Finiteness and Skolem Closure of Ideals for Non-Unibranched Domains

Abstract

A one-dimensional, Noetherian, local domain D with maximal ideal 𝔪 and finite residue field was known to be an almost strong Skolem ring if analytically irreducible. It was unknown whether this condition is necessary. We show that it is at least necessary for D to be unibranched. After introducing a general notion of equalizing ideal, we show that, for k large enough, the ideals of the form 𝔐 _{k, a}  = {f ∈ Int(D) | f(a) ∈ 𝔪 ^k }, for a ∈ D, are distinct. This allows to show that the maximal ideals 𝔐 _a  = {f ∈ Int(D) | f(a) ∈ 𝔪}, although not necessarily distinct, are never finitely generated.

Key Words:

• Analytically irreducible domain
• Integer-valued polynomials
• Skolem properties
• Unibranched domain

2010 Mathematics Subject Classification:

• 13F20

1. INTRODUCTION

The classical ring of integer-valued polynomials is the ring Int(ℤ) = {f ∈ ℚ[X] | f(ℤ) ⊆ ℤ} of polynomials with rational coefficients taking integer values on the integers. This ring has many interesting properties. First of all, it is probably the simplest and most natural example of a non-Noetherian domain. Among other properties, Skolem [6] pointed out that, given finitely many integer-valued polynomials g ₁,…, g _k such that g ₁(n),…, g _k (n) are relatively prime for each integer n, the ideal generated by these polynomials in Int(ℤ) is the whole ring Int(ℤ) (while a similar property does not hold for polynomials with coefficients in ℤ).

More generally, for a domain D with quotient field K, one can study the properties of the ring of integer-valued polynomials

The reader can refer to [1] for a survey. We aim in this paper to answer some questions on the Noetherian and Skolem properties that remained open from the time of publication of this book. Usually, the main focus is the case where D is Noetherian. In this case, Int(D)_𝔭 = Int(D _𝔭) for every prime 𝔭 of D [1, Theorem I.2.3]. For most properties, one can thus assume D to be a local ring with maximal ideal 𝔪 (however, the Skolem property does not localize, but we come back, in more details, to this question later on in this introduction). One can also assume that the residue field D/𝔪 is finite, lest Int(D) = D[X] [1, Corollary I.3.7]. Finally, one can assume that the Krull dimension of D is one; otherwise, one has the containment D[X] ⊆ Int(D) ⊆ D′[X], where D′ denotes the integral closure of D [1, Corollary IV.4.10], a case that is not entirely trivial (unless D is integrally closed) but, in some sense, less interesting (for instance, it may occur in this case that Int(D) is Noetherian).

The first question we address deals with the prime spectrum of Int(D) and the finiteness of some maximal ideals. In the case of a one-dimensional local domain D, there are two types of prime ideals: those above (0) and those above the maximal ideal 𝔪 of D. Let us focus on the last ones (the first ones are easily described, whatever the domain D [1, Corollary V.1.2]). Recall that, given a maximal ideal 𝔪 of any domain D, we have for each a ∈ D a prime ideal of Int(D) above 𝔪 of the form

These primes are clearly maximal. In the case of a one-dimensional, Noetherian, local domain D with maximal ideal 𝔪, the polynomials with coefficients in K are continuous functions in the 𝔪-adic topology [1, Proposition III.2.1]. Considering the 𝔪-adic completion of D, we thus have also, for each , maximal ideals of the form , where is the topological closure of 𝔪 in . If moreover the residue field D/𝔪 is finite, all the prime ideals of Int(D) above 𝔪 are of the form 𝔐_α [1, Proposition V.2.2]. Yet these ideals are not necessarily distinct and it may happen that, in fact, they are all of the form 𝔐 _a for some a ∈ D. More precisely, as summarized in [1, Chapter V], if D is a discrete valuation domain, or more generally, if D is analytically irreducible, that is, is a domain, there is a one-to-one correspondence between and the prime ideals of Int(D) above 𝔪 (to corresponds 𝔐_α). This nice property is linked with the Stone–Weierstrass theorem (if D is analytically irreducible the ring of integer-valued polynomials is dense in the ring of continuous functions from to ). Under the weaker hypothesis that D is unibranched, that is, the integral closure D′ of D is local, this is no longer the case, but it remains at least true that the ideals 𝔐 _a , for a ∈ D, are all distinct (one must consider the 𝔪′-adic completion of D, where 𝔪′ is the maximal ideal of D′, rather than the 𝔪-adic completion, to obtain a one-to-one correspondence). Finally, in the non-unibranched case, there are only finitely many prime ideals of Int(D) above 𝔪, all of them of the form 𝔐 _a (a ∈ D), these primes being in one-to-one correspondence with the residue classes of D modulo some nonzero ideal 𝔮_𝔪 of D, called the equalizing ideal of 𝔪 in Int(D).

In case the ideals 𝔐 _a are all distinct, there is a very easy argument to show that these ideals are not finitely generated: assume by way of contradiction that 𝔐 _a  = (f ₁,…, f _n ). Thus f _i (a) ∈ 𝔪 for all i and by continuity, f _i (b) ∈ 𝔪 for b close enough to a. Hence f ₁,…, f _n are in 𝔐 _b , and one would have 𝔐 _a  ⊆ 𝔐 _b in contradiction with the fact that the ideals 𝔐 _a are maximal and all distinct. This argument cannot be used if the ideals 𝔐 _a are not distinct, that is, if D is not unibranched. However, we show in this paper that, unibranched or not unibranched, the ideals 𝔐 _a are never finitely generated.

The second question deals with the Skolem properties. If 𝔄 is an ideal of Int(D), then for each a ∈ D, the set 𝔄(a) = {f(a) | f ∈ 𝔄} is clearly an ideal of D called the ideal of values of 𝔄 at a. The property pointed out by Skolem in the case of Int(ℤ) (the Skolem property) is that, given a finitely generated ideal 𝔄 of Int(D), if 𝔄(a) = D for each a ∈ D, then 𝔄 = Int(D). In fact, Int(ℤ) satisfies even a stronger property (the strong Skolem property): given two finitely generated ideals 𝔄 and 𝔅, if 𝔄(a) = 𝔅(a) for each a ∈ D, then 𝔄 = 𝔅. Now the Skolem property (let alone the strong Skolem property) cannot hold if D is a local ring: the polynomial f = 1 + tX with t ∈ 𝔪 is clearly such that f(a) is a unit for each a ∈ D, whereas the ideal generated by f is not the whole ring Int(D) (indeed, it fails to contain any nonzero constant). In fact, it turns out that, to study the Skolem properties, one can split the ideals of Int(D) in two categories: the ideals which contain nonzero constants, called the unitary ideals, and the ideals 𝔄 such that 𝔄 ∩ D = (0) (in the case of a one-dimensional local ring with maximal ideal 𝔪, the prime ideals above 𝔪 are thus the unitary prime ideals). Finally, it is convenient to phrase the Skolem properties in terms of Skolem closure. The reader can find a survey in [1, Chapter VII]), and we summarize the needed definitions as follows.

Definition 1.1

Let 𝔄 be an ideal of Int(D). We call

the Skolem closure of 𝔄. We further say that 𝔄 is Skolem closed, if 𝔄 = 𝔄^☆.

The Skolem closure of 𝔄 is the largest ideal 𝔅 of Int(D) such that 𝔄 and 𝔅 have the same ideals of values, that is, 𝔄(a) = 𝔅(a) for each a ∈ D.

Definition 1.2

1.	We say that D is a Skolem ring (resp. an almost Skolem ring) or that Int(D) satisfies the Skolem property (resp. the almost Skolem property), if for each finitely generated ideal (resp. for each finitely generated unitary ideal) 𝔄 of Int(D), 𝔄☆ = Int(D) implies 𝔄 = Int(D).
2.	We say that D is a strong Skolem ring (resp. an almost strong Skolem ring) or that Int(D) satisfies the strong Skolem property (resp. the almost strong Skolem property), if each finitely generated ideal (resp. finitely generated unitary ideal) 𝔄 of Int(D) is Skolem closed.

The “almost” properties are thus the Skolem properties restricted to the (finitely generated) unitary ideals. As for the non-unitary ideals, they are linked to a natural divisibility property [4]: D is said to be a d-ring if each integer-valued rational function on D is in fact an integer-valued polynomial (equivalently for each nonconstant polynomial f with coefficients in D, there exists a ∈ D such that f(a) is not a unit in D, equivalently also, for each finitely generated non-unitary ideal 𝔄, there exists a ∈ D with 𝔄(a) ≠ D). Finally, it turns out that a ring is a (strong) Skolem ring if and only if it is an almost (strong) Skolem d-ring [1, Proposition VII.2.14].

The almost (strong) Skolem property does localize and hence, to study this property in the Noetherian case, one may assume, as above, that D is a one-dimensional, Noetherian, local domain with finite residue field. Under these hypotheses, D is always an almost Skolem ring [1, Lemma VII.4.2], and this is because, as said above, all the prime ideals of Int(D) above 𝔪 are of the form 𝔐_α for some . The almost strong Skolem property holds in case D is analytically irreducible [1, Corollary VII.3.9] and this, as for the first question considered here, thanks to the Stone–Weierstrass theorem. Yet, it was not known whether it is necessary that D is analytically irreducible to be an almost strong Skolem ring [1, Remark VII.3.10]. Our second result shows that D must at least be unibranched. (Recall that, under our hypotheses, D is analytically irreducible if and only if D is unibranched and the integral closure D′ is finitely generated as a D-module.) This result had been announced in a survey on Skolem properties [2] in 1999 but has not been proved since then.

To reach our goal, we generalize in the next section the notion of equalizing ideal, considering the equalizing ideal 𝔮 _I of an arbitrary ideal I of D. In the particular case where I = 𝔪 ^k (for some integer k), we consider, for each a ∈ D the ideals of Int(D) of the form

and then have 𝔐 _{k, a} = 𝔐 _{k, b} if and only if a ≡ bmod𝔮_{𝔪 ^k} . We show that for k large enough, 𝔮_{𝔪 ^k} is trivial, that is, 𝔮_{𝔪 ^k}  = (0); equivalently, we thus show that for k large enough, the ideals 𝔐 _{k, a} are distinct. This holds whether D is unibranched or not, but simply, if D is unibranched, it holds for each k ≥ 1, whereas if D is not unibranched, one must take k > 1. This can be considered as the main result of this paper, as we can then derive easily, in the next and final section, the two results we were aiming at.

For instance, it immediately follows that the ideals 𝔐 _{k, a} are not finitely generated with the same easy argument as for the ideals 𝔐 _a (as recalled above, in case D is unibranched). We could derive of course that Int(D) is not Noetherian (yet, this was already known for every one-dimensional, Noetherian domain D, unless Int(D) is trivial, that is, Int(D) = D[X] [1, Corollary VI.2.6]). More interestingly, we can prove also that the quotient ring Int(D)/𝔪 ^k Int(D) is not Noetherian (using the fact that the ideals 𝔐 _{k, a} /𝔪 ^k Int(D) are not finitely generated) and finally can conclude that the ideals of the form 𝔐 _a are never finitely generated.

It is now also easy to prove that an almost strong Skolem ring must be unibranched. Indeed, in case there are only finitely many ideals of the form 𝔐 _a (that is, in case D is not unibranched), it is easy to produce a finitely generated ideal 𝔄 containing 𝔪 and contained in one and only one ideal of the form 𝔐 _a . It follows that the ideals of values of 𝔄 and 𝔐 _a are the same (that is, 𝔄(x) = 𝔐 _a (x) for each x ∈ D). As 𝔐 _a is not finitely generated, there is obviously a finitely generated ideal 𝔅 such that 𝔄 ⊊︀ 𝔅 ⊊︀ 𝔐 _a . Clearly, 𝔄 and 𝔅 must then have the same ideal of values, they are both finitely generated, they both contain 𝔪, and thus are both unitary, but 𝔄 ≠ 𝔅. We can therefore conclude that D is not an almost strong Skolem ring.

2. GENERALIZED EQUALIZING IDEALS

In this section, we generalize the notion of equalizing ideal. First, for an arbitrary ideal I of a domain D, we set

Clearly, ℑ _{I, a} is the kernel of the map sending a polynomial f ∈ Int(D) on the class of f(a) in the quotient ring D/I, and thus is an ideal of Int(D) above I (that is, ℑ _{I, a} ∩ D = I). The quotient Int(D)/ℑ _{I, a} is thus isomorphic to D/I. Hence ℑ _{I, a} is a prime ideal (resp. a maximal ideal) of Int(D) if and only if I is a prime ideal (resp. a maximal ideal) of D (as already observed, in a more general setting, in [1, Lemma V.1.3], for polynomials that are integer-valued on a subset of the domain D). Obviously, if I = 𝔪 is a maximal ideal, the ideals ℑ_{𝔪, a} are nothing else than the classical ideals 𝔐 _a described above. In the special case where I = 𝔪 ^k , we thus simply denote ℑ_{𝔪 ^k , a} by 𝔐 _{k, a} .

Raising the question whether the ideals of the form ℑ _{I, a} are distinct or not, we introduce the equalizing ideal of I similarly to the equalizing ideal 𝔮_𝔪 of 𝔪, introduced by Cahen and Chabert in [1] as the set of elements a ∈ D such that 𝔐 _a = 𝔐₀. We rephrase this definition and prove the results of [1, Proposition V.3.5], in this more general setting, as follows.

Proposition 2.1

Let I be an ideal of a domain D, and consider the subset 𝔮 _I of D defined by

Then the following statements hold:

i.	a − b ∈ 𝔮 I if and only if, for all f ∈ Int(D), f(a) − f(b) ∈ I;
ii.	𝔮 I is an ideal of D;
iii.	𝔮 I  ⊆ I.

Proof

(i) By definition, a − b ∈ 𝔮 _I if and only if, for all f ∈ Int(D), we have f(a − b) − f(0) ∈ I. Considering the isomorphism of Int(D) onto itself, sending f on the polynomial g such that g(X) = f(X − b), we see that f(a − b) − f(0) ∈ I (for all f ∈ Int(D)) if and only g(a) − g(b) ∈ I (for all g ∈ Int(D)).

(ii) Let a, b ∈ 𝔮 _I . By definition, for all f ∈ Int(D) we have f(a) − f(0) ∈ I and f(b) − f(0) ∈ I, and thus f(a) − f(b) ∈ I. By (i), it follows that a − b ∈ 𝔮 _I . Next, let a ∈ 𝔮 _I and λ ∈D. Setting h(X) = f(λX), then h ∈ Int(D), and thus, f(λa) − f(0) = h(a) − h(0) ∈ I. Therefore, λa ∈ 𝔮 _I . We can conclude that 𝔮 _I is an ideal.

(iii) As f = X belongs to Int(D), a ∈ 𝔮 _I implies a = f(a) − f(0) ∈ I. Thus 𝔮 _I  ⊆ I.

With these notations, we then set the following definition.

Definition 2.2

The ideal 𝔮 _I is called the equalizing ideal of I in Int(D).

The equalizing ideal (as its name suggests) allows to determine whether two ideals of the form ℑ _{I, a} are distinct or even whether they are comparable. Yet, as the ideals ℑ _{I, a} are not maximal in general (unless I is maximal), the containment ℑ _{I, a}  ⊆ ℑ _{I, b} does not imply a priori the equality ℑ _{I, a} = ℑ _{I, b} , contrary to the case of the ideals 𝔐 _a . Nevertheless, this implication turns out to be true.

Corollary 2.3

Let I be an ideal of the domain D and 𝔮 _I be the corresponding equalizing ideal. Then the following assertions are equivalent:

i.	ℑ I, a = ℑ I, b ;
ii.	ℑ I, a  ⊆ ℑ I, b ;
iii.	a ≡ bmod𝔮 I .

Proof

That (i) implies (ii) is trivial.

Deny (iii). Then a − b ∉ 𝔮 _I . By definition, there is a polynomial f ∈ Int(D) such that f(a − b) − f(0) ∉ I. Set g(X) = f(a − X) − f(0). Then g ∈ Int(D), g(a) = 0, and g(b) = f(a − b) − f(0). Thus g(a) ∈ I, that is, g ∈ ℑ _{I, a} , while g(b) ∉ I, that is, g ∉ ℑ _{I, b} . Hence ℑ _{I, a}  ⊈ ℑ _{I, b} . Therefore, (ii) implies (iii).

Suppose (iii), that is, a − b ∈ 𝔮 _I . It follows from the first assertion of Proposition 2.1, that, for all f ∈ Int(D), we have f(a) − f(b) ∈ I. Thus f(a) ∈ I if and only if f(b) ∈ I, that is ℑ _{I, a} = ℑ _{I, b} .

Another way to phrase Corollary 2.3 is to say there is a one-to-one correspondence between the ideals of the form ℑ _{I, a} and the residue classes of D modulo 𝔮 _I . In particular, the ideals ℑ _{I, a} are all distinct if and only if the equalizing ideal is trivial, that is, 𝔮 _I  = (0). Moreover, it is known that, for every nonzero ideal I of a one-dimensional, Noetherian, local domain D with finite residue field, the residue ring D/I is finite. In this case, when the equalizing ideal is not trivial, there are but finitely many ideals of the type ℑ _{I, a} .

We next easily show that equalizing ideals preserve inclusion.

Proposition 2.4

Let I, J be two ideals of the domain D such that I ⊆ J. Then 𝔮 _I  ⊆ 𝔮 _J .

Proof

Assume a ∉ 𝔮 _J : there exists f ∈ Int(D) such that f(a) − f(0) ∉ J. A fortiori f(a) − f(0) ∉ I, and hence a ∉ 𝔮 _I .

To ensure that 𝔮 _I is trivial for some ideal I it is then enough to find a larger ideal J with a trivial equalizing ideal 𝔮 _J . We shall use this argument below, but first consider some easy examples with the following remark.

Remark 2.5

Given a D-algebra B such that X ∈ B, B ⊆ K[X], and the property that, for each f ∈ B and each a ∈ D, f(a − X), f(X − a), f(aX) ∈ B, we could more generally consider the ideals of the form ℑ _{I, a} , for every ideal I of D, and the corresponding equalizing ideal 𝔮 _I , with the same definitions and the same properties as for Int(D).

For instance, for B = D[X] (the classical ring of polynomials with coefficients in D), we know that, for f ∈ D[X] and a, b ∈ D, a − b divides f(a) − f(b) in D. It follows that, for every ideal I, a ∈ I implies f(a) − f(0) ∈ I, that is, a ∈ 𝔮 _I . As 𝔮 _I is always contained in I, we then have 𝔮 _I  = I. In particular the equalizing ideal 𝔮 _I of every nonzero ideal I is not trivial and the ideals of the form ℑ _{I, a} are not all distinct (however, that does not necessarily mean there are finitely many such ideals, since the residue ring D/I is not finite in general). The same may happen for Int(D) (let alone because it may happen that Int(D) = D[X]).

On the opposite, for B = Int(D), where D is a unibranched, one-dimensional, Noetherian, local domain with maximal ideal 𝔪 and finite residue field, as the maximal ideals of the form 𝔐 _a are all distinct in Int(D) (as recalled above, [1, Remark V.3.2]), it follows that 𝔮_𝔪 = (0) and hence, a fortiori, that 𝔮 _I  = (0) for every nonzero ideal I of D.

Back to the case of a one-dimensional, Noetherian, local domain D with maximal ideal 𝔪 and finite residue field, unibranched or not unibranched, we want to show there is a power 𝔪 ^k of 𝔪 such that the corresponding equalizing ideal is trivial for Int(D).

By the Krull–Akizuki theorem [5], the integral closure D′ of D is a semilocal Dedekind domain. We denote by the maximal ideals of D′ and by ν₁,…, ν _r the corresponding valuations (thus D is unibranched in case r = 1). We can choose a ∈ D′, which is contained in exactly one of the maximal ideals of D′. For simplicity, we let a be such that and (if D is unibranched, a can thus be any non-unit of D′). The ring R = D[a] is such that D ⊆ R ⊆ D′ and is a finitely generated D-module, since a is integral over D. Hence the conductor (D: R) is not trivial and there exists a nonzero element d ∈ D such that dR ⊆ D. With these notations, the following technical lemma exhibits a nonzero ideal I with a trivial equalizing ideal 𝔮 _I .

Lemma 2.6

Let D be a one-dimensional, Noetherian, local domain with finite residue field. With the previous notations let k = ν₁(d) +1 and set . Then 𝔮 _I  = (0).

Proof

The intersection is a maximal ideal of R = D[a]. The localization R _η is one-dimensional, Noetherian, obviously local, and its residue field is finite. Moreover, by the choice of a, is the only maximal ideal of D′ lying over η, and hence, R _η is unibranched (its integral closure is ). As recalled already in several instances, it follows that, for two elements r ≠ s ∈ R _η there exists a polynomial f ∈ Int(R _η) such that f(r) ∈ ηR _η and f(s) ∉ ηR _η. The same holds a fortiori for r, s ∈ D. Moreover, as R is Noetherian, we have Int(R _η) = Int(R)_η [1, Theorem I.2.3], and there exists some element b ∈ R∖η such that g = bf ∈ Int(R). Thus g(r) ∈ η but g(s) ∉ η, that is, ν₁(g(r)) ≥1 and ν₁(g(s)) = 0. Now, as d ∈ (D: R), the polynomial h = dg belongs to Int(D) and ν₁(h(r)) = ν₁(d) + ν₁(g(r)) ≥ k, while ν₁(h(s)) = ν₁(d) = k − 1. In other words, , while . As both h(r) and h(s) are in D and , we thus have h(r) ∈ I but h(s) ∉ I; that is, h ∈ ℑ _{I, r} but h ∉ ℑ _{I, s} . Since r and s are arbitrary (but distinct) elements of D, we can conclude that 𝔮 _I  = (0).

Still with the same notations, the ideal clearly contains 𝔪 ^k . By Proposition 2.4, it follows that the equalizing ideal of 𝔪 ^k is trivial. Thus we can conclude this section with the following theorem.

Theorem 2.7

Let D be a one-dimensional, local, Noetherian domain with maximal ideal 𝔪 and finite residue field. Then there exists a power 𝔪 ^k of 𝔪, such that, for a ≠ b ∈ D, 𝔐 _{k, a}  ≠ 𝔐 _{k, b} .

3. NONFINITENESS AND SKOLEM CLOSURE

The last result of the previous section provides the tools to answer both questions raised in the introduction: for a one-dimensional, Noetherian, local domain D with maximal ideal 𝔪 and finite residue field, 1) the maximal ideals of Int(D) of the form 𝔐 _a for a ∈ D are never finitely generated, and 2) if D is an almost strong Skolem ring, then D is unibranched.

For the nonfiniteness of the ideals 𝔐 _a , we could first easily establish (as observed in the introduction) that the ideals 𝔐 _{k, a} for a ∈ D are not finitely generated. Now each 𝔐 _{k, a} clearly contains the ideal 𝔪 ^k Int(D), of Int(D) and we prove the following (somewhat stronger) lemma.

Lemma 3.1

Let D be a one-dimensional, Noetherian, local domain with finite residue field. Then, for some power 𝔪 ^k of the maximal ideal 𝔪, the ideals 𝔐 _{k, a} /𝔪 ^k Int(D) of the quotient ring Int(D)/𝔪 ^k Int(D) are not finitely generated.

Proof

From Theorem 2.7, there is a power 𝔪 ^k of 𝔪 such that 𝔮_{𝔪 ^k}  = (0), that is, the ideals of the form 𝔐 _{k, a} are all distinct. By way of contradiction, suppose that 𝔐 _{k, a} /𝔪 ^k Int(D) is finitely generated, say by the classes in the quotient ring Int(D)/𝔪 ^k Int(D) of polynomials f ₁,…, f _n ∈ 𝔐 _{k, a} . Now let g ∈ 𝔐 _{k, a} . Then, , with each h _i an integer-valued polynomial and h ∈ 𝔪 ^k Int(D). As f _i ∈ 𝔐 _{k, a} , we have, by definition, f _i (a) ∈ 𝔪 ^k . Also, as the polynomials f _i are continuous, there exists a neighborhood U of a such that f _i (b) ∈ 𝔪 ^k for all b ∈ U, that is, f _i ∈ 𝔐 _{k, b} , and thus, a fortiori, h _i f _i ∈ 𝔐 _{k, b} . Finally, as 𝔐 _{k, b} contains 𝔪 ^k Int(D), we have h ∈ 𝔐 _{k, b} . Thus g ∈ 𝔐 _{k, b} . Therefore 𝔐 _{k, a}  ⊆ 𝔐 _{k, b} , and hence, from Corollary 2.3, 𝔐 _{k, a} = 𝔐 _{k, b} . We obtain a contradiction for a ≠ b.

It follows immediately that Int(D)/𝔪 ^k Int(D) is not Noetherian (and a fortiori, neither is Int(D)). This allows to conclude with the following theorem.

Theorem 3.2

Let D be a one-dimensional, Noetherian, local domain with finite residue field. Then the maximal ideals 𝔐 _a of Int(D) are not finitely generated.

Proof

From Lemma 3.1, the ideals 𝔐 _{k, a} /𝔪 ^k Int(D) of Int(D)/𝔪 ^k Int(D) are not finitely generated for some power 𝔪 ^k of 𝔪, and hence, this quotient ring is not Noetherian. If D is unibranched, we can choose k = 1, and the theorem is proved. Thus we can assume that D is not unibranched and hence that all the prime ideals of Int(D) above 𝔪 are of the form 𝔐 _a for a ∈ D. As a prime containing 𝔪 ^k must contain 𝔪, and hence, be above 𝔪, it follows that all the prime ideals of the quotient ring Int(D)/𝔪 ^k Int(D) are of the form 𝔐 _a /𝔪 ^k Int(D). From a theorem by Cohen (see [3]), one of the primes 𝔐 _a /𝔪 ^k Int(D) of the non-Noetherian ring Int(D)/𝔪 ^k Int(D) is not finitely generated. A fortiori, the corresponding maximal ideal 𝔐 _a of Int(D) is not finitely generated. From the isomorphism , we can conclude that, in fact, none of the maximal ideals 𝔐 _a for a ∈ D is finitely generated.

Addressing now the second question, we close this paper with the proof that, under our running hypotheses (of a one-dimensional, Noetherian, local domain D with finite residue field), it is necessary that D be unibranched for Int(D) to satisfy the almost strong Skolem property.

Theorem 3.3

Let D be a one-dimensional, Noetherian, local domain with maximal ideal 𝔪 and finite residue field D/𝔪. If D is an almost strong Skolem ring, then D is unibranched.

Proof

By way of contradiction, suppose that D is not unibranched. As already mentioned above (see also [1, Proposition V.3.10]), there are finitely many maximal ideals above 𝔪, all of the form 𝔐 _a . Let us denote these ideals by 𝔐₁,…, 𝔐 _r . We can choose f ∈ 𝔐₁ with . For a ∈ D, we then have f(a) ∈ 𝔪 if and only if 𝔐 _a = 𝔐₁.

Let 𝔄 = (𝔪, f) be the ideal of Int(D) generated by 𝔪 and f. We then have 𝔪 ⊆ 𝔄(a) for all a ∈ D. Thus, for a ∈ D, either 𝔐 _a = 𝔐₁, and then f(a) ∈ 𝔪, and hence, 𝔄(a) = 𝔪, or 𝔐 _a  ≠ 𝔐₁, then f(a) ∉ 𝔪, and hence 𝔄(a) = D. In other words, 𝔄(a) = 𝔐₁(a), for all a ∈ D.

By Theorem 3.2, we know that 𝔐₁ is not finitely generated. We can thus choose g ∈ 𝔐₁∖𝔄. Let now 𝔅 = (𝔪, f, g). For all a ∈ D, we clearly have 𝔄(a) ⊆ 𝔅(a) ⊆ 𝔐₁(a), and hence, 𝔄(a) = 𝔅(a). But 𝔄 ≠ 𝔅, whereas both ideals are finitely generated and unitary. We can conclude that, if D is not unibranched, then Int(D) does not satisfy the almost strong Skolem property.

Recalling that the condition that D is analytically irreducible is sufficient for Int(D) to satisfy the almost strong Skolem property ([1, Theorem VII.3.7]), the question to find a condition that is both necessary and sufficient remains open. (Note that, under our hypotheses, D is analytically irreducible if and only if it is unibranched and moreover the integral closure D′ of D is finitely generated as a D-module, cf. [1, Proposition III.5.2].)

FUNDING

Supported by Austrian Science Fund (FWF): P23245-N18.

Notes

Communicated by S. Bazzoni.