J. reine angew. Math. (Crelle) 536 (2001) 31–42

NULLSTELLENSATZ AND SKOLEM PROPERTIES

FOR INTEGER-VALUED POLYNOMIALS

Sophie Frisch

Abstract. Skolem and Nullstellensatz properties are analogues of the weak Nullstellensatz and Hilbert’s Nullstellensatz, respectively, for the ring of integervalued polynomials in several indeterminates Int(D^{n}) = {f ∈ K[x_{1}, …, x_{n}] | f(D^{n}) ⊆ D}, where D is a domain and K its quotient field. We show their equivalence when D is a Noetherian domain and extend the criterion of Brizolis and Chabert for Int(D^{n}) to have the Nullstellensatz property to all Noetherian domains D.

I. Introduction and Definitions

Hilbert’s Nullstellensatz states for an algebraically closed field F that if f_{1}, …, f_{m} and f are in F [x_{1}, …, x_{n}] such that f(\underline{a}) = 0 for all those \underline{a} ∈ F^{n} for which f_{1}(\underline{a}) = … = f_{m}(\underline{a}) = 0, then f ∈ \sqrt{(f_{1}, …, f_{m})}. In this paper we investigate analogues of this theorem that hold when we replace F by a Noetherian domain D and F [x_{1}, …, x_{n}] by the ring of integer-valued polynomials in n indeterminates over D. For a survey of this topic, see chapters VII and XI of [2], and for related recent research [3, 8, 9, 5]. Throughout this paper, D denotes an integral domain with quotient field K.

Definition. The ring of integer-valued polynomials in n indeterminates over D is defined as Int(D^{n}) = \{f ∈ K[x_{1}, …, x_{n}] | f(\underline{a}) ∈ D for all \underline{a} ∈ D^{n}\}. We write Int(D) for Int(D^{1}), and define Int(D^{0}) = D.

Brizolis [1] in 1975 proposed the following “Nullstellensatz property” and characterized among Dedekind domains with perfect residue fields those D for which Int(D^{n}) had the property (including, for instance, the ring of integers of every number field).

MSC 2000: Primary 13F20; Secondary 13B25, 11C08, 14A25. Keywords: Hilbert rings, integer-valued polynomials, Jacobson rings, Nullstellensatz, Skolem properties.

NULLSTELLENSATZ PROPERTIES

Definition. Int(D^{n}) has the Nullstellensatz property, if for all polynomials f_{1}, …, f_{m} and f ∈ Int(D^{n}) satisfying f(\underline{a}) ∈ (f_{1}(\underline{a}), …, f_{m}(\underline{a})) for all \underline{a} ∈ D^{n}, it is true that f ∈ \sqrt{(f_{1}, …, f_{m})}.

We also consider a slightly stronger property, which we accordingly call the stronger Nullstellensatz property. For an ideal I in a commutative ring R, we denote by J (I) the Jacobson radical of I, i.e., the intersection of all maximal ideals of R containing I.

Definition. We say that Int(D^{n}) has the stronger Nullstellensatz property, if for all polynomials f_{1}, …, f_{m} and f ∈Int(D^{n}) satisfying f(\underline{a})∈J ((f_{1}(\underline{a}), …, f_{m}(\underline{a}))) for all \underline{a} ∈ D^{n}, it is true that f ∈ \sqrt{(f_{1}, …, f_{m})}.

In 1936, Skolem [16] had shown an analogue for integer-valued polynomials of the so called “weak Nullstellensatz”, the theorem stating for an algebraically closed field F that if f_{1}, …, f_{m} generate a proper ideal of F [x_{1}, …, x_{n}] then there exists \underline{a} ∈ F^{n} with f_{1}(\underline{a}) = … = f_{m}(\underline{a}) = 0. Skolem gave a construction of g_{1}, …, g_{m} ∈ Int(ℤ^{n}) with f_{1}g_{1} + … + f_{m}g_{m} = 1 for any f_{1}, …, f_{m} ∈ Int(ℤ^{n}) satisfying f_{1}(\underline{a}) ℤ + … + f_{m}(\underline{a}) ℤ = ℤ for all \underline{a} ∈ ℤ^{n}, thereby effectively proving for Int(ℤ^{n}) the condition since known as the Skolem property:

Definition. Int(D^{n}) has the Skolem property, if for every finite set of polynomials f_{1}, …, f_{m} ∈ Int(D^{n}) satisfying (f_{1}(\underline{a}), …, f_{m}(\underline{a})) = D for all \underline{a} ∈ D^{n}, it is true that (f_{1}, …, f_{m}) = Int(D^{n}). D is called a Skolem ring if the ring of integer-valued polynomials in one indeterminate, Int(D), has the Skolem property.

Clearly, the Nullstellensatz property for Int(D^{n}) implies the Skolem property for Int(D^{n}). Also, it is easy to see that each of the properties for Int(D^{n}) implies the same property for Int(D^{m}) for all m ≤ n ([2] Lemma XI.3.2). In particular, if Int(D^{n}) satisfies the Skolem property for some n then D is a Skolem ring.

Remark 1.1. We must bear in mind that the Skolem property of Int(D^{n}) depends not only on the ring Int(D^{n}), but also on the manner of its elements being interpreted as functions. As a ring, Int(D^{n+1}) is canonically isomorphic to Int(Int(D^{n})) for every infinite domain D ([2] Prop. XI.1.1), but the Skolem property of Int(D^{n+1}) (concerning functions in n + 1 variables f: D^{n+1} → D) is different from the Skolem property of Int(Int(D^{n})) (concerning functions in one variable f: Int(D^{n}) → Int(D^{n})). It is, however, easy to see that the former implies the latter (cf. Remark 4.3).

Of the various necessary and sufficient conditions for Skolem and Nullstellensatz properties proved by Brizolis, Chabert and McQuillan, the most general so far is Chabert’s result [7] that a certain list of properties, which we will call the “Chabert-Brizolis criterion” for short (condition 5 of Theorem 4.2 below), characterizes the domains D for which Int(D^{n}) has the Nullstellensatz property for all n, both among Noetherian domains with dim(D)>1 and among Noetherian domains of dim(D) = 1 and char(D) = 0. This characterization implies the equivalence of Skolem and (a priori stronger) Nullstellensatz properties for those Noetherian domains for which it holds, since the criterion is known to be necessary for D to be a Skolem ring.

In fact, it is known that a Noetherian domain satisfies the Chabert-Brizolis criterion if and only if it is a Skolem ring. This previous result of Chabert [6] and McQuillan [15], together with the equivalence of Skolem and Nullstellensatz properties in one variable (Chabert [7] and Brizolis [1]), shows that the characterization we wish to prove for several variables is true for one variable. Due to the fact that K[x] is a principal ideal domain, the one variable case can be proved with comparatively little effort (cf. [2], Theorems VII.5.1 and VII.5.9).

In this paper we show for an arbitrary Noetherian domain D the equivalence of the Chabert-Brizolis criterion to the Nullstellensatz property of Int(D^{n}) for all n, thereby closing the gap at dim(D) = 1, char(D) ≠ 0. Our proof takes its simplest form in the case of a one-dimensional domain, but works for any Noetherian domain with the addition of a few technicalities (Remark 2.3, Lemma 3.3, Lemma 3.4) that can be skipped when assuming dim(D) = 1.

We use a two-step approach showing first the equivalence of Skolem and Nullstellensatz property (Theorem 2.4) and then the sufficiency of the criterion for the Skolem property (Theorem 3.5). For the first step, we adapt Rabinowitsch’s argument that the weak Nullstellensatz for F [x_{1}, …, x_{n+1}] implies the Nullstellensatz for F [x_{1}, …, x_{n}], to show that the Skolem property for Int(D^{n+1}) implies the Nullstellensatz property for Int(D^{n}).

To formulate the Chabert-Brizolis criterion, we must define two ring-theoretic entities, the well-known Hilbert ring (a.k.a. Jacobson ring), and the lesser known d-ring. The criterion then consists of these two properties, together with the requirement that every maximal ideal M of D either have an algebraically closed residue field or be of height 1 with a finite residue field.

Definition. A domain D is a Hilbert ring (Goldman [11]), if the radical of every ideal I of D is equal to J (I), the intersection of all maximal ideals containing I. (Krull called such rings Jacobson rings [14].)

Although Hilbert rings play an important part in the investigation of Nullstellensatz analogues for integer-valued polynomials, so far no analogue has been proposed for Goldman’s and Krull’s proof of the Nullstellensatz ([13] §1.3) based on the fact that a domain R is a Hilbert ring if and only if R[x] is a Hilbert ring. While in principle it is possible to get to Int(D^{n}) from D inductively by repeatedly passing from R to Int(R) (by the isomorphism Int(Int(D^{n})) ≃ Int(D^{n+1})), no condition comparable to the Hilbert ring property is known to pass from R to Int(R). In particular, R being a Hilbert ring does not imply the same for Int(R) ([1] Example 4.3).

Brizolis [1] and Gunji and McQuillan [12] simultaneously introduced d-rings in 1975. This is the concept needed to show the Skolem property with respect to ideals of Int(D^{n}) lying over (0) in D.

Definition. A domain D is a d-ring if for every non-constant polynomial f ∈ D[x] there exists a maximal ideal M of D and an element d ∈ D such that f(d) ∈ M.

Alternatively, d-rings can be characterized as those domains, for which every integer-valued rational function is an integer-valued polynomial. We summarize a few facts about d-rings that we’ll need (cf. [2], §VII.2).

Fact 1.2 ([1] Lemma 1.3, [12] Prop. 1). The following are equivalent: (i) D is a d-ring. (ii) For every non-constant f ∈ D[x], if S is the set of maximal ideals M of D,

such that f has a zero mod M, then ⋂_{M∈S} M = (0). (iii) For every non-constant f ∈ Int(D), there exists a maximal ideal M of D

and an element d ∈ D such that f(d) ∈ M.

Fact 1.3 ([12] Prop. 3, Corollary 2). Let D ⊆ R be domains, and D a d-ring.

(a) If R is integral over D, then R is a d-ring.

(b) If R is finitely generated as a ring over D, then R is a d-ring.

Note that Fact 1.2 (ii) implies that (0) in a d-ring is always an intersection of maximal ideals, and that therefore every one-dimensional d-ring is a Hilbert ring.

II. Equivalence of Nullstellensatz and Skolem properties

Let M be a maximal ideal of D. For every \underline{a} ∈ D^{n}, there is a maximal ideal of Int(D^{n}) lying over M (with the same residue field as M) given by

M_{\underline{a},M} = \{f ∈ Int(D^{n}) | f(\underline{a}) ∈ M\}.

Int(D^{n}) has the Skolem property if and only if every finitely generated proper ideal of Int(D^{n}) is contained in a maximal ideal of the form M_{\underline{a},M}, for some maximal ideal M of D and \underline{a} ∈ D^{n}. Similarly, Int(D^{n}) has the stronger Nullstellensatz property if and only if the radical of every finitely generated ideal I of Int(D^{n}) is an intersection of maximal ideals of the form M_{\underline{a},M}, with M a maximal ideal of D and \underline{a} ∈ D^{n}.

Since prime ideals play a major rôle in the following, we briefly describe Spec(Int(D^{n})). First recall that if P is a prime ideal of infinite index in D, then Int(D^{n}) ⊆ D_{P} [x_{1}, …, x_{n}] ([2], Corollary XI.1.11) and therefore

(D \ P)^{−1}Int(D^{n}) = D_{P} [x_{1}, …, x_{n}],

which implies a bijective correspondence (given by lying over) between all prime ideals of D_{P} [x_{1}, …, x_{n}] and those prime ideals of Int(D^{n}) whose intersection with D is contained in P.

Fact 2.1. Let D be a domain and P a prime ideal of infinite index in D. The prime ideals Q of Int(D^{n}) with Q ∩ D = P are in bijective correspondence, via

Q = Q ∩ Int(D^{n}), (Q prime ⊴ D_{P} [x_{1}, …, x_{n}], P D_{P} [x_{1}, …, x_{n}] ⊆ Q),

to the prime ideals Q of D_{P} [x_{1}, …, x_{n}] containing P D_{P} [x_{1}, …, x_{n}], and therefore in bijective correspondence to the prime ideals Q̃ of (D_{P} /P D_{P})[x_{1}, …, x_{n}], via

Q = π^{−1}(Q̃) ∩ Int(D^{n}), (Q̃ prime ⊴(D_{P} /P D_{P})[x_{1}, …, x_{n}]),

where π: D_{P} [x_{1}, …, x_{n}] → (D_{P} /P D_{P})[x_{1}, …, x_{n}] is the canonical projection.

The prime ideals of Int(D^{n}) above a maximal ideal M of finite index in D are not as easily characterized, but if ht(M) = 1, they are known. First note that for every ideal M of D, every polynomial in Int(D^{n}) is uniformly continuous in M -adic topology. For Noetherian D, this is shown in [2], Prop. III.2.3, by means of the Artin-Rees Lemma; an elementary proof for arbitrary domains can be found in [4] Prop. 1.4 (cf. also Remark 1.5). Both proofs are for one variable, but obviously carry over to several. Therefore, if M is a maximal ideal of D, and D̂_{M} the M -adic completion of D, then every f ∈ Int(D^{n}) induces a continuous function f: D̂_{M}^{n} → D̂_{M}.

Fact 2.2 ([2], Corollary XI.1.8 and Lemma XI.2.8). Let D be a Noetherian domain and M a maximal ideal of finite index in D with ht(M) = 1. By D̂_{M} denote the M -adic completion of D with maximal ideal M̂. Then every prime ideal of Int(D^{n}) above M is maximal and of the form

M_{\underline{a},M̂} = \{f ∈ Int(D^{n}) | f(\underline{a}) ∈ M̂\},

for some \underline{a} ∈ D̂_{M}^{n}.

Remark 2.3 ([2] Lemma VII.4.2). If M is a maximal ideal in a Noetherian Skolem ring, then D/M is algebraically closed or ht(M) = 1. Proof. Suppose otherwise. (This implies D is not a field.) Since ht(M) > 1, there exists a maximal ideal M′ of the integral closure D′ of D with M′ ∩ D = M and ht(M′) > 1. D being Noetherian, D′ is a Krull domain and so is D′_{M′}, such that D′_{M′} = ⋂_{P′⊆M′} D′_{P′}, the intersection being over all height 1 prime ideals of D′ contained in M′. Each such P′ is properly contained in M′, so P = P′ ∩ D is properly contained in M and therefore of infinite index in D, such that Int(D) ⊆ D_{P} [x] ⊆ D′_{P′}[x]. We have shown Int(D) ⊆ D′_{M′}[x].

As D/M is not algebraically closed, there exists a polynomial f ∈ D[x], not a constant in (D/M)[x], that has no zero mod M, wherefore by the Skolem property, (M, f)Int(D) = Int(D). This implies (M, f)D′_{M′}[x] = D′_{M′}[x], making f a constant in (D′_{M′}/M′D′_{M′})[x], contrary to assumption. □

Theorem 2.4. Let D be a Noetherian domain and n ∈ ℕ. If Int(D^{n+1}) has the Skolem property, then Int(D^{n}) has the stronger Nullstellensatz property.

Proof. Given f_{1}, …, f_{m}, f ∈ Int(D^{n}) such that for every maximal ideal M of D, and every \underline{a} ∈ D^{n}, whenever f_{1}(\underline{a}), …, f_{m}(\underline{a}) ∈ M then f(\underline{a}) ∈ M, we must show for every prime ideal Q of Int(D^{n}): if f_{1}, …, f_{m} ∈ Q then f ∈ Q.

First consider a prime ideal Q lying over a prime ideal of infinite index in D. By Fact 2.1, Q=Int(D^{n})∩Q, where Q is a prime ideal of R[x_{1}, …, x_{n}] for an overring R of D, such that Int(D^{n}) ⊆ R[x_{1}, …, x_{n}] and Int(D^{n+1}) ⊆ R[x_{1}, …, x_{n+1}].

The polynomials f_{1}, …, f_{m}, 1 − x_{n+1}f ∈ Int(D^{n+1}) satisfy the hypothesis of the Skolem property of Int(D^{n+1}), so there exist h_{1}, …, h_{m}, h ∈ Int(D^{n+1}) with

f_{1}h_{1} + … + f_{m}h_{m} + (1 − x_{n+1}f)h = 1.

We now substitute 1/f for x_{n+1} and multiply both sides by f^{r}, where r is the maximal degree in x_{n+1} of any h_{i}. This gives f_{1}g_{1} + … + f_{m}g_{m} = f^{r}, where g_{1}, …, g_{m} ∈R[x_{1}, …, x_{n}], because h_{i} is of the form ∑ c_{k_{1},...,k_{n+1}}x^{k_{1}} … x^{k_{n+1}} , with c_{k_{1},...,k_{n+1}} ∈ R and therefore g_{i} = ∑ c_{k_{1},...,k_{n+1}}x^{k_{1}} … x^{k_{n}}f^{r−k_{n+1}} , with c_{k_{1},...,k_{n+1}} ∈ R and f^{r−k_{n+1}} ∈ R[x_{1}, …, x_{n}]. Therefore f_{1}, …, f_{m} ∈ Q implies f^{r} ∈ Q, and, Q being prime, f ∈ Q. We conclude that f ∈ Q = Int(D^{n}) ∩ Q.

Now consider a prime ideal Q of Int(D^{n}) above a maximal ideal M of finite index in D. By Fact 2.2 and Remark 2.3, Q = M_{\underline{a},M̂} = \{g ∈ Int(D^{n}) | g(\underline{a}) ∈ M̂\}, for some \underline{a} ∈ D̂_{M}^{n}. If f_{1}, …, f_{m} ∈ M_{\underline{a},M̂}, then, since the functions f_{i}: D̂_{M}^{n} → D̂_{M} are continuous, some M̂ -adic neighborhood U of \underline{a} gets mapped into M̂ by all f_{i}. In particular, f_{1}(\underline{b}), …, f_{m}(\underline{b}) ∈ M for all \underline{b} ∈ U ∩ D^{n}, and therefore, by the hypothesis of the stronger Nullstellensatz property, f(\underline{b}) ∈ M for all \underline{b} ∈ U ∩ D^{n}. Since D is dense in D̂_{M}, the continuity of f: D̂_{M}^{n} → D̂_{M} implies f(\underline{a}) ∈ M̂, i.e. f ∈ M_{\underline{a},M̂}. □

Corollary. If D is a Noetherian domain then Int(D^{n}) has the stronger Nullstellensatz property for all n ∈ ℕ if and only if Int(D^{n}) has the Skolem property for all n ∈ ℕ.

The Rabinowitsch-type argument from the first case of the above proof also yields a simpler proof of the following fact, which forms part of the argument for the necessity of the Chabert-Brizolis criterion:

Remark 2.5 ([2], Thm. VII.5.1). Every Noetherian Skolem ring is a Hilbert ring; more generally, the radical of every finitely generated ideal in a Skolem ring is an intersection of maximal ideals. Proof. Given d_{1}, …, d_{m} and d ∈ D, such that d_{1}, …, d_{m} ∈ M ⟹ d ∈ M, for every maximal ideal M of D, and a non-maximal prime ideal P of D with d_{1}, …, d_{m} ∈ P, we use the Skolem property of Int(D) to show d ∈ P:

Since [D:P] is infinite, Int(D)⊆D_{P} [x]. The polynomials d_{1}, …, d_{m} and 1−dx in Int(D) satisfy the hypothesis of the Skolem property, so there exist g_{1}, …, g_{m} and g ∈ Int(D) with d_{1}g_{1}+…+d_{m}g_{m}+(1−dx)g = 1. We substitute d^{−1} for x and multiply both sides by d^{r}, where r = max deg g_{i}, and get d_{1}h_{1} + … + d_{m}h_{m} = d^{r}, with h_{i} ∈ K. Since g_{i}(x) = ∑_{k} a_{k}x^{k} with a_{k} ∈ D_{P}, however, h_{i} = ∑_{k} a_{k}d^{r−k} is actually in D_{P}. Therefore, d_{1}, …, d_{m} ∈ P implies d^{r} ∈ P D_{P}, and we conclude d ∈ P D_{P} ∩ D = P. □

III. Sufficiency of Criterion for Skolem Property

We proceed to show for Noetherian domains of arbitrary characteristic that the Chabert-Brizolis criterion implies the Skolem property in several indeterminates (Theorem 3.5). The nontrivial part of the argument concerns maximal ideals of Int(D^{n}) lying over (0) in D (Lemma 3.2). To work around the problems presented by non-separable extensions of K we will replace K by its prefect closure F, whose algebraic extensions are all separable. This necessitates replacing D by a finite ring extension inside F, but, owing to the restrictions on the residue fields of D, this ring extension turns out to be harmless for our purposes, since it involves only few nontrivial residue field extensions. (Lemma 3.1).

Recall that for every field K there exists in the algebraic closure K̄ a unique smallest perfect field containing K, called the perfect closure of K. If char(K) = 0, then K is its own prefect closure; if char(K) = p, then the perfect closure is obtained from K by adjoining the p^{n}-th roots (for all n ∈ ℕ) of all elements of K. This is a consequence of the fact that a field E of characteristic p is perfect if and only if the Frobenius endomorphism φ(x) = x^{p} on E, φ: E → E, is surjective.

The following two lemmata are mainly relevant in characteristic p, as the contingency they provide for, a maximal ideal M of Int(D^{n}) with M∩D = (0), is known not to occur for a Noetherian d-ring of characteristic 0 ([2], Lemma XI.3.3, Prop. XI.3.4). Allowing char(D) = 0, however, does not add any difficulty and keeps our proof self-contained.

Lemma 3.1. Let D be a domain with quotient field K, F the perfect closure of K, and R = D[c_{1}, …, c_{k}] ⊆ F a domain finitely generated as a ring over D. Let S denote the set of prime ideals P of D, such that D_{P} /P D_{P} is a perfect field and for some prime ideal Q of R with Q ∩ D = P the embedding of residue fields D_{P} /P D_{P} ↪ R_{Q}/QR_{Q} is not surjective. Then ⋂_{P ∈S} P ≠ (0).

Proof. First assume char(K) = p. Since R ⊆ F, there exists an n ∈ ℕ such that c_{i}^{p^{n}} ∈ K for 1 ≤ i ≤ k. Write c_{i}^{p^{n}} = a_{i}b_{i}^{−1} with a_{i}, b_{i} ∈ D and set b = b_{1} · . . . · b_{k}. We claim that b ∈ ⋂_{P ∈S} P. Suppose P is a prime ideal of D such that b ∉ P and D_{P} /P D_{P} is perfect. Let Q be a prime ideal of R with Q ∩ D = P; we must show that D_{P} /P D_{P} ↪ R_{Q}/QR_{Q} is surjective.

The residue classes of c_{1}, …, c_{k} generate R_{Q}/QR_{Q} over D_{P} /P D_{P}, and c_{i} is a p^{n}-th root of a_{i}b_{i}^{−1} ∈ D_{P}. Now a_{i}b_{i}^{−1} + P D_{P} has a p^{n}-th root d_{i} + P D_{P} in D_{P} /P D_{P}, since this field is perfect. By the injectivity of the Frobenius homomorphism, c_{i}+QR_{Q} =d_{i}+QR_{Q}, i.e., c_{i}+QR_{Q} is in the image of D_{P} /P D_{P} ↪ R_{Q}/QR_{Q}.

If char(K) = 0, then R ⊆ K and c_{i} = a_{i}b_{i}^{−1} with a_{i}, b_{i} ∈ D. Let b = b_{1} · . . . · b_{k}. Then c_{1}, …, c_{k} ∈ D_{P} for every prime ideal P of D with b ∉ P, i.e., any residue field extension D_{P} /P D_{P} ↪ R_{Q}/QR_{Q} with b ∉ P is trivial. □

Lemma 3.2. Let D be a d-ring, such that the residue field of every maximal ideal of D is perfect. If I is a finitely generated ideal of Int(D^{n}) contained in an ideal of the form M = P ∩ Int(D^{n}), where P = {f ∈ K[x_{1}, …, x_{n}] | f(u_{1}, …, u_{n}) = 0}, u_{1}, …, u_{n} ∈ K̄, then I ⊆ M_{\underline{a},M} for some maximal ideal M of D and \underline{a} ∈ D^{n}.

Proof. Let F be the perfect closure of K, and u∈K̄ such that F [u_{1}, …, u_{n}]=F [u]. By multiplying u with a suitable constant in D, if necessary, we may assume u integral over D. For 1≤i≤n, let g_{i}(x)∈F [x], such that g_{i}(u)=u_{i}. If c_{1}, …, c_{l} ∈F is a list of the coefficients of g_{1}, …, g_{n}, let d ∈ D such that dc_{i} = c′_{i} is integral over D for 1 ≤ i ≤ l and set E = K[c_{1}, …, c_{l}] = K[c′_{1}, …, c′_{l}].

Let f(x) = x^{k} + α_{k−1}x^{k−1} + … + α_{0} be the minimal polynomial of u over E. Its coefficients α_{0}, …, α_{k−1} are integral over D, since u is. Define R = D[c′_{1}, …, c′_{l}, α_{0}, …, α_{k−1}], then R is a domain with quotient field E, finitely generated integral over D, and a d-ring by Fact 1.3.

Let f_{1}, …, f_{m} be generators of I, and define f_{i}^{∗} = f_{i}(g_{1}(x), …, g_{n}(x)) ∈ E[x]. Since f_{i}^{∗}(u) = f_{i}(u_{1}, …, u_{n}) = 0, there exist h_{i} ∈ E[x] with f_{i}^{∗}(x) = f(x)h_{i}(x). Let β ∈ R be a common denominator of the coefficients (written as fractions of elements in R) of all the polynomials g_{j}, f_{i}, f_{i}^{∗} and h_{i} for 1 ≤ j ≤ n, 1 ≤ i ≤ m.

Let b ∈ D be a non-zero element (existing by Lemma 3.1) in the intersection of all maximal ideals M of D such that D_{M}/MD_{M} ↪ R_{P} /P R_{P} is not surjective for some prime ideal P of R with P ∩ D = M.

Using Fact 1.2 (ii), let Q be a maximal ideal of R with bβ ∉ Q, and r ∈ R with f(r) ∈ Q. Let M = Q ∩ D; then M is a maximal ideal of D, since R is integral over D and Q is maximal. We claim that f_{1}, …, f_{m} have a simultaneous zero mod M in D^{n}.

Since β ∉ Q, all coefficients of all the polynomials g_{j}, f_{i}, f_{i}^{∗}, h_{i} are in R_{Q} (and those of f are in R, by construction). Let s_{j} = g_{j}(r) (1 ≤ j ≤ n), then

f_{i}(s_{1}, …, s_{n}) = f_{i}^{∗}(r) = f(r)h_{i}(r) ∈ QR_{Q}.

Let t_{j} ∈ D_{M} with t_{j} + QR_{Q} = s_{j} + QR_{Q} (exists since b ∉ M) and let a_{j} ∈ D with a_{j} + MD_{M} = t_{j} + MD_{M} (exists since M is maximal). Then for 1 ≤ i ≤ m, f_{i}(a_{1}, …, a_{n}) ∈ QR_{Q} ∩ D = M, or in other words, f_{i} ∈ M_{\underline{a},M}. □

The following two lemmata are needed to accommodate domains of dim(D)>1. We show that for those domains, under certain hypotheses (which are always satisfied by a Skolem ring), every maximal ideal of Int(D^{n}) lies over a maximal ideal of D. Actually, more is true: under the same hypotheses, every G-ideal of Int(D^{n}) lies over a maximal ideal of D ([2], proof of Prop. XI.3.7). (Recall that a G-ideal, or Goldman ideal, is a prime ideal that is not an intersection of prime ideals strictly containing it.)

Lemma 3.3. Let D be a Noetherian Hilbert ring with dim(D) > 1, such that every maximal ideal of finite index in D is of height 1. Then every maximal ideal of Int(D^{n}) lies either over (0) or over a maximal ideal in D.

Proof. Let P be a prime ideal of Int(D^{n}) not lying over (0) or a maximal ideal of finite index in D. We will show that if P does not lie over a maximal ideal of D[x_{1}, …, x_{n}], then it is itself not maximal. Since Hilbert rings are characterized by the fact that every maximal ideal of D[x_{1}, …, x_{n}] lies over a maximal ideal of D, this does it. Suppose P′ = P ∩ D[x_{1}, …, x_{n}] is not maximal. Let P′ ⊂ Q′ be a proper containment of prime ideals, and Q = Q′ ∩ D. Then [D : Q] is infinite.

We have (D \ Q)^{−1}Int(D^{n}) = D_{Q}[x_{1}, …, x_{n}] = (D \ Q)^{−1}D[x_{1}, …, x_{n}], so the prime ideals of D_{Q}[x_{1}, …, x_{n}] mediate a bijective correspondence between the prime ideals whose intersection with D is contained in Q of Int(D^{n}) on one hand and of D[x_{1}, …, x_{n}] on the other hand. Therefore P′ ⊂ Q′ corresponds to a proper containment of prime ideals P ⊂ Q in Int(D^{n}) and P is not maximal. □

The next lemma is only needed to show the redundancy of the d-ring property for domains of dim(D) > 1 in Theorem 4.2. If we did not care about this redundancy, we could skip the lemma and show Theorem 3.5 (by the same proof) under the hypothesis “Noetherian d-ring and, if dim(D) > 1, also Hilbert ring”.

Lemma 3.4 ([2], proof of Prop. XI.3.7) Let D be a Noetherian Hilbert ring with dim(D) > 1, such that every maximal ideal of finite index is of height 1. Then no maximal ideal of Int(D^{n}) lies over (0) in D.

Proof. Let M be a maximal ideal of Int(D^{n}) and suppose M∩D = (0). Let Q be a prime ideal of D with ht(Q) > 1. Then [D: Q] is infinite and therefore Int(D^{n}) ⊆ D_{Q}[x_{1}, …, x_{n}]. By the correspondence between prime ideals of D_{Q}[x_{1}, …, x_{n}] = (D\Q)^{−1}Int(D^{n}) and those prime ideals P of Int(D^{n}) with P∩D ⊆Q, M=M̃∩D for some maximal ideal M̃ of D_{Q}[x_{1}, …, x_{n}]. Now M̃ ∩ D_{Q} = (0), making (0) a G-ideal of D_{Q} ([10], proof of Thm. 31.8), but (0) never is a G-ideal in a Noetherian domain of dimension greater 1 ([13] Thm. 146), a contradiction. □

Theorem 3.5. Let D be a Noetherian domain and either a d-ring of dim(D) ≤ 1 or a Hilbert ring of dim(D) > 1. If for every maximal ideal M of D, either D/M is finite and ht(M) = 1 or D/M is algebraically closed, then Int(D^{n}) has the Skolem property for all n ∈ ℕ.

Proof. Given polynomials f_{1}, …, f_{m} ∈ Int(D^{n}) that generate a proper ideal of Int(D^{n}), we wish to show that they are contained in a maximal ideal of the form M_{\underline{a},M} = \{f ∈ Int(D^{n}) | f(\underline{a}) ∈ M\} for some \underline{a} ∈ D^{n} and a maximal ideal M of D. Let M be a maximal ideal of Int(D^{n}) with f_{1}, …, f_{m} ∈ M and let P = M ∩ D. By Lemma 3.3, P is either a maximal ideal of D or (0).

First case: [D : P] finite. By Fact 2.2, M = M_{\underline{b},P̂}, for some \underline{b} ∈ D̂_{P}^{n}. By continuity of the functions f_{i}: D̂_{P}^{n} → D̂_{P}, there is a neighborhood U of \underline{b}, such that f_{i}(U) ⊆ P̂ for 1 ≤ i ≤ m. Pick \underline{a} ∈ U ∩ D^{n} (which exists since D^{n} is dense in D̂_{P}^{n}), then f_{i}(\underline{a}) ∈ P̂ ∩ D = P, i.e. f_{i} ∈ M_{\underline{a},P} for 1 ≤ i ≤ m.

Second case: [D: P] infinite, P ≠ (0). Int(D^{n}) ⊆ D_{P} [x_{1}, …, x_{n}], since [D: P] is infinite. P is maximal, so by Fact 2.1, M = Q∩Int(D^{n}) for some maximal ideal Q of D_{P} [x_{1}, …, x_{n}] with P D_{P} [x_{1}, …, x_{n}] ⊆ Q. As D_{P} /P D_{P} is algebraically closed, there exist a_{1}, …, a_{n} in D_{P}, such that Q=(x_{1}−a_{1}, …, x_{n}−a_{n})+P D_{P} [x_{1}, …, x_{n}]. As the a_{i} only matter mod P D_{P} and P is maximal, they can be chosen to lie in D, and we see that M = M_{\underline{a},P} for some \underline{a} = (a_{1}, …, a_{n}) ∈ D^{n}.

Third case: P = (0). By the hypothesis of the theorem and Lemma 3.4, D is a d-ring in this case. By Fact 2.1 and the weak Nullstellensatz, M is exactly the kind of maximal ideal treated in Lemma 3.2, so we are done. □

IV. Nullstellensatz criterion for Noetherian domains

We now have all the necessary components for the characterization (among all Noetherian domains) of those domains that satisfy the Nullstellensatz property for integer-valued polynomials in any number of variables, except that the necessity of part of the residue field condition hasn’t been explained yet.

Remark 4.1 ([2] Prop. VII.4.2). If M is a finitely generated maximal ideal in a Skolem ring D, then D/M is either algebraically closed or finite. Proof. Suppose otherwise. D/M not being algebraically closed, there exists a polynomial f ∈ D[x], whose residue class in (D/M)[x] is not a constant, that has no zero in D mod M. Also, D/M being infinite, Int(D) ⊆ D_{M}[x]. By the Skolem property, (M, f)Int(D) = Int(D), and therefore (M, f)D_{M}[x] = D_{M}[x]. This would make f a constant in (D_{M}/MD_{M})[x], contrary to assumption. □

In the case of a Noetherian domain D with dim(D) > 1 ([7] Thm. 2, [2] Prop. XI.3.7) and in the case of a Noetherian domain with dim(D) = 1 and char(D) = 0 ([7], Thm. 3, [2] Thm. XI.3.6), the following result is due to Chabert, several special cases (mostly concerning Dedekind domains) having previously been shown by Brizolis. Our approach also provides a new proof of the known cases.

If D is a field, the theorem below reduces to the statement that Hilbert’s Nullstellensatz holds if and only if the weak Nullstellensatz holds, which in turn is the case if and only if the field is algebraically closed. Note, however, that we did not prove this, since we took the weak Nullstellensatz for algebraically closed fields for granted in the proof of Theorem 3.5.

Theorem 4.2. Let D be a Noetherian domain. The following are equivalent:

(1) For all n ∈ ℕ, Int(D^{n}) has the stronger Nullstellensatz property.

(2) For all n ∈ ℕ, Int(D^{n}) has the Nullstellensatz property.

(3) For all n ∈ ℕ, Int(D^{n}) has the Skolem property.

(4) D is a Skolem ring.

(5) D is a d-ring and a Hilbert ring, and for every maximal ideal M of D, either D/M is finite and ht(M) = 1 or D/M is algebraically closed.

If dim(D) = 1 then “Hilbert ring” in (5) is redundant. If dim(D) > 1 then “d-ring” in (5) is redundant.

Proof. The implications (1 ⇒ 2), (2 ⇒ 3) and (3 ⇒ 4) are evident.

(4 ⇒ 5) The d-ring property of D is just a special case of the Skolem property of Int(D) (Fact 1.2 (iii)), Skolem ring implies Hilbert ring by Remark 2.5, and the residue field condition is explained in Remarks 2.3 and 4.1.

(5 ⇒ 3) is Theorem 3.5; and (3 ⇒ 1) is a corollary of Theorem 2.4.

Finally, the redundancy of either d-ring or Hilbert ring, depending on the dimension of D, is evident from the hypothesis of Theorem 3.5. □

Since d-rings feature prominently in the above characterization, we should mention that they occur in abundance (cf. [2] §VII.2). For instance, ℤ is a d-ring, and so is every integral or finitely generated ring extension of a d-ring (Fact 1.3). In particular, every ring consisting of integers in a number field is a d-ring. Also, every ring R with D[x] ⊆ R ⊆ K[x] for a domain D with quotient field K is a d-ring. No semi-local ring, however, is a d-ring (Fact 1.2 (ii)).

Remark 4.3. Another equivalent property that we could insert in Theorem 4.2 is (3\frac{1}{2}) For all n ≥ 0, Int(D^{n}) is a Skolem ring. This is the statement that for all n ≥ 0, Int(Int(D^{n})), the ring of integer-valued polynomials in one indeterminate over Int(D^{n}), has the Skolem property. For a finite field, this and all other items of Theorem 4.2 are trivially false, so let D be an infinite domain. Obviously, (3\frac{1}{2} ⇒ 4); to see (3 ⇒ 3\frac{1}{2}), note that the Skolem property of Int(D^{n+1}) implies the Skolem property of Int(Int(D^{n})). Indeed, under the customary identification of Int(Int(D^{n})) with Int(D^{n+1}) ([2] Prop. XI.1.1), the Skolem property of Int(Int(D^{n})) says that every finitely generated proper ideal of this ring is contained in a maximal ideal of the form M_{a,Q}, with a ∈ Int(D^{n}) and Q a maximal ideal of Int(D^{n}), while the Skolem property of Int(D^{n+1}) says the same thing with the additional requirements that a ∈ D ⊆ Int(D^{n}) and Q be of the form Q = M_{\underline{b},M}, with \underline{b} ∈ D^{n} and M a maximal ideal of D.

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