negative (see [17],[5] and [4, Example 3.8]); we investigate in Section 3 arithmetical rings for which the answer to this question is in the affirmative for all regular ideals. (An ideal A of R is regular if it contains a regular element, i.e., an element that is not a zero-divisor.) This is also of special interest, because an ideal of an arithmetical ring is primal exactly if it is irreducible (Theorem 1.8 of [4]).

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 3

In Section 3 we examine conditions on an arithmetical ring R in order that each maximal prime divisor of a regular ideal A of R be a Krull associated prime of A. It follows from Theorem 2.3 that a necessary condition for this to hold is that R satisfy the strong separation property: for regular prime ideals P ⊂ Q and regular element r ∈ P, there exists s ∈ Q such that P ⊂ (r, s)R ⊆ Q. We conclude that Prüfer domains such as the ring of entire functions and the ring Int(ℤ) of integer-valued polynomials contain ideals A having the property that there exists a maximal prime divisor of A that is not a Krull associated prime of A.

For a regular ideal A of an arithmetical ring, we consider in Theorem 3.7 conditions in order that the set of maximal prime divisors of A is precisely the set of prime ideals that are maximal among the Krull associated primes of A. We deduce a complete characterization of the QR-domains with this property. (By contrast, ideals of Noetherian rings always exhibit equality between these two sets of prime ideals.)

The results that lay the groundwork for this theorem touch on several interesting technical aspects of arithmetical rings and Prüfer domains. In particular, there is a connection between our problem and that of when End(X) localizes, that is, when End(X)_{M} = End(X_{M}) for a submodule X of the quotient field of R and maximal ideals M of R.

Acknowledgement. We thank Gabriel Picavet and David Rush for helpful conversations on the topics of this paper and [4], and for showing us the connections of our work to the literature on Krull associated primes.

1. Krull associated primes

In this section we briefly review the notion of a Krull associated prime of a proper ideal of a ring. We became interested in the Krull associated primes of an ideal because of their connection to the primal isolated components of the ideal. For a more complete treatment of these primes and their relation to issues involving the primal decompositions of an ideal, see [4].

Let A be a proper ideal of the ring R. Following [12], we define a prime ideal P of R to be a Krull associated prime of the ideal A if for every element x ∈ P, there exists y ∈ R such that x ∈ (A : y) ⊆ P. A prime ideal P of R is called a weak-Bourbaki associated prime of A if it is a minimal prime divisor of (A : x) for some x ∈ R\A. Following [11, page 279], we call P a Zariski-Samuel associated prime of A if P = \sqrt{(A : x)} for some x ∈ R \ A. If R is a Noetherian ring, then

4 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

all three of these notions of an associated prime coincide, but for non-Noetherian rings these notions are in general distinct (see [11], for example).

It is clear that a Zariski-Samuel associated prime is a weak-Bourbaki associated prime. Moreover, it is noted in Lemma 2.1 of [4] that P is a Krull associated prime of A if and only if P is a set-theoretic union of weak-Bourbaki primes of A. It is this characterization of Krull associated primes that we shall use in what follows.

We denote by Ass(A) the set of Krull associated primes of A. Notice that if A is a proper ideal of R, then Ass(A) is nonempty. By contrast, Nakano [19] gives an example of a Prüfer domain such that no finitely generated ideal has a Zariski-Samuel associated prime.

The set Ass(A) behaves well with respect to localization:

Lemma 1.1. (Lemma 2.4(ii) of [4]) Let A be an ideal of a ring R and P be a prime ideal of R containing A. Then P ∈ Ass(A) if and only if P_{M} ∈ Ass(A_{M}) for some (or equivalently every) maximal ideal M of R containing P.

It is not hard to see that if A is a proper ideal of a ring R, then every member of Ass(A) is contained in a maximal member of Ass(A). We define X_{A} to be the maximal members of Ass(A), that is, X_{A} consists of the maximal Krull associated primes of A. There exist examples of rings with ideals A such that X_{A} is infinite. (See Example 2.6 of [4] for example, or use Lemma 1.5 below.) These examples are necessarily non-Noetherian.

In Lemma 2.3 of [4], we observe that S(A) = ⋃_{P ∈X_{A}} P. In particular, ⋃_{P ∈X_{A}} P = ⋃_{Q∈Max(A)} Q. Despite this close connection between X_{A} and Max(A), one cannot conclude in general that X_{A} ⊆ Max(A) or Max(A) ⊆ X_{A}. If Max(A) has only one member, then it is easy to see Max(A) = X_{A}. However, in Example 3.8 of [4] we construct a ring R with Noetherian prime spectrum and an ideal A of R such that Max(A) has only two elements, but neither maximal prime divisor of A is in X_{A}. We also construct in Example 2.9 of [4] a 2-dimensional Prüfer domain R for which Spec(R) is Noetherian and for which there exist an ideal A and a finitely generated maximal ideal M such that M is a maximal prime divisor of A and yet M ∉ Ass(A).

When R is arithmetical, the set Ass(A) has some striking properties. In the next sections we will need the following lemmas.

Lemma 1.2. (Proposition 2.7 of [4]) Let R be an arithmetical ring and let A be a proper ideal of R. If P ∈ Ass(A), then Q ∈ Ass(A) for every prime ideal Q of R such that A ⊆ Q ⊆ P.

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 5

Corollary 1.3. Let A be a proper ideal of an arithmetical ring R. Then X_{A} = Max(A) if and only if every prime ideal of R that contains A and is not prime to A is a Krull associated prime of A.

Proof. Assume that X_{A} = Max(A). If P contains A and is not prime to A, then P is contained in some Q ∈ X_{A}. Hence by Lemma 1.2, Q ∈ Ass(A). The converse is clear. □

Let A be a nonzero proper ideal of a Prüfer domain R and let Q be a prime ideal of R with A ⊆ Q. Since R_{Q} is a valuation domain, End(AR_{Q}) = R_{P} for some P ∈ Spec R with P ⊆ Q. Define

E_{A} = {P ∈ Spec R : R_{P} = End(AR_{Q}) for some prime ideal Q containing A}.

Lemma 1.4. (Proposition 2.8 of [4]) Let A be a nonzero proper ideal of a Prüfer domain R. Then Ass(A) = E_{A}. In particular, if P ∈ X_{A}, then there exists a maximal ideal M of R such that End(A_{M}) = R_{P}.

Lemma 1.5. Let A be an ideal of an arithmetical ring R. If A_{M} is a nonzero finitely generated ideal of R_{M} for each maximal ideal M of R, then Ass(A) is precisely the set of prime ideals of R containing A.

Proof. First assume that R is a valuation ring and A is a nonzero finitely generated proper ideal of R. Then A = yR is a principal ideal of R. Let P be the maximal ideal of R. If A = P, then clearly P ∈ Ass(A), so suppose there exists x ∈ P \ A. Since R is a valuation ring, A ⊂ Rx and A = (A : x)xR. If (A : x) = A, then A = xA and y = xya for some a ∈ A. But this means y(1 − xa) = 0. Since 1 − xa is a unit of R, this implies y = 0, a contradiction to our assumption that A ≠ 0. Therefore each x ∈ P is non-prime to A, so A is a primal ideal with S(A) = P and P ∈ Ass(A). In the general case where R is an arithmetical ring, we conclude from Lemma 2.1 and the preceding argument that every maximal ideal P of R that contains A is in Ass(A). Hence by Lemma 1.2, every prime ideal of R containing A is in Ass(A). □

2. The case X_{A} ⊆ Max(A)

In this section we characterize when X_{A} ⊆ Max(A) for every regular ideal A of an arithmetical ring. To do this, we first recall the notion of separated prime ideals, but we reformulate this definition to include rings with zero-divisors in such a way that our definition agrees with that of the separation property for domains (see pp. 91-92 of [3]). We define a ring R to have the separation property if for each

6 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

pair of distinct comparable regular prime ideals P ⊂ Q of R, there exists a finitely generated ideal A such that P ⊂ A ⊆ Q. The ring R has the strong separation property if for each pair of comparable regular prime ideals P ⊂ Q of R and regular element r ∈ P, there exists s ∈ Q such that P ⊂ (r, s)R ⊆ Q.

Clearly every one-dimensional domain has the strong separation property. By the Altitude Theorem of Krull [18, page 26] or [13, page 110], a Noetherian domain of dimension greater than two has the separation property but not the strong separation property. By contrast, Prüfer domains need not possess either separation property. The Prüfer domains that have the separation property have been well-studied (see, for example, pp. 91-92 of [3] and Lemma 2.7 of [20]). Note that a Bézout domain R has the separation property if and only if R has the strong separation property. More generally, a Prüfer domain with the QR-property ( so in particular a Prüfer domain with torsion Picard group ) has the separation property if and only if it has the strong separation property.

In the following lemma, we collect some technical characterizations of the separation property for Prüfer domains that will be needed in the next section.

Lemma 2.1. The following statements are equivalent for a Prüfer domain R. (i) R has the separation property. (ii) For each nonzero prime ideal P of R, P is a maximal prime ideal of End(P). (iii) For each nonzero prime ideal P of R, End(P_{M}) = End(P)_{M} for every

maximal ideal M of R. (iv) For each nonzero nonmaximal prime ideal P of R, if {M_{i}} is the collection

of maximal ideals of R not containing P, then R_{P} ⊆ (⋂_{i} R_{M_{i}})R_{M} for each maximal ideal M of R containing P.

(v) For each nonzero ideal A of R, if M is a maximal ideal with A ⊆ M and P is a prime ideal such that End(A)_{M} = R_{P}, then no element of P is prime to A.

Proof. (i) ⇔ (ii) Lemma 4.2.38 of [3].

(i) ⇔ (iii) Lemma 2.7 of [20].

(iii) ⇒ (iv) By Theorem 3.2.6 of [3] End(P) = R_{P} ∩ (⋂_{i} R_{M_{i}}) and End(P_{M}) = R_{P}. Thus by (iii) R_{P} = End(P_{M}) = End(P)_{M} = R_{P} ∩ (⋂_{i} M_{i})R_{M}, and (iv) follows.

(iv) ⇒ (v) Let A be a nonzero ideal of R and M a maximal ideal with A ⊆ M. Let P be a prime ideal such that End(A)_{M} = R_{P}. It follows from Lemma 1.4 that

End(A) = ( ⋂_{Q∈X_{A}} R_{Q}) ∩ (⋂_{N} R_{N}),

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 7

where N ranges over the maximal ideals of R that do not contain A. Let Q′ be the unique member of X_{A} that is contained in M. By Lemma 1.4, End(A_{M}) = R_{Q′}. If Q′ is a maximal ideal of R, then Q′ = M and End(A_{M}) = R_{M}, so R_{P} ⊆ R_{M} implies that M = P = Q′ ∈ Ass(A), and the claim is clear. It remains to consider the case where Q′ is a non-maximal prime ideal of R. Now

R_{P} = End(A)_{M} = ( ⋂_{Q∈X_{A}} R_{Q})R_{M} ∩ (⋂_{N} R_{N})R_{M}.

We claim that ⋂_{Q∈X_{A}} R_{Q} ⊆ R_{P}. If this is not the case then since R_{M} is a valuation domain it must be that R_{P} ⊊ (⋂_{Q∈X_{A}} R_{Q})R_{M}. Hence from the above representation of End(A)_{M} we deduce that since R_{P} is a valuation domain, R_{P} = (⋂_{N} R_{N})R_{M}. Thus (⋂_{N} R_{N})R_{M} ⊊ (⋂_{Q∈X_{A}} R_{Q})R_{M}. By (iv), R_{Q′} ⊆ (⋂_{N} R_{N})R_{M} since no N contains Q′. However Q′ ∈ X_{A}, so this implies R_{Q′} ⊊ R_{Q′}R_{M}, but since M contains Q′, R_{Q′}R_{M} = R_{Q′}. This contradiction implies that ⋂_{Q∈X_{A}} R_{Q} ⊆ R_{P}, so every element r ∈ P is contained in some Q ∈ X_{A}. Consequently, no element of P is prime to A.

(v) ⇒ (iii) Suppose P is a non-maximal prime ideal of R, and let M be a maximal ideal of R containing P. Let Q be a prime ideal of R such that End(P)_{M} = R_{Q}. (Since End(P)_{M} is a valuation domain such a prime Q must exist). Then by (v), the elements of Q are not prime to P. Consequently, P = Q, and (iii) follows. □

Using the strong separation property, we can characterize when X_{A} ⊆ Max(A) for every regular ideal A of an arithmetical ring. In fact, the strong separation property is always sufficient, regardless of whether R is arithmetical, to guarantee that X_{A} ⊆ Max(A):

Lemma 2.2. If A is a regular ideal of a ring R having the strong separation property, then every P ∈ X_{A} is a maximal prime divisor of A, i.e., X_{A} ⊆ Max(A).

Proof. Let P ∈ X_{A}. There exists Q ∈ Max(A) such that P ⊆ Q. We show P = Q. If X_{A} consists only of P, then A is P -primal and P = Q. Suppose X_{A} \ {P } is nonempty, and let {P_{i}} = X_{A} \ {P }. Since Q ∈ Max(A), Q ⊆ P ∪ (⋃_{i} P_{i}). Assume P is properly contained in Q. Since R has the strong separation property and A is a regular ideal, there exists x ∈ Q \ P such that P ⊂ A + (x) ⊆ Q. It follows that x is in one of the P_{i}, so P ⊂ A + (x) ⊆ P_{i}. This contradicts the assumption that P ∈ X_{A} is maximal among the Krull associated primes of A. Thus P = Q. □

Theorem 2.3. Let R be an arithmetical ring. The following statements are equivalent.

(i) R has the strong separation property.

(ii) For each proper regular ideal B of R, X_{B} ⊆ Max(B).

Proof. By Lemma 2.2, (i) implies (ii). To prove that (ii) implies (i), suppose P ⊂ Q are regular prime ideals of R and that a is a regular element with a ∈ P. Define B = aP and note that the maximal ideals of R that contain B are precisely those that contain aR. We first show that X_{B} is the set union of {P } and the set of maximal ideals of R that contain B but not P.

Since a is a regular element, we have P = B : a, so P ∈ Ass(B). We claim that P ∈ X_{B}. Suppose L ∈ X_{B} and P ⊆ L. Let M be a maximal ideal containing L. By Lemma 1.7 of [4], (BR_{M})_{(P R_{M})} is a P_{M}-primal ideal of R_{M} since BR_{M} is a finitely generated ideal of R_{M}. We claim BR_{M} = (BR_{M})_{(P R_{M})}. To verify that this is the case, we need only show that BR_{M} : y = BR_{M} for all y ∈ R_{M} \ P_{M}. To this end, observe first that P R_{M} = yP R_{M} for all y ∈ R_{M} \ P_{M}. This is because R_{M} is a chained ring, so if y ∈ R_{M} \ P_{M}, then P R_{M} ⊆ yR_{M}, and if p ∈ P R_{M}, then p = yu for some u ∈ R_{M}. Since y ∉ P R_{M}, it follows that u ∈ P R_{M} and p ∈ yP_{M}. Thus P R_{M} = yP R_{M}. Hence, with y ∈ R_{M} \ P R_{M}, we have that BR_{M} : y = yBR_{M} : y = BR_{M}, since y ∉ P R_{M} implies that P R_{M} ⊆ yR_{M} and hence that y is a regular element of R_{M}. Therefore BR_{M} = (BR_{M})_{P R_{M}}, and we conclude that BR_{M} is a P R_{M}-primal ideal. However, by Lemma 1.1 LR_{M} ∈ Ass(BR_{M}), so this forces LR_{M} ⊆ P R_{M}. Hence L ⊆ P, and we may conclude that P = L ∈ X_{B}.

Now suppose N is a maximal ideal of R that contains B but not P. Then B_{N} = aR_{N} ≠ 0, so N is in X_{B} by Lemma 1.5 and Lemma 1.1. This proves that the set X_{B} contains P and the maximal ideals of R that contain B but not P. The reverse inclusion follows from the fact that any prime ideal of R that is a member of X_{B} must contain B.

Let {N_{i}} be the set of members of X_{B} distinct from P. As noted, each N_{i} is a maximal ideal of R, and {N_{i}} is precisely the set of maximal ideals of R that contain aR but not P. Since P ∈ X_{B} and, by (ii), X_{B} ⊆ Max(B), we have Q ⊈ S(B) = P ∪ (⋃_{i} N_{i}), for otherwise P, as a proper subset of Q, would not be in Max(A). In particular, there exists x ∈ Q such that x is not contained in P nor in any of the N_{i}. We have (a, x)R ⊆ Q, and to complete the proof we show that P ⊂ (a, x)R. It is enough by Theorem II.3.1 of [1, page 88] to show that P_{M} ⊂ (A, x)R_{M} for all maximal ideals M of R that contain (A, x). Observe that if M is such a maximal ideal, then since by design x is not an element of any maximal ideal of R that contains Ra but not P, it must be that M contains P. Now since R_{M} is valuation ring, the ideals P_{M} and (a, x)R_{M} are comparable, so if P_{M} ⊈ (a, x)R_{M}, then it must be that (a, x)R_{M} ⊆ P_{M}. But then x ∈ P_{M} and

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 9

since P ⊆ M, it is the case that P_{(M)} = P, and we have x ∈ P, a contradiction to the choice of x. Thus we conclude that P ⊆ (a, x)R ⊆ Q, and R has the strong separation property. □

It follows from Theorem 2.3 that if R is an arithmetical ring that does not have the strong separation property, then there exist a proper regular ideal B of R and P ∈ X_{B} such that P ∉ Max(B). By definition of Max(B), this means there exists Q ∈ Max(B) such that P is properly contained in Q and by definition of X_{B}, this means Q ∉ Ass(B). Therefore the strong separation property is a necessary condition on an arithmetical ring in order that maximal prime divisors (of regular ideals) always be Krull primes.

Discussion 2.4. Some important examples of Prüfer domains do not have the separation property, and so by Theorem 2.3, these Prüfer domains have ideals B such that X_{B} ⊈ Max(B). Thus there exist maximal prime divisors of B that are not in Ass(B). For example, neither the ring of entire functions nor the ring of integer-valued polynomials has the separation property. Indeed, both these rings are completely integrally closed integral domains of Krull dimension greater than one. If P is a nonzero nonmaximal prime ideal of such a Prüfer domain R, then End(P) = R and P is not maximal in End(P). Hence by Lemma 2.1 (ii), such domains do not have the separation property. On the other hand, every Prüfer domain having Noetherian maximal spectrum has the separation property (apply Theorem 4.2.39 of [3] and [22]).

3. The case X_{A} = Max(A)

Our main objective in this section is to investigate conditions on an arithmetical ring R in order that X_{A} = Max(A) for each regular ideal A of R. We recall that an integral domain R is a QR-domain if every overring of R is a localization of R with respect to some multiplicatively closed subset of R. It is well-known that QR-domains are necessarily Prüfer, and that a Prüfer domain with torsion Picard group (e.g. a Bézout domain) is a QR-domain. More generally, a Prüfer domain R is a QR-domain if and only if the radical of every finitely generated ideal of R is the radical of a principal ideal of R [21]. There exist QR-domains having nontorsion Picard group [9]. Thus the condition on a Prüfer domain R that the radical of every finitely generated ideal is the radical of a principal ideal does not imply R has torsion Picard group. However, it is clear that a QR-domain that has the separation property also has the strong separation property.

10 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

Lemma 3.1. Let A be an ideal of a Prüfer domain R. Suppose Q is a prime ideal of R that contains A, and P is a prime ideal such that End(A)_{Q} = R_{P}. If P ∈ Ass(A), then End(A)_{Q} = End(A_{Q}).

Proof. Since P ∈ Ass(A), A_{(P)} is a primal ideal with adjoint prime P, and it follows that A_{P} is a P_{P} -primal ideal. By Lemma 1.4, End(A_{P}) = R_{P}. Thus End(A_{P}) = End(A)_{Q}, so A End(A_{P}) = A End(A)_{Q} implies A_{P} = A_{Q}. Consequently, End(A_{Q}) = End(A_{P}) = R_{P} = End(A)_{Q}. □

Lemma 3.2. Let R be a Prüfer domain with field of fractions F, let X be an Rsubmodule of F, and let M be a maximal ideal of R. Then End(X)_{M} = R_{P} for some P ∈ Spec R with P ⊆ M. Assume that P is the union of prime ideals P_{i}, where each P_{i} is the radical of a finitely generated ideal. Then End(X)_{Q} = End(X_{Q}) for all prime ideals Q such that P ⊆ Q ⊆ M.

Proof. Since R_{M} ⊆ End(X)_{M} and R_{M} is a valuation domain, End(X)_{M} = R_{P} for some prime ideal P ⊆ M. If End(X)_{M} = F, then clearly End(X)_{M} = End(X_{M}), so we assume End(X)_{M} ≠ F and thus P ≠ (0). Let Q be a prime ideal of R such that P ⊆ Q ⊆ M. Since End(X)_{M} = R_{P}, we have End(X)_{Q} = R_{P}. Now R_{P} = End(X)_{Q} ⊆ End(X_{Q}) ⊆ End(X_{P}), so to prove Lemma 3.2, it suffices to show that End(X_{P}) ⊆ R_{P}.

Let S = End(X). Now P S ⊆ P R_{P}, so P S ≠ S. Since S is an overring of the Prüfer domain R, S is a flat extension of R, so P S is a prime ideal of S and S_{P S} = R_{P}. Also, P S is the union of the prime ideals P_{i}S, and each P_{i}S is the radical of a finitely generated ideal of S.

Let L be a prime ideal of S such that L ⊆ P S and such that L = \sqrt{I}, where I is a finitely generated ideal of S. We prove there exists a nonzero q ∈ F such that qX_{L} is an ideal of S_{L} that is primary for L_{L}. The invertible ideal I^{2} of S is an intersection of principal fractional ideals of S. Since End(X) = S, each principal fractional ideal of S is an intersection of S-submodules of F of the form qX, q ∈ F. Since I^{2} ⊆ L, I^{2} is an intersection of ideals of S of the form L ∩ qX, where q ∈ F. Since I^{2} ⊊ I ⊆ L, there exists q ∈ F such that I^{2} ⊆ L ∩ qX ⊊ L. Hence there exists a maximal ideal N of S with I^{2} ⊆ N such that I_{N}^{2} ⊆ L_{N} ∩ qX_{N} ⊊ L_{N}. Since S_{N} is a valuation domain, the S_{N}-modules qX_{N} and L_{N} are comparable and I_{N}^{2} ⊆ L_{N} ∩ qX_{N} ⊊ L_{N} implies I_{N}^{2} ⊆ qX_{N} ⊊ L_{N}. Now \sqrt{I^{2}} = \sqrt{I} = L and I^{2} ⊆ N implies L ⊆ N. Thus I_{L}^{2} ⊆ qX_{L} ⊆ L_{L}, and we conclude that \sqrt{qX_{L}} = L_{L}.

We observe next that X_{P} ≠ F. Since P ≠ 0, there exists a nonzero L = P_{i}S ⊆ P S such that L = \sqrt{I}, where I is finitely generated. As we have established in the

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 11

paragraph above, there exists a nonzero q ∈ F such that qX_{L} is an ideal of S_{L}. Thus qX_{P} ⊆ qX_{L} ⊆ S_{L}, so it is not possible that X_{P} = F.

Fix some member L of the chain {P_{i}S}. Since X_{P} ≠ F, L ⊆ P S and R_{P} is a valuation domain, there exists a nonzero element s of S such that sX ⊆ L_{L}. Since End(X_{P}) = End(sX_{P}) and we wish to show that End(X_{P}) ⊆ R_{P} we may assume without loss of generality that s = 1; that is, we assume for the rest of the proof that X ⊆ L_{L}. Define A = X ∩ S. Then A is an ideal of S. Moreover A is contained in L since A_{L} ⊆ X_{L} ⊆ L_{L}.

With the aim of applying Lemma 3.1, we observe that P S ∈ Ass(A). For each i define L_{i} = P_{i}S. It suffices to show each L_{i} with L ⊆ L_{i} ⊆ P S is in Ass(A), since this implies that P S = ⋃_{L_{i}⊇L} L_{i} is a union of members of Ass(A). Let i be such that L ⊆ L_{i}. Since L_{i} is the radical of a finitely generated ideal of S, there exists (as we have established above) a nonzero q ∈ F such that qX_{L_{i}} is an ideal of S_{L_{i}} that is primary for (L_{i})_{L_{i}}. Now A_{L_{i}} = X_{L_{i}} ∩ S_{L_{i}}. Since S_{L_{i}} is a valuation domain, A_{L_{i}} = X_{L_{i}} or S_{L_{i}} ⊆ X_{L_{i}}. By assumption, X ⊆ L_{L}. Since L ⊆ L_{i}, X_{L_{i}} ⊆ L_{L}, and it is impossible that S_{L_{i}} ⊆ X_{L_{i}}. Thus A_{L_{i}} = X_{L_{i}}. Consequently, qX_{L_{i}} = qA_{L_{i}} and qA_{L_{i}} is an ideal of S_{L_{i}} that is primary for (L_{i})_{L_{i}}. Since S_{L_{i}} is a valuation domain, it follows that qA_{L_{i}} = A_{L_{i}} : s for some s ∈ S. Thus (L_{i})_{L_{i}} ∈ Ass(A_{L_{i}}), so by Lemma 1.1, L_{i} ∈ Ass(A). This proves P S ∈ Ass(A).

Since A = X ∩ S is an ideal of S, S ⊆ End(A). For each maximal ideal N of S, either A_{N} = X_{N} or A_{N} = S_{N} It follows that End(A) ⊆ End(X) = S, so End(A) = S. Thus End(A)_{P} = S_{P} = R_{P}, and by Lemma 3.1, End(A_{P}) = R_{P}. (We have used here that S_{SP} = R_{P}.) Now A_{P} = X_{P} ∩ S_{P} = X_{P} ∩ R_{P}. Since R_{P} is a valuation domain, A_{P} = X_{P} or R_{P} ⊆ X_{P}. The latter case is impossible since X_{P} ⊆ X_{L} ⊆ L_{L}. Thus A_{P} = X_{P}. We conclude that End(X_{P}) = End(A_{P}) = R_{P}. □

A prime ideal P is branched if there exists a P -primary ideal different from P. If P is a nonzero prime ideal of a Prüfer domain R, then P is branched if and only if P is not the union of the prime ideals properly contained in P, and in this case if P fails to be branched, then the valuation domain R_{P} is infinite dimensional and there is no maximal element among the prime ideals properly contained in P.

Lemma 3.3. Let R be a Prüfer domain having the separation property. If there exists a finitely generated ideal J of R having infinitely many minimal primes, then there exists a submodule X of the field of fractions F of R and a maximal ideal M containing J such that End(X)_{M} ≠ End(X_{M}).

12 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

Proof. Let P_{1}, P_{2}, P_{3}, … be countably many distinct minimal primes of J. Define S = ⋂_{i} R_{P_{i}}. Then S is a Prüfer overring of R and the minimal primes of JS are of the form QS, where Q ∈ Spec R is a minimal prime of J. In particular, P_{i}S is a minimal prime of JS for each i ≥ 1. Since R has the separation property, each P_{i}S is a maximal ideal of S. This is because if j is any positive integer, then End(P_{j}) = R_{P_{j}} ∩ (⋂_{N} R_{N}) by Theorem 3.2.6 of [3], so End(P_{j}) ⊆ S since the P_{i}’s are comaximal. By Lemma 2.1(ii) P_{j} is a maximal ideal of End(P_{j}), and since R is a Prüfer domain, either P_{j} extends to a maximal ideal SP_{j} of S or SP_{j} = S. The latter case is impossible since S ⊆ R_{P_{j}}. Thus SP_{j} is a maximal ideal of S.

Since J is a nonzero finitely generated ideal, each minimal prime of J is branched. Since JS has infinitely many minimal primes, Theorem 1.6 of [7] implies there exists a minimal prime QS of JS that is not the radical of a finitely generated ideal. If Q = P_{j} for some j, then by [6, Theorem 2], we have S = ⋂_{i≠j} R_{P_{i}}. Therefore, by relabeling if necessary, we may assume that Q ∉ {P_{i}}_{i=1}^{∞}.

Define A = JR_{Q} ∩ R. Then AS = JR_{Q} ∩ S is QS-primary. In particular, QS is the unique minimal prime of AS and A ⊈ P_{i}S ∩ R = P_{i} for each i ≥ 1.

Let Q_{1} be the prime ideal of R just below Q (such a prime Q_{1} exists because Q is branched). Since R has the separation property, there is a finitely generated ideal B of R such that Q_{1} ⊂ B ⊆ Q. Local verification shows that Q_{1} ⊂ B^{n} ⊆ Q for each positive integer n.

For each i ≥ 1, let

x_{i} ∈ A^{i} \ (P_{1} ∪ ⋯ ∪ P_{i} ∪ A^{i+1}).

(By Prime Avoidance (see for example [2, Lemma 3.3]) such an element x_{i} exists.) Define

J_{i} = B^{i+1} + x_{i}R.

Then Q_{1} ⊆ J_{i} and J_{i} ⊈ (P_{1} ∪ ⋯ ∪ P_{i}). Define X = ∑_{i>1} J_{i}^{−1}S.

Let N = QS and for each i > 0, define N_{i} = P_{i}S. As previously demonstrated, N and the ideals N_{i}, i > 0, are maximal ideals of S. We show first that End(X_{N}) ≠ End(X)_{N}.

For each i ≥ 1,

X_{N_{i}} = ∑_{k=1}^{i−1} J_{k}^{−1}S_{N_{i}},

since by design, J_{k}^{−1}S_{N_{k}} = S_{N_{k}} for all k ≥ i. So as a finitely generated fractional ideal of S_{N_{i}}, X_{N_{i}} is isomorphic to S_{N_{i}}. Thus End(X_{N_{i}}) = S_{N_{i}} for each i. We have

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 13

that for each i, S_{N_{i}} = R_{P_{i}}; hence

End(X) ⊆ ⋂_{i} End(X_{N_{i}}) = S.

Therefore, End(X) = S. In particular, End(X)_{N} = S_{N}.

Thus to prove the claim that End(X)_{N} ≠ End(X_{N}), it suffices to show that N End(X_{N}) = End(X_{N}). To this end we verify that X_{N} = NX_{N}. Since N = SQ, we have S_{N} = R_{Q}, so for each i ≥ 1, A^{i+1}S_{N} ⊂ A^{i}S_{N}. In particular, x_{i+1}S_{N} ⊂ x_{i}S_{N} since x_{i} ∈ A^{i} \ A^{i+1} and A^{i+1}S_{N} ∩ R = A^{i+1}R_{Q} ∩ R = A^{i+1}. It follows that (J_{i+1})S_{N} ⊂ (J_{i})S_{N}. Since J_{i} is a finitely generated ideal of S, it follows that (J_{i+1})S_{N} ⊆ N(J_{i})S_{N}. Consequently, J_{i}^{−1} ⊆ N(J_{i+1}^{−1})S_{N} for each i ≥ 1. Therefore X ⊆ NX_{N}, as claimed.

For S-submodules Y and W of F, let [W : Y] = {q ∈ F : qY ⊆ W }. Since X_{N} = NX_{N} and End(N_{N}) = S_{N}, we have

[S_{N} : X_{N}] = [N_{N} : NX_{N}] = [N_{N} : X_{N}].

Moreover, [S_{N} : X_{N}] = ⋂_{i} J_{i}S_{N} and

Q_{1}S_{N} ⊆ ⋂_{i} J_{i}S_{N} ⊆ ⋂_{i} A^{i}S_{N} = Q_{1}S_{N}.

The latter equality follows from the fact that since S_{N} is a valuation domain and AS_{N} is a principal ideal of S_{N}, the intersection ⋂_{i} A^{i}S_{N} is a prime ideal. Since Q_{1} is the largest prime ideal of R properly contained in Q, A ⊆ Q and A ⊈ Q_{1}, we have ⋂_{i} A^{i}S_{N} = Q_{1}S_{N}.

It follows that [N_{N} : X_{N}] = [S_{N} : X_{N}] = (SQ_{1})_{N}. Since R_{N} is a valuation domain, the maximal ideal N_{N} of S_{N} is m-canonical (see [10]), that is,

X_{N} = [N_{N} : [N_{N} : X_{N}]] = [N_{N} : (SQ_{1})_{N}].

Since SQ_{1} is a nonmaximal prime ideal of S, [N_{N} : (SQ_{1})_{N}] = S_{SQ_{1}} [3, Theorem 4.1.21]. Thus X_{N} = S_{SQ_{1}}, and we have End(X_{N}) = S_{SQ_{1}} = R_{Q_{1}}. In particular, End(X_{N}) ≠ R_{Q}. Now X_{N} = X_{SQ} = X_{Q}, and if M is a maximal ideal of R containing Q, then S_{SQ} = S_{M}, so X_{M} = X_{Q}. Thus

R_{Q_{1}} = End(X_{N}) = End(X_{M}),

while

End(X)_{M} = S_{M} = R_{Q}.

Therefore End(X_{M}) ≠ End(X)_{M}. □

Theorem 3.4. Let R be a Prüfer domain with field of fractions F. The following statements are equivalent.

14 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

(i) For each finitely generated ideal A of R, every weak-Bourbaki associated prime of A is a Zariski-Samuel associated prime of A.

(ii) Every finitely generated ideal of R has only finitely many minimal prime

ideals. (iii) Every principal ideal of R has only finitely many minimal prime ideals. (iv) Every branched prime ideal of R is the radical of a finitely generated ideal.

(v) For each R-submodule X of F, End(X_{M}) = End(X)_{M}, for every maximal ideal M of R.

Proof. (i) ⇒ (ii) Let A be a finitely generated ideal of R. Every minimal prime P of A is a weak-Bourbaki prime of A and therefore by (i) a Zariski-Samuel prime of A and hence of the form \sqrt{(A : x)}. Since R is Prüfer, (A : x) is finitely generated, see, for example, [6, Lemma 2]. Thus each minimal prime of R/A is the radical of a finitely generated ideal of R/A. By Theorem 1.6 in [7], R/A has finitely many minimal primes. This proves (ii).

(ii) ⇒ (iii) is obvious.

(iii) ⇒ (iv) Let P be a branched prime of R. Then there exists a prime ideal Q of R such Q ⊂ P and there are no other prime ideals between Q and P. Let x ∈ P \Q. Since the prime ideals contained in P are linearly ordered, P is a minimal prime ideal of xR. By (iii), xR has only finitely many minimal prime ideals. Since R is Prüfer, if B is the intersection of the other minimal primes of xR, then B + P = R. Hence there exists y ∈ P such that B + yR = R. It follows that P is the radical of (x, y)R. This proves (iv).

(iv) ⇒ (v) Since every nonzero prime ideal of R is the union of branched prime ideals, this follows from Lemma 3.2.

(v) ⇒ (i) Let A be a finitely generated ideal of R and let P be a weak-Bourbaki associated prime of A. Then P is a minimal prime of (A : x) for some x ∈ R. Since R is Prüfer, (A : x) is finitely generated. Lemma 3.3 implies that (A : x) has only finitely many minimal primes. Thus by Lemma 5.10 of [4] each minimal prime of (A : x) is a Zariski-Samuel associated prime of (A : x), and therefore a Zariski-Samuel associated prime of A. This proves (i). □

There is an interesting connection between the conditions of Theorem 3.4 and “trace properties” of Prüfer domains. T. Lucas shows in [16, Theorem 23] that a Prüfer domain R satisfies (iv) of Theorem 3.4 if and only if R has the radical trace property, namely, for every ideal A of R, AA^{−1} is either R or a radical ideal of R.

Corollary 3.5. The following statements are equivalent for a Prüfer domain R.

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 15

(i) Every branched prime ideal of R of height greater than one is the radical of a finitely generated ideal of R.

(ii) For each nonzero ideal A of R, End(A_{M}) = End(A)_{M} for all maximal

ideals M of R. (iii) For each ideal A of R, Ass(A) is precisely the set of prime ideals P such

that A ⊆ P and P End(A) ≠ End(A).

Proof. (i) ⇒ (ii) Suppose (i) holds, and let A be a nonzero ideal of R. Let M be a maximal ideal of R. If M has height one then R_{M} is a one-dimensional valuation domain, so End(A_{M}) = R_{M}. Consequently, End(A)_{M} = End(A_{M}). If M has height greater than 1, then M is the union of branched prime ideals of height greater than 1, so by (i) and Lemma 3.2, End(A_{M}) = End(A)_{M}. This proves (ii).

(ii) ⇒ (i) Let P be a branched prime ideal of R such that there is a nonzero prime ideal Q of R with Q ⊂ P. From (ii) and Lemma 2.1, it follows that R has the separation property. Thus Q is a maximal ideal of End(Q) and End(Q)/Q is the quotient field of R/Q. From (ii) it follows that if X is an R-submodule of End(Q) with Q ⊆ X, then

End((X/Q)_{M/Q}) = End(X/Q)_{M/Q}

for all maximal ideals M of R containing Q. By Theorem 3.4, the branched prime ideal P/Q of R/Q is the radical of a finitely generated ideal. Hence P is the radical of an ideal I + Q, where I is a finitely generated ideal of R. Since R has the separation property, there is a finitely generated ideal J such that Q ⊂ J ⊆ P. Thus P is the radical of the finitely generated ideal I + J.

(ii) ⇒ (iii) Let A be an ideal of R, and suppose P ∈ Ass(A). Then there exists Q ∈ X_{A} such that P ⊆ Q. By Lemma 1.4, R_{Q} = End(A_{M}) for some maximal ideal M of R. Thus by (ii), R_{Q} = End(A)_{M}, and we have Q End(A) ≠ End(A). It follows that P End(A) ≠ End(A).

Now suppose P is a prime ideal containing A such that P End(A) ≠ End(A). Then End(A)_{M} ≠ P End(A)_{M} for some maximal ideal M of R, and it follows that End(A)_{M} ⊆ R_{P}. By (ii), End(A_{M}) ⊆ R_{P}. If End(A_{M}) = R_{Q} for some prime ideal Q, then by Lemma 1.2 and 1.4, it follows that Q ∈ Ass(A). Since A ⊆ P ⊆ Q, Lemma 1.2 implies P ∈ Ass(A). This proves (iii).

(iii) ⇒ (ii) Let A be a proper ideal of R and M be a maximal ideal of R containing A. If End(A)_{M} = R_{P} for some prime ideal P of R, then P End(A) ≠ End(A), so by (iii), P ∈ Ass(A). By Lemma 3.1, End(A)_{M} = End(A_{M}). This proves (ii). □

16 LASZLO FUCHS, WILLIAM HEINZER, AND BRUCE OLBERDING

Remark 3.6. Theorem 3.4 shows that conditions for good behavior with respect to localization for endomorphism rings of submodules of the fraction field involve all the primes of R. On the other hand, Corollary 3.5 shows that the corresponding property for ideals involves only conditions on the primes of height greater than one. This phenomenon is illustrated in the article Goeters-Olberding [8], where the statement of Corollary 3.5 that End(A) commutes with localizations is of central importance. Our proof of Lemma 3.3 is a refinement of the argument given in Theorem 3.7 of [8].

Theorem 3.7. Consider the following statements for an arithmetical ring R. (i) X_{A} = Max(A) for each proper regular ideal A of R. (ii) Every branched prime ideal of R that properly contains a regular prime ideal

is the radical of a finitely generated ideal. (iii) Every branched prime ideal of R that properly contains a regular prime ideal

is the radical of a principal ideal. Then (iii) ⇒ (i) ⇒ (ii). Moreover, if R is a QR-domain, then all three statements are equivalent.

Proof. (i) ⇒ (ii) We consider first the case where R is a domain. Then to prove (ii), it suffices by Corollary 3.5 to show that for every ideal A of R and maximal ideal M containing A, End(A)_{M} = End(A_{M}). Let A be a proper nonzero ideal of R and let M be a maximal ideal such that A ⊆ M. Since R_{M} is a valuation domain, there exists a prime ideal P of R such that End(A)_{M} = R_{P}. We claim first that the elements of P are not prime to A. Let {M_{α}} denote the set of maximal ideals of R that contain A, and let {N_{β}} denote the set of maximal ideals of R that do not contain A. For each α, there exists a prime ideal P_{α} such that End(A_{M_{α}}) = R_{P_{α}}. Moreover, by Lemmas 1.2 and 1.4, we have X_{A} ⊆ {P_{α}} ⊆ Ass(A). Since End(A) = (⋂_{α} End(A_{M_{α}})) ∩ (⋂_{β} End(A_{N_{β}})), we have

R_{P} = End(A)_{M} = (⋂_{α} R_{P_{α}})R_{M} ∩ (⋂_{β} R_{N_{β}})R_{M}.

Since R has the separation property, Lemma 2.3(iv) implies that ⋂_{β} R_{N_{β}} ⊈ R_{P}. Thus because R_{P} is a valuation domain, it must be that ⋂_{α} R_{P_{α}} ⊆ R_{P}, and it follows that P ⊆ ⋃_{α} P_{α}. Since S(A) = ∪_{α}P_{α}, the elements of P are not prime to A, as claimed. Therefore there exists Q ∈ Max(A) such that P ⊆ Q. By (i), Q ∈ X_{A}, so by Lemma 1.2, P ∈ Ass(A) since A ⊆ P ⊆ Q. Thus by Lemma 3.1, End(A)_{M} = End(A_{M}). This proves that (i) implies (ii) in the case R is a domain.

Now consider the general case where R is not necessarily a domain. Observe that since statement (i) holds for R, statement (i) holds for R/P for all regular

MAXIMAL PRIME DIVISORS IN ARITHMETICAL RINGS 17

prime ideals P of R. This follows from the observation that if A is an ideal of R that contains P then for all x ∈ R,

(A/P :_{R/P} x + P) = ((A : x) + P)/P.

In particular, Ass(A/P) = {Q/P : Q ∈ Ass(A), P ⊆ Q}. Also, Max(A/P) = {Q/P : Q ∈ Max(A), P ⊆ Q}. Thus statement (ii) holds for R/P for all regular prime ideals P of R. Let Q be a branched prime ideal of R containing at least one regular prime ideal. Then there exists a regular prime ideal P of R such that P ⊂ Q. By Theorem 2.3, there exists a finitely generated ideal B of R such that P ⊆ B ⊂ Q. Moreover, since (ii) holds for R/P, there exists a finitely generated ideal C of R such that Q is the radical of C + P. It follows that Q is the radical of the finitely generated ideal C + B.

(iii) ⇒ (i) Let A be a proper regular ideal of R. For each P ∈ X_{A} there exists M ∈ Max(A) such that P ⊆ M and then P = M if and only if M ∈ Ass(A). Thus it suffices to prove that M ∈ Max(A) implies M ∈ Ass(A). This is clear if M/A has height one as prime ideal of R/A. Assume that ht(M/A) ≥ 2. Then M is the union of a chain of branched prime ideals P_{i} of R, where A ⊆ P_{i} and ht(P_{i}/A) ≥ 2. Since A is a regular ideal, so is each P_{i}. Thus by (iii), each P_{i} is the radical of a principal ideal of R. Since P_{i} ⊆ M, the elements in P_{i} are non-prime to A. Since P_{i} is the radical of a principal ideal and S(A) = ⋃_{Q∈X_{A}} Q, P_{i} is contained in some member of X_{A}. By Lemma 1.2, each P_{i} ∈ Ass(A). Therefore ⋃_{i} P_{i} = M ∈ Ass(A).

Finally, we conclude that statements (i),(ii) and (iii) are equivalent for a QRdomain R since in a QR-domain every prime ideal that is the radical of a finitely generated ideal is the radical of a principal ideal. □

Corollary 3.8. If R is a Prüfer domain having the property that every maximal prime divisor of an ideal A of R is a Krull associated prime of A, then every branched prime ideal of R of height greater than one is the radical of a finitely generated ideal.

Proof. The corollary is a consequence of Theorem 3.7 and Corollary 1.3. □

Example 3.9 illustrates the fact that it is possible for a 2-dimensional Prüfer domain to satisfy the hypotheses of Theorem 2.3, but not Theorem 3.7.

Example 3.9. There exists a Prüfer domain R that has the strong separation property and yet contains an ideal A such that X_{A} is a proper subset of Max(A). Moreover, R does not satisfy any of the conditions (i), (ii) or (iii) of Theorem 3.7. Let D be an almost Dedekind domain (that is, D_{M} is a DVR for all maximal ideals M of D) that is not Dedekind. (See [3, page 281] for an example of such a domain.)

Let F denote the field of fractions of D, let x be an indeterminate over F, and

let P = xF [x]_{(x)}. Define R = D + P. Then R is a Prüfer domain and P is the

unique nonzero nonmaximal prime ideal of R. Moreover, if M is a maximal ideal

of R, then M contains P and if r ∈ M \ P, then P ⊆ Rr ⊆ M, so R has the

strong separation property. In particular, by Theorem 2.3, every ideal A of R has

the property that X_{A} ⊆ Max(A). Since D is almost Dedekind, but not Dedekind,

there exists a maximal ideal Q of D that is not the radical of a finitely generated

ideal. Consequently, the (height 2) maximal ideal Q + P of R is not the radical of

a finitely generated ideal of R. By Theorem 3.7, there exists an ideal A of R such

that X_{A} is a proper subset of Max(A).

In Example 2.9 of [4], we construct a 2-dimensional Prüfer domain with Noether-

ian spectrum having an ideal A for which there exists a maximal ideal M such that

M is in Max(A) but M is not a Krull associated prime of A. Since this domain R has

Noetherian spectrum, R satisfies statement (ii) of Theorem 3.7. Hence condition

(ii) of Theorem 3.7 does not imply condition (i), even when R is a 2-dimensional

Prüfer domain. This leaves the question of whether in general (i) implies (iii) in

Theorem 3.7:

Question 3.10. Does there exist an arithmetical ring R such that R satisfies state-

ment (i) but not statement (iii) of Theorem 3.7?

References

[1] N. Bourbaki, Commutative Algebra, Chapters 1 - 7, Springer-Verlag, 1989.

[2] D. Eisenbud, Commutative algebra with a view toward algebraic geometry, Springer-Verlag, 1994.

[3] M. Fontana, J. Huckaba and I. Papick, Prüfer domains, Marcel Dekker, 1998.

[4] L. Fuchs, W. Heinzer and B. Olberding, Commutative ideal theory without finiteness conditions: primal ideals, to appear.

[5] R. Gilmer, A counterexample to two conjectures in ideal theory, Amer. Math. Monthly, 74 (1967), 195-197.

[6] R. Gilmer and W. Heinzer, Overrings of Prüfer domains II, J. Algebra 7 (1967), 281-302.

[7] R. Gilmer and W. Heinzer, Primary ideals with finitely generated radical in a commutative ring, Manuscripta Math. 78 (1993), 201-221.

[8] H. P. Goeters and B. Olberding, Extension of ideal-theoretic properties of a domain to its quotient field, J. Algebra 237 (2001), 14-31.

[9] W. Heinzer, Quotient overrings of integral domains, Mathematika 17 (1970), 139-148.

[10] W. Heinzer, J. Huckaba and I. Papick, m-Canonical ideals in integral domains, Comm. Algebra 26 (1998), 3021-3043.

[11] W. Heinzer and J. Ohm, Locally Noetherian commutative rings, Trans. Amer. Math. Soc., 158 (1971), 273-284.

[12] J. Iroz and D. Rush, Associated prime ideals in non-Noetherian rings, Can. J. Math. 36 (1984), 344-360.

[13] I. Kaplansky, Commutative Rings (University of Chicago Press, 1974).

[14] W. Krull, Idealtheorie in Ringen ohne Endlichkeitsbedingung, Math. Ann. 101 (1929), 729-744.

[15] W. Krull, Idealtheorie, Ergebnisse d. Math. (Berlin, 1935).

[16] T. Lucas, The radical trace property and primary ideals, J. Algebra 184 (1996), 1093-1112.

[17] M. Nagata, Some remarks on prime divisors, Memoirs College of Science, Kyoto University, Series A, 33 (1960) 297-299.

[18] M. Nagata, Local Rings (Interscience, 1962).

[19] N. Nakano, Idealtheorie in einem speziellen unendlichen algebraischen Zahlkörper, J. Sci. Hiroshima Univ. Ser. A 16 (1953), 425-439.

[20] B. Olberding, Globalizing local properties of Prüfer domains, J. Algebra 205 (1998), 480-504.

[21] R. Pendleton, A characterization of Q-domains, Bull. Amer. Math. Soc. 72 (1966), 499-500.

[22] D. Rush and L. Wallace, Noetherian maximal spectrum and coprimely packed localizations of polynomial rings, Houston J. Math. 28 (2002) 437-448.

Department of Mathematics, Tulane University, New Orleans, Louisiana 70118

  E-mail address: fuchs@tulane.edu

Department of Mathematics, Purdue University, West Lafayette, Indiana 47907

  E-mail address: heinzer@math.purdue.edu

Department of Mathematical Sciences, New Mexico State University, Las Cruces, New Mexico 88003-8001

  E-mail address: olberdin@emmy.nmsu.edu