1 Introduction

Let D be an integral domain with quotient field K.

We recall that a Krull domain is an integral domain D such that \(D=\cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}\), where \(X^1(D)\) denotes the set of all height-one prime ideals of D, \(D_{\mathfrak {p}}\) is a DVR for each \({\mathfrak {p}}\in X^1(D)\) and the intersection \(\cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}\) is locally finite (i.e., each nonzero element of D belongs to only finitely many ideals \({\mathfrak {p}}\in X^1(D)\)).

The notion of weakly-Krull domain was first introduced by Anderson et al. in [1, 5] as follows: an integral domain D is weakly-Krull if \(D=\cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}\) and this intersection is locally finite. Obviously Krull domains are weakly-Krull.

For the convenience of the reader we begin by recalling some definitions and notation necessary for the comprehension of the discussion about weakly-Krull property. Let \({\mathcal {F}}(D)\) (resp., \({\mathcal {F}}_{\mathrm {f.g.}}(D)\)) be the set of all nonzero fractional ideals (resp., nonzero finitely generated fractional ideals) of D. For an ideal \(I\in {\mathcal {F}}(D),\) we set \(I^{-1}:=\{x\in K;\;xI\subseteq D\}.\) The v-operation is defined on \({\mathcal {F}}(D)\) by \(I_v :=(I^{-1})^{-1},\) the t-operation is defined by \(I_t:=\cup \{J_v;\;J\in {\mathcal {F}}_{\mathrm {f.g.}}(D)\) and \(J\subseteq I\},\) and the w-operation is defined by \(I_w:=\{x\in K;\;xJ\subseteq I\) for some \(J\in {\mathcal {F}}_{\mathrm {f.g.}}(D)\) with \(J^{-1}=D\}.\) It is straightforward that \(I \subseteq I_w \subseteq I_t \subseteq I_v.\) An ideal \(I \in {\mathcal {F}}(D)\) is a v-ideal (or divisorial) (resp., t-ideal, w-ideal) if \(I_v=I\) (resp., \(I_t=I\), \(I_w=I\)). A prime ideal that is also a t-ideal is called a t-prime ideal, and an ideal maximal among integral t-ideals is called a t-maximal ideal (and it is a prime ideal). We let t-\(\mathrm {Max}(D)\) denote the set of all t-maximal ideals of D. It is well-known that for any integral domain D, \(D=\cap _{{\mathfrak {p}}\in t\text {-}\mathrm {Max}(D)}D_{\mathfrak {p}}.\) We say that an integral domain D has t-dimension one if each t-prime ideal of D is of height-one (we then write t-\(\dim (D)=1\)). Notice that if t-\(\dim (D)=1\) then t-\(\mathrm {Max}(D)=X^1(D).\) Lastly, an integral domain D is said to be of t-finite character (resp., finite character) if every nonzero element of D belongs to only finitely many t-maximal (resp., maximal) ideals of D.

Representations of domains D as locally finite intersections of a family \(\{D_{\mathfrak {p}}\}_{{\mathfrak {p}}\in {\mathcal {P}}}\) (\({\mathcal {P}} \subseteq \mathrm{Spec}(D)\)) of its localization overrings may correspond to some factorization properties for ideals of D. For instance, Krull domains are exactly the domains for which each principal ideal is a  t-product of prime ideals ([5, Theorem 3.2]) and weakly-Krull domains are exactly the domains in which each principal ideal is a t-product of primary ideals ([5, Theorem 3.1]).

In [5, Lemma 2.1(1)] it is showed that weakly-Krull domains are exactly the one t-dimensional domains with t-finite character.

The polynomials with coefficients in K that take values from D into D itself form a commutative D-algebra denoted by \(\mathrm {Int}(D).\) More precisely \(\mathrm {Int}(D):=\{f\in K[X];\;f(D)\subseteq D\}\) is the ring of integer-valued polynomials over D.

Some Krull-like properties for \(\mathrm{Int}(D)\) have been investigated in past literature. For instance, if D is a Krull domain, then \(\mathrm{Int}(D)\) is a Krull domain if and only if \(\mathrm{Int}(D)\) is Mori if and only if \(\mathrm{Int}(D)=D[X]\) ([9, Corollary 2.7]). This result has been somehow generalized in [16] by showing that for a domain D of w-dimension one, \(\mathrm{Int}(D)\) is Strong Mori if and only if D is Strong Mori and \(\mathrm{Int}(D)=D[X]\).

It is well-known that \({\mathbb {Z}}\) is a (weakly-)Krull domain and \(\mathrm {Int}({\mathbb {Z}})\) is a two-dimensional Prüfer domain (see, for instance, [7]). But \(\mathrm {Int}({\mathbb {Z}})\) is not weakly-Krull since it is of t-dimension two. Thus the weakly-Krull property is not preserved, in general, upon passage from D to \(\mathrm{Int}(D).\) This led us to ask when \(\mathrm{Int}(D)\) is a weakly-Krull domain.

2 Results and applications

By definition, weakly-Krull property is based on a locally finite intersection representation of D. Hence, we start analyzing more generally when \(\mathrm{Int}(D)\) has a representation as locally finite intersection of a family of its localizations \(\{\mathrm{Int}(D)_{\mathfrak {P}}; {\mathfrak {P}}\in {\mathcal {P}}\}\), for \({\mathcal {P}} \subseteq \mathrm{Spec}(\mathrm{Int}(D))\).

Proposition 1.1

Let D be an integral domain with quotient field K. If \(\mathrm{Int}(D)\) is a locally finite intersection of a family of its localizations then D so is.

Proof

Assume that \(\mathrm{Int}(D)= \cap _{{\mathfrak {P}}\in {\mathcal {P}}}\mathrm{Int}(D)_{\mathfrak {P}}\), where \({\mathcal {P}} \subseteq \mathrm{Spec}(\mathrm{Int}(D))\) is the representation of \(\mathrm{Int}(D)\) as given in the statement and set \(\mathcal {P'} := \{{\mathfrak {P}}\cap D; {\mathfrak {P}}\in {\mathcal {P}}\} \subseteq \mathrm{Spec}(D)\). Then \(D = \mathrm{Int}(D)\cap K = \cap _{{\mathfrak {P}}\in {\mathcal {P}}}(\mathrm{Int}(D)_{\mathfrak {P}}\cap K) \supseteq \cap _{{\mathfrak {p}}\in \mathcal {P'}}D_{\mathfrak {p}}\supseteq D\) (in fact, \(\mathrm{Int}(D)_{\mathfrak {P}}\cap K \supseteq D_{{\mathfrak {P}}\cap D}\)). Thus \(D = \cap _{{\mathfrak {p}}\in \mathcal {P'}}D_{\mathfrak {p}}\). Obviously this intersection is locally finite because the intersection \(\cap _{{\mathfrak {P}}\in {\mathcal {P}}}\mathrm{Int}(D)_{\mathfrak {P}}\) is locally finite. \(\square \)

As mentioned in the introduction, weakly-Krull domains are exactly the one t-dimensional domains with t-finite character. By [3, Corollary 3.4] an integral domain D has t-finite character if and only if D[X] has the t-finite character. About the (t-) finite character for \(\mathrm{Int}(D)\) we have the following corollary.

Corollary 1.2

Let D be an integral domain. If \(\mathrm{Int}(D)\) has the t-finite (resp., finite) character then D has it too.

Proof

We can apply Proposition 1.1 where \({\mathcal {P}}\) is the set of t-maximal ideals of \(\mathrm{Int}(D)\). By [20, Corollary 2.2] we have that the set \({\mathcal {P}}'\) of Proposition 1.1 is exactly the set of t-maximal ideals of D and so D has the t-finite character. If \({\mathcal {P}} = \mathrm{Max}(\mathrm{Int}(D))\) (i.e., \(\mathrm{Int}(D)\) has the finite character), then \({\mathcal {P}}' = \mathrm{Max}(D)\), hence D has the finite character too. \(\square \)

Remarks 1.3

 

  1. (a)

    An alternative (direct) proof of Corollary 1.2 can be given, by contraposition, as follows:

    Suppose that D has not the t-finite character. Then there is a nonzero element of D which is contained in infinitely many t-maximal ideals of D. By [20, Corollary 2.2], each t-maximal ideal of D is the contraction of a t-maximal ideal of \(\mathrm{Int}(D)\). Then \(\mathrm{Int}(D)\) has not the t-finite character. Now, if D has not the finite character, then there is a nonzero element \(x \in D\) that is contained in infinitely many maximal ideas of D. Then, the same x is contained in infinitely many maximal ideals of \(\mathrm{Int}(D)\): for any \({\mathfrak {m}}\in \mathrm{Max}(D)\), choose the ideal \({\mathfrak {M}}_0 := \{f \in \mathrm{Int}(D); f(0) \in {\mathfrak {m}}\}\). Thus \(\mathrm{Int}(D)\) has not the finite character.

  2. (b)

    t -finite character

    It is well-known that D[X] has the t-finite character if and only if D has it ([3, Corollary 3.4]). This is not true, in general, for \(\mathrm{Int}(D)\). In fact, if D has the t-finite character \(\mathrm{Int}(D)\) may not have it. For instance \(\mathrm {Int}({\mathbb {Z}})\) does not have the t-finite character (and \({\mathbb {Z}}\) has it). Indeed \(\mathrm {Int}({\mathbb {Z}})\) is Prüfer, so each ideal is a t-ideal and the t-finite character is equivalent to the finite character on maximal ideals. A Prüfer domain with the finite character is Krull-type and, from [18, Theorem 2.30], we would have that \(\mathrm {Int}({\mathbb {Z}}) = {\mathbb {Z}}[X]\) which is not true. This example shows that t-dimension one and t-finite character, properties that combined together characterize weakly-Krull domains, maybe both verified in D but not in \(\mathrm{Int}(D)\).

    Conversely, there are also examples of nontrivial integer-valued polynomial rings \(\mathrm{Int}(D)(\ne D[X]\)) that have the t-finite character. We recall that an integral domain is Mori if it satisfies the ascending chain condition on integral divisorial ideals. In [9] the authors give an example of a one-dimensional Mori domain D such that \(\mathrm{Int}(D)\) is Mori and with \(\mathrm{Int}(D)\ne D[X]\). It is known that Mori domains have the t-finite character ([5, Lemma 2.1(1)]), hence \(\mathrm{Int}(D)\) has it.

  3. (c)

    finite character

    It is known ([2, Proposition 18]) that the polynomial ring D[X] never has the finite character on maximal ideals, unless D is a field. We can see, however, that \(\mathrm{Int}(D)\) may have the finite character on maximal ideals. Indeed, consider a one-dimensional, local, non unibranched Noetherian domain D as given in [7, §  5, page 110]. If \({\mathfrak {m}}\) is the maximal ideal of D, then the prime spectrum of \(\mathrm{Int}(D)\) is made of the primes above \({\mathfrak {m}}\) and the primes above (0). In this case the prime ideals of \(\mathrm{Int}(D)\) above \({\mathfrak {m}}\) are finitely many. The set of nonzero primes above (0) has the finite character (since they correspond to the nonzero primes of K[X] which is Dedekind). So, we have that \(\mathrm{Int}(D)\) has the finite character on maximal ideals.

We now recall that a prime ideal \({\mathfrak {p}}\) of an integral domain D is called int prime if \(\mathrm{Int}(D)\not \subseteq D_{\mathfrak {p}}[X]\) and it is called polynomial prime if \(\mathrm{Int}(D)\subseteq D_{\mathfrak {p}}[X]\). If \({\mathfrak {p}}\) is a polynomial prime we also have that \(\mathrm{Int}(D)_{\mathfrak {p}}= D_{\mathfrak {p}}[X]\) (where \(\mathrm{Int}(D)_{\mathfrak {p}}:= \mathrm{Int}(D)_{D \backslash {\mathfrak {p}}}\)), by [7]. We remark that if \({\mathfrak {p}}\) has infinite residue field (for instance, if it is not maximal), \({\mathfrak {p}}\) is a polynomial prime ([7, Proposition I.3.4]).

The following two lemmas can be found in [18] but, for the sake of completeness, we include their proofs.

Lemma 1.4

Let D be a weakly-Krull domain. Then each int prime ideal of D is of height-one.

Proof

Let \({\mathfrak {m}}\) be an int prime ideal of D. By way of contradiction, assume that \({\mathfrak {m}}\) is of height at least two. Thus \({\mathfrak {m}}\not \in X^1(D)\). By the local finiteness of the intersection \(\cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}\), it follows from [19, Lemma 1.5] that \(D_{\mathfrak {m}}=\cap _{{\mathfrak {p}}\in X^1(D)}(D_{\mathfrak {p}})_{\mathfrak {m}}\). For each \({\mathfrak {p}}\in X^1(D)\), we have \({\mathfrak {m}}\ne {\mathfrak {p}}\) and \(\mathrm {Int}((D_{\mathfrak {p}})_{{\mathfrak {m}}})=(D_{\mathfrak {p}})_{{\mathfrak {m}}}[X]\). Moreover, \((D_{\mathfrak {p}})_{\mathfrak {m}}= \cap _{{\mathfrak {q}}\subseteq {\mathfrak {p}}\cap {\mathfrak {m}}}D_{\mathfrak {q}}\) where the primes \({\mathfrak {q}}\) are polynomial primes (\({\mathfrak {q}}\subsetneq {\mathfrak {m}}\) and it is not maximal). Thus, by [8, Corollaires (3), page 303], we have:

$$\begin{aligned} \mathrm {Int}(D_{\mathfrak {m}})= & {} \bigcap _{{\mathfrak {p}}\in X^1(D)}\mathrm {Int}((D_{\mathfrak {p}})_{{\mathfrak {m}}})= \bigcap _{{\mathfrak {p}}\in X^1(D)}\left( \bigcap _{{\mathfrak {q}}\subseteq {\mathfrak {p}}\cap {\mathfrak {m}}}\mathrm{Int}(D)_{\mathfrak {q}}\right) = \bigcap _{{\mathfrak {p}}\in X^1(D)}\left( \bigcap _{{\mathfrak {q}}\subseteq {\mathfrak {p}}\cap {\mathfrak {m}}}D_{\mathfrak {q}}[X]\right) \\= & {} \bigcap _{{\mathfrak {p}}\in X^1(D)}(D_{\mathfrak {p}})_{\mathfrak {m}}[X]=D_{\mathfrak {m}}[X]. \end{aligned}$$

Therefore \(\mathrm {Int}(D)_{\mathfrak {m}}= \mathrm {Int}(D_{\mathfrak {m}})=D_{\mathfrak {m}}[X]\), which contradicts the hypothesis that \({\mathfrak {m}}\) is an int prime. \(\square \)

Lemma 1.5

Let D be an integral domain. If t-\(\dim (\mathrm {Int}(D))=1\) then D is either a field or of t-dimension 1.

Proof

Suppose that D is not a field and, by way of contradiction, that t-\(\dim (D)>1\). Then, there exist at least two nonzero t-primes of D, \({\mathfrak {q}}\) and \({\mathfrak {p}}\), such that \((0)\subsetneq {\mathfrak {q}}\subsetneq {\mathfrak {p}}\). Obviously, the ideal \({\mathfrak {q}}\) is not maximal and so it is a polynomial prime. Hence \({\mathfrak {q}}D_{\mathfrak {q}}[X] \cap \mathrm{Int}(D)\) is a t-prime of \(\mathrm{Int}(D)\), because it is the contraction of the t-prime \({\mathfrak {q}}D_{\mathfrak {q}}[X]\) (where \(\mathrm{Int}(D)_{\mathfrak {q}}= D_{\mathfrak {q}}[X]\)).

If \({\mathfrak {p}}\) is a polynomial prime then we argue similarly as done for \({\mathfrak {q}}\) and we have the chain of t-primes in \(\mathrm{Int}(D)\):

$$\begin{aligned} (0)\subsetneq {\mathfrak {q}}D_{\mathfrak {q}}[X]\cap \mathrm{Int}(D)\subsetneq {\mathfrak {p}}D_{\mathfrak {p}}[X]\cap \mathrm{Int}(D). \end{aligned}$$

If \({\mathfrak {p}}\) is an int prime, as \({\mathfrak {P}}_0:=\{f\in \mathrm{Int}(D);\;f(0)\in {\mathfrak {p}}\}\) contains \(\mathrm{Int}(D,{\mathfrak {p}}):=\{f\in \mathrm{Int}(D);\;f(D)\subseteq {\mathfrak {p}}\}\), it follows from [10, Propositions 1.2 and 1.4] that \({\mathfrak {P}}_0\) is a t-prime ideal of \(\mathrm{Int}(D)\). Thus, from the inclusions \({\mathfrak {q}}D_{\mathfrak {q}}[X]\cap \mathrm{Int}(D)\subseteq \mathrm{Int}(D,{\mathfrak {q}})\subsetneq \mathrm{Int}(D,{\mathfrak {p}})\subsetneq {\mathfrak {P}}_0,\) we have the chain of t-primes in \(\mathrm{Int}(D)\):

$$\begin{aligned} (0)\subsetneq {\mathfrak {q}}D_{\mathfrak {q}}[X] \cap \mathrm{Int}(D)\subsetneq {\mathfrak {P}}_0. \end{aligned}$$

Therefore, in each case, we have obtained chain of t-primes in \(\mathrm{Int}(D)\) of length 2 and this contradicts the fact that t-\(\dim (\mathrm {Int}(D))=1\). \(\square \)

We recall that D is a UMT-domain if every nonzero prime ideal of D[X] which contracts to zero in D is a t-maximal ideal. UMT domains were introduced and studied by E.G. Houston and M. Zafrullah in [14]. Later, they have been used to characterize when the polynomial rings D[X] are weakly-Krull.

Lemma 1.6

([4, Proposition 4.11]) Let D be an integral domain. Then D[X] is weakly-Krull if and only if D is a weakly-Krull UMT-domain.

In the following Theorem we show that \(\mathrm{Int}(D)\) may be weakly-Krull only in the trivial case.

Theorem 1.7

Let D be an integral domain. The ring \(\mathrm {Int}(D)\) is weakly-Krull if and only if \(\mathrm {Int}(D)=D[X]\) and D is a weakly-Krull UMT-domain.

Proof

Assume that \(\mathrm {Int}(D)\) is weakly-Krull. From Lemma 1.6, it is sufficient to show that \(\mathrm {Int}(D)=D[X]\). If \(\mathrm{Int}(D)\ne D[X]\), from [7, Lemma I.3.6] there exists a maximal ideal \({\mathfrak {m}}\) of D such that \(\mathrm {Int}(D)_{\mathfrak {m}}\ne D_{\mathfrak {m}}[X]\), thus \({\mathfrak {m}}\) is an int prime ideal of D, whence it is a t-ideal ([10, Propositon 1.2]). It is well-known that weakly Krull domains have t-dimension one, then t-\(\dim (\mathrm {Int}(D))=1\), t-\(\dim (D)=1\) (Lemma 1.5) and \({\mathfrak {m}}\) is height-one. Set \({\mathfrak {M}}_0:=\{f\in \mathrm {Int}(D);\;f(0)\in {\mathfrak {m}}\}\). Since \({\mathfrak {M}}_0\) contains \(\mathrm {Int}(D,{\mathfrak {m}})\), it follows from [10, Proposition 1.4] that \({\mathfrak {M}}_0\) is an int prime of \(\mathrm {Int}(D)\) and it is of height-one (Lemma 1.4).

Now, consider the prime ideal \({\mathfrak {Q}}:=\{f\in \mathrm {Int}(D);\;f(0)=0\}\). It is easily seen that \((0)\subsetneq {\mathfrak {Q}}\subsetneq {\mathfrak {M}}_0\). Hence, \({\mathfrak {M}}_0\) has height at least two, which is a contradiction. Thus D has not int prime ideals, so \(\mathrm {Int}(D)=D[X]\).

The converse follows from Lemma 1.6. \(\square \)

We investigate when \(\mathrm{Int}(D)=D[X]\) in the case D is weakly-Krull.

Proposition 1.8

Let D be a weakly-Krull domain. Then \(\mathrm{Int}(D)= D[X]\) if and only if \(\mathrm{Int}(D_{\mathfrak {p}}) = D_{\mathfrak {p}}[X]\), for each \({\mathfrak {p}}\in X^1(D)\).

Proof

We observe that since the intersection \(\cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}= D\) is locally finite, \(\mathrm{Int}(D_{\mathfrak {p}})=\mathrm{Int}(D)_{\mathfrak {p}}\) for each \({\mathfrak {p}}\in X^1(D)\) (cf. [20, Proposition 2.3]). Then if \(\mathrm{Int}(D)=D[X]\), obviously \(\mathrm{Int}(D)_{\mathfrak {p}}=D_{\mathfrak {p}}[X]\) and the thesis follows from the equality \(\mathrm{Int}(D)= \cap _{{\mathfrak {p}}\in X^1(D)}\mathrm{Int}(D_{\mathfrak {p}})\).

On the contrary, if \(\mathrm{Int}(D_{\mathfrak {p}}) = D_{\mathfrak {p}}[X]\) for each \({\mathfrak {p}}\in X^1(D)\), then \(\mathrm{Int}(D)=\cap _{{\mathfrak {p}}\in X^1(D)}\mathrm{Int}(D_{\mathfrak {p}}) = \cap _{{\mathfrak {p}}\in X^1(D)}D_{\mathfrak {p}}[X]=D[X]\). \(\square \)

Thus to have \(\mathrm{Int}(D)=D[X]\) for a weakly-Krull domain D, it is sufficient to know when a one-dimensional local domain D is such that \(\mathrm{Int}(D)=D[X]\). This happens, for instance, when the residue field of D over the maximal ideal is infinite (cf. [7, Proposition I.3.4]) or when D is a valuation domain with non-principal maximal ideal (cf. [7, Proposition I.3.16]).

It is well-known that a Noetherian domain D is a UMT-domain if and only if it has t-dimension one (cf. [14, Theorem 3.7]). Thus, from Theorem 1.7 and [4, Corollary 4.12] we get a similar result to [9, Corollary 2.7], which states that \(\mathrm{Int}(D)\) is a Krull domain if and only if D is Krull and \(\mathrm{Int}(D)= D[X]\).

Corollary 1.9

Let D be a Noetherian domain. Then \(\mathrm{Int}(D)\) is weakly-Krull if and only if D is weakly-Krull and \(\mathrm{Int}(D)=D[X]\).

Corollary 1.10

Let D be a Noetherian domain. Then \(\mathrm{Int}(D)\) is weakly-Krull if and only if D is weakly-Krull and the residue fields of D over the height-one primes are infinite.

Proof

By [7, Corollary I.3.15], if D is a one-dimensional Noetherian domain, then \(\mathrm{Int}(D)=D[X]\) if and only if its residue field is infinite. The thesis follows from Proposition 1.8 and Corollary 1.9. \(\square \)

Krull domains are exactly the domains which are locally finite intersection of localizations that are DVR (or, equivalently PID). Like weakly-Krull domains, a natural generalization of Krull domains are the infra-Krull domains which are locally finite intersections of one-dimensional Noetherian domains that are localizations. Infra-Krull domains were introduced by M. Martin and M. Zafrullah in [15] and it is easily seen that they are weakly-Krull. Conversely, weakly-Krull domains are not necessarily infra-Krull: for instance, consider \(D = {\mathbb {Q}}+X{\mathbb {R}}[[X]]\). In this case D is a local, Mori non-Noetherian domain of dimension one ([6, Corollary 3.5]). Since Mori domains have the t-finite character and D is one-dimensional, then D is weakly-Krull. But D is not infra-Krull because it is not Noetherian.

By definition, an integrally closed infra-Krull domain is Krull (since a one-dimensional, local, Noetherian, integrally closed domain is DVR). Then, any non integrally closed, one-dimensional Noetherian domain is infra-Krull and not Krull.

We recall that an integral domain D is Strong Mori if it satisfies the ascending chain condition on integral w-ideals (see [12]). Note that the class of Strong Mori domains includes Noetherian domains and Krull domains and it is well-known that an infra-Krull domain is exactly a Strong Mori domain of t-dimension one.

Corollary 1.11

Let D be an integral domain. Then \(\mathrm{Int}(D)\) is infra-Krull if and only if D is infra-Krull and \(\mathrm{Int}(D)=D[X]\).

Proof

Assume that \(\mathrm{Int}(D)\) is infra-Krull. Then, by Theorem 1.7, \(\mathrm{Int}(D)=D[X]\) because any infra-Krull domain is weakly-Krull. Since infra-Krull is Strong Mori of t-dimension one, it follows from [11, Theorem 2.2] and Lemma 1.5 that D is a Strong Mori domain of t-dimension one, i.e., D is infra-Krull.

The converse follows from [17, Theorem 4.3]. \(\square \)

We remark that the notions “infra-Krull” and “weakly-Krull” for \(\mathrm{Int}(D)\) coincide when D is an infra-Krull domain.

Corollary 1.12

Let D be an infra-Krull domain. Then the following statements are equivalent.

  1. (i)

    \(\mathrm {Int}(D)\) is an infra-Krull domain;

  2. (ii)

    \(\mathrm {Int}(D)\) is a weakly-Krull domain;

  3. (iii)

    \(\mathrm{Int}(D)=D[X]\).

Another interesting class of weakly-Krull domains are the generalized Krull domains, described by R. Gilmer in [13, Section 43]. These are weakly-Krull domains such that their localizatons \(D_{\mathfrak {p}}\) are valuation domains for each \({\mathfrak {p}}\in X^1(D)\), or equivalently, weakly-Krull PvMDs (recall that a PvMD is a domain whose localizations at t-maximal ideals are valuation domains).

The following result that characterizes when \(\mathrm{Int}(D)\) is generalized Krull was obtained by the current authors in [18] in the context of Krull-type domains. Here we get it as a corollary of Theorem 1.7.

Corollary 1.13

Let D be an integral domain. Then \(\mathrm{Int}(D)\) is generalized Krull if and only if D is generalized Krull and \(\mathrm{Int}(D)=D[X]\).

Proof

Suppose that \(\mathrm{Int}(D)\) is a generalized Krull domain. Then, by Theorem 1.7, \(\mathrm {Int}(D)=D[X]\) and hence D[X] is generalized Krull. Thus, as cited above, D[X] is a weakly-Krull PvMD and therefore it follows from Lemma 1.6 and [21, Corollary 4] that D is a weakly-Krull PvMD, i.e., D is generalized Krull. The converse follows from [13, Theorem 43.11(3)]. \(\square \)

In the following Corollary we give a description of when an infra-Krull or generalized Krull domain D is such that \(\mathrm{Int}(D)=D[X]\).

Corollary 1.14

Let D be an integral domain.

  1. (a)

    If D is infra-Krull, then \(\mathrm {Int}(D)=D[X]\) if and only if each height-one prime ideal of D has infinite residue field.

  2. (b)

    If D is generalized-Krull, then \(\mathrm {Int}(D)=D[X]\) if and only if \(D_{\mathfrak {p}}\) has infinite residue field or non-principal maximal ideal, for each \({\mathfrak {p}}\in X^1(D)\).