DECIDABILITY OF THE THEORY OF MODULES OVER PRÜFER DOMAINS WITH DENSE VALUE GROUPS

LORNA GREGORY, SONIA L’INNOCENTE, AND CARLO TOFFALORI

Abstract. We provide algebraic conditions ensuring the decidability of the theory of modules over effectively given Prüfer (in particular Bézout) domains whose localizations at maximal ideals have dense value groups. For Bézout domains, these conditions are also necessary.

1. Introduction

This paper contributes to a body of work characterizing when the theory of modules of a Prüfer domain is decidable. This direction of research was initiated by Puninskaya, Puninski and the third author in [13], where it was shown that all effectively given valuation domains with dense Archimedean value group have decidable theory of modules. Confirming a conjecture in [13], the first author proved, in [4], that the theory of modules of an effectively given valuation domain V is decidable if and only if the set of pairs (a, b) ∈ V^{2} such that a ∈ rad(bV) is recursive. This work was picked up in [6], where a complete characterization of effectively given Bézout domains R with infinite residue fields and decidable theory of modules was given in terms of the recursivity of a certain subset, DPR(R), of R^{4}, generalizing the prime radical relation (see later in this introduction for a definition of this set). Analogous sufficient conditions were given for the theory of modules of an effectively given Prüfer domain with infinite residue fields to be decidable.

In this article we characterize effectively given Bézout domains R with decidable theory of modules under the assumption that the value groups of all localizations of R at maximal ideals are dense. We also give a sufficient condition for the theory of modules of a Prüfer domain to be decidable under the same assumption.

Thanks to the Baur-Monk theorem, if R is a recursive ring then the theory of R-modules is decidable if and only if there exists an algorithm which, given φ_{1}/ψ_{1}, …, φ_{h}/ψ_{h} pairs of pp-formulae and intervals [n_{1}, m_{1}], …, [n_{h}, m_{h}] ⊆ ℕ ∪ {∞}, answers whether there exists an R-module M such that, for all 1 ≤ i ≤ h, | φ_{i}(M)/ψ_{i}(M) | ∈ [n_{i}, m_{i}]. A standard argument means we may assume that M is a finite direct sum of indecomposable pure injective R-modules.

2010 Mathematics Subject Classification. 03C60 (primary), 03C98, 03B25, 13F05.

Key words and phrases. Prüfer domain, Bézout domain, Dense value group, Decidability.

The second and third authors thank the Italian GNSAGA-INdAM for its support.

Incidentally, the assumption “R recursive”, as well as the stronger one “R is effectively given”, are necessary to guarantee that the decidability problem of the theory of R-modules makes sense. We will recall both of them in § 2.

Modern proofs of decidability for theories of modules roughly split into two steps. The first step of such a proof gives an algorithm which decides whether one Ziegler basic open set is contained in a finite union of other Ziegler basic open sets. Equivalently, it gives an algorithm, as in the previous paragraph, but where the intervals [n_{i}, m_{i}] are either [1, 1] or [2, ∞]. When the Baur-Monk invariants |φ(M)/ψ(M)| of all R-modules M are either infinite or 1 then this is enough to show that the theory of R-modules is decidable.

However, when there exists R-modules with finite Baur-Monk invariants different from 1, more work is required. This, the second step, usually amounts to a fine detailed analysis of the indecomposable pure injective R-modules N such that there exists a pair of pp-formulae φ/ψ with |φ(N)/ψ(N)| finite but not equal to 1. This analysis is then used to reduce to the case of the first step.

For Bézout and Prüfer domains, the first step was dealt with in [6]. Note that, if R is a Prüfer domain with all residue fields infinite then for all R-modules M and pairs of pp-formulae φ/ψ, | φ(M)/ψ(M) | is either 1 or infinite.

On the other hand it was shown, in [13], that if V is valuation domain with finite residue field and dense value group then for each pp-pair φ/ψ there are at most finitely many indecomposable pure injective V -modules N with | φ(N)/ψ(N) | finite but not equal to 1. This makes the combinatorial problem of dealing with finite invariants sentences somewhat easier. In turn, since every indecomposable pure injective over a Prüfer domain R is the restriction of an indecomposable pure injective over some localization of R at a maximal ideal, our assumption that R_{𝔪} has dense value group for all maximal ideals 𝔪, makes it easier to deal with finite invariants sentences in this case too.

Before explaining the content of this article in more detail, we fix some notation. For R a ring, let L_{R} denote the language of R-modules, and T_{R} the theory of R-modules. Let ℕ be the set of positive integers and ℕ_{0} the set of the non-negative integers; ℙ is the set of prime numbers (in ℕ). For k_{1}, …, k_{s} ∈ ℕ, Span_{ℕ_{0}}{k_{1}, …, k_{s}} denotes the set of linear combinations of k_{1}, …, k_{s} with coefficients in ℕ_{0}.

Our characterization of Bézout domains with decidable theory of modules is based on two key sets.

• The first is DPR(R), introduced in [6, §6], that is, the set of 4-tuples (a, b, c, d) in R^{4} such that, for every choice of prime ideals 𝔭, 𝔮 of R with 𝔭 + 𝔮 ≠ R, either a ∈ 𝔭, b ∉ 𝔭, c ∈ 𝔮 or d ∉ 𝔮. We call DPR(R) the double prime radical relation because of its similarity to the prime radical relation in R, a ∈ rad(bR) with a, b ∈ R. Note that, a ∈ rad(bR) if and only if, for every proper prime ideal 𝔭 of R, if b ∈ 𝔭, then a ∈ 𝔭, that is, either a ∈ 𝔭 or b ∉ 𝔭. Hence a ∈ rad(bR) if and only if (a, b, a, b) ∈ DPR(R).

• The second set is inspired by the characterization of the (effectively given) commutative regular rings with decidable theory of modules given by Point and Prest in [9]. For this reason we denote it PP(R). The set PP(R) consists of the 4-tuples (p, n, c, d) ∈ ℙ × ℕ × R^{2} such that there exist positive integers s, k_{1}, …, k_{s} and maximal ideals 𝔪_{1}, …, 𝔪_{s} of R for which n ∈ Span_{ℕ_{0}}{k_{1}, …, k_{s}} and for all i = 1, …, s,

(1) |R/𝔪_{i}| = p^{k_{i}},

(2) c ∈ 𝔪_{i},

(3) d ∉ 𝔪_{i}.

By definition, see §2, the elements of an effectively given Prüfer domain R come equipped with an enumeration, so it makes sense to say that a subset of R^{n} for some n ∈ ℕ or ⋃_{n∈ℕ} R^{n} is recursive.

The main result for Bézout domains of this article is the following.

Theorem 6.2. Let R be an effectively given Bézout domain such that each localization of R at a maximal ideal has dense value group. Then T_{R} is decidable if and only if both DPR(R) and PP(R) are recursive.

For Prüfer domains the situation is more complicated. This time we introduce, for every l ∈ ℕ, two sets DPR_{l}(R) ⊆ R^{2l+2} and PP_{l}(R) ⊆ ℙ × ℕ × R^{l+1}.

• DPR_{l}(R) is the set of (2l + 2)-tuples (a, b_{1}, …, b_{l}, c, d_{1}, …, d_{l}) ∈ R^{2l+2} such that, for every choice of prime ideals 𝔭, 𝔮 of R with 𝔭+𝔮 ≠ R, either a ∈ 𝔭, c ∈ 𝔮, b_{j} ∉ 𝔭 or d_{j} ∉ 𝔮 for some 1 ≤ j ≤ l. Note that DPR_{1}(R) = DPR(R).

We put DPR^{⋆}(R) = ⋃_{l∈ℕ} DPR_{l}(R). Note that, when R is Bézout and 𝔭 is a proper prime ideal of R, then b_{j} ∉ 𝔭 for some 1 ≤ j ≤ l if and only if the greatest common divisor of b_{1}, …, b_{l} is not in 𝔭. Thus, for R an effectively given Bézout domain, DPR^{⋆}(R) is recursive if and only if DPR(R) is.

Recall that the sets DPR_{l}(R) were already considered in [6, §7] in order to partially extend the main decidability theorem there from Bézout to Prüfer domains with infinite residue fields. As observed in [6, Theorem 7.1], an algorithm deciding membership of the DPR_{l}(R) uniformly in l ensures, under the infinite residue fields assumption, that T_{R} itself is recursive^{1}. The existence of an algorithm deciding membership of the DPR_{l}(R) uniformly in l is equivalent to DPR^{⋆}(R) being recursive.

• For every l ∈ ℕ, let PP_{l}(R) consist of the tuples (p, n, c_{1}, …, c_{l}, d) ∈ ℙ × ℕ × R^{l+1} such that there exist positive integers s, k_{1}, … k_{s} and maximal ideals 𝔪_{1}, …, 𝔪_{s} of R for which n ∈ Span_{ℕ_{0}}{k_{1}, …, k_{s}} and for all i = 1, …, s,

(1) |R/𝔪_{i}| = p^{k_{i}},

(2) c_{j} ∈ 𝔪_{i} for 1 ≤ j ≤ l,

(3) d ∉ 𝔪_{i}.

Moreover put PP^{⋆}(R) = ⋃_{l∈ℕ} PP_{l}(R). Once again, if R is Bézout and effectively given, then PP^{⋆}(R) is recursive if and only if PP(R) = PP_{1}(R) is. This is again because, if 𝔪 is a maximal ideal of R, then c_{1}, …, c_{l} ∈ 𝔪 if and only if the greatest common divisor of c_{1}, …, c_{l} is in 𝔪.

As for Bézout domains, [6, 6.4], if R is an effectively given Prüfer domain with decidable theory of modules then DPR(R) is recursive. However, for Prüfer domains, we don’t know if T_{R} decidable implies DP R_{l}(R) is recursive for any l ≥ 2. In particular, we don’t know if T_{R} decidable implies that DPR^{⋆}(R) is recursive.

Theorem 3.2. Let R be an effectively given Prüfer domain. If T_{R} is decidable, then PP^{⋆}(R) is recursive.

The main result for Prüfer domains of this article is the following.

Theorem 6.1. Let R be an effectively given Prüfer domain such that each localization of R at a maximal ideal has dense value group. If both DPR^{⋆}(R) and PP^{⋆}(R) are recursive, then T_{R} is decidable.

Section 2 provides some basic information about model theory of modules over Prüfer and Bézout domains. In §3 we prove Theorem 3.2. Sections 4 and 5 prepare for the proof of the main theorems. The proofs of 6.1 and 6.2 are contained in §6.

We assume some familiarity with basic model theory of modules, as illustrated in Prest’s fundamental books [10], [11] and in the capital paper [17]. Prüfer domains and in particular Bézout domains are treated in [2] and [3]. Other recent papers dealing with decidability of modules over Bézout domains or related questions include [14], [8] and [7], while [15] provides a general treatment of the model theory of modules over Bézout domains.

Domains are assumed to be commutative with unity, and modules right unital.

1There is an omission in the statement of the published version of [6, 7.1]; the algorithms deciding membership of the various DPR_{l}(R) need to be uniform in l and this is also what is explicitly used in the proof.

2. Preliminaries

Recall that a domain R is Prüfer if all its localizations at maximal ideals, and consequently at non-zero prime ideals, are valuation domains, and Bézout if every 2-generated ideal (and consequently every finitely generated ideal) is principal. Thus R is Bézout if and only if the so called Bézout identity holds: for every 0 ≠ a, b ∈ R there are c, u, v, g, h ∈ R such that au+bv = c and cg = a, ch = b hold. Then c is called a greatest common divisor of a and b and is unique up to a multiplicative unit. Bézout domains are Prüfer.

We will make frequent use of the following result of Tuganbaev [16].

Fact 2.1. If R is a Prüfer domain then for all a, b ∈ R there exist α, r, s ∈ R such that bα = as and a(α − 1) = br.

When proving decidability results about T_{R}, for R a Prüfer domain, we will work under the hypothesis that R is effectively given. A Prüfer domain R is effectively given if it is countable and its elements can be listed as a_{0} = 0, a_{1} = 1, a_{2}, … (possibly with repetitions) so that suitable algorithms effectively execute the following, when m, n range over natural numbers.

(1) Deciding whether a_{m} = a_{n} or not.

(2) Producing a_{m} + a_{n} and a_{m} · a_{n}, or rather indices of these elements in the list.

(3) Establishing whether a_{m} divides a_{n}. This is a natural assumption to ensure that the decidability problem of T_{R} makes sense. In particular, if (1) and (2) hold then T_{R} is recursively axiomatizable. Moreover, in order for the theory of R-modules to be decidable, conditions (1) − (3) must hold. For instance, a_{m} divides a_{n} if and only if the sentence ∀x(xa_{m} = 0 → xa_{n} = 0) is in T_{R}.

Fact 2.2. Every pp-1-formula over a Prüfer domain is equivalent to a finite sum of formulae ∃y(ya = x ∧ yb = 0), with a, b ∈ R, and also to a finite conjunction of formulae c|xd, with c, d ∈ R (see [15, 2.2]). Over a Bézout domain R a stronger result holds, since every pp-1-formula is equivalent to a finite sum of formulae of the form a|x ∧ xb = 0, a, b ∈ R, and also to a finite conjunction of formulae c|x + xd = 0 with c, d ∈ R (see for instance [15, 2.3]).

We will denote a pp-1-pair, that is, an ordered pair of pp-1-formulae φ, ψ, by φ/ψ. Of course we will be mainly interested in the cases when φ, ψ have one of the forms in Fact 2.2.

For φ, ψ pp-1-formulae and t a positive integer, let |φ / ψ| ≥ t be the sentence of L_{R} saying that the index of the pp-subgroup defined by φ∧ψ inside that defined by φ is at least t. We call such sentences invariants sentences and for N an R-module, we call the values of |φ(N) : φ(N) ∩ ψ(N)| (either finite or ∞) the Baur-Monk invariants of N. In the following we will abbreviate |φ(N) : φ(N) ∩ ψ(N)| by |φ/ψ(N)|. We will say that an R-module N opens a pp-1-pair φ/ψ when |φ/ψ(N)| > 1. The already mentioned Baur-Monk theorem, [10, 2.15] asserts that, relative to T_{R}, every sentence in L_{R} is equivalent to a boolean combination of invariants sentences. Moreover, [10, 2.18], an R-module N is determined up to elementary equivalence by its Baur-Monk invariants.

We will need the following well-known argument: Let σ be a boolean combination of invariants sentences. Since, [10, 4.36], every R-module is elementary equivalent to a direct sum of indecomposable pure injective modules, σ is satisfied by some R-module if and only if σ is satisfied by a direct sum of indecomposable pure injective R-modules. Essentially because solution sets of pp-formulae commute with direct sums, if σ is satisfied by a direct sum of indecomposable pure injectives then σ is satisfied by a finite direct sum of indecomposable pure injectives.

Thus, in order to prove that an effectively given Prüfer domain R has decidable theory of modules, it is enough to show that there is an algorithm which, given a conjunction of invariants sentences and negations of invariants sentences σ, answers whether there exists a finite direct sum of indecomposable pure injective R-modules satisfying σ.

Fact 2.3. If R is a Prüfer domain and N is an indecomposable pure injective R-module then N is pp-uniserial, i.e. its lattice of pp-definable subgroups is totally ordered.

This follows from [12, 3.3], recalling that the lattice of pp-1-formulae over a Prüfer domain is distributive [1, 3.1]. Recall also that a module N is said to be uniserial if the lattice of all its submodules is totally ordered.

For every R-module N we put

Ass N := {r ∈ R | there exists m ∈ N\{0} with mr = 0}

and

Div N := {r ∈ R | r |̸ m for some m ∈ N}.

If N is an indecomposable pure injective module over a Prüfer domain R, then Ass N and Div N and their union Ass N ∪ Div N are (proper) prime ideals of R (see [6, Lemma 2.7]).

The Ziegler spectrum of a ring R, Zg_{R}, is a topological space whose points are (isomorphism classes of) indecomposable pure injective R-modules, and whose topology is given by basic open sets of the form (φ/ψ) where φ and ψ range over pp-1-formulae of L_{R}. Recall that an open set (φ/ψ) consists of the R-modules N in Zg_{R} such that φ(N) strictly includes its intersection with ψ(N).

For any commutative ring R, if N ∈ Zg_{R_{𝔪}} for some maximal ideal 𝔪 of R then N restricted to R is an indecomposable pure injective R-module. This gives, [11, 5.53], a homeomorphic embedding of Zg_{R_{𝔪}} into Zg_{R} as a closed subset. Moreover, for all N ∈ Zg_{R}, there exists a maximal ideal 𝔪 of R such that N is the restriction of an indecomposable pure injective R_{𝔪}-module (see for instance [5, 6.4]).

Let us also recall the correspondence, over a valuation domain V, between ordered pairs of proper ideals of V and indecomposable pp-1-types over V. The indecomposable pp-1-type associated to an ordered pair (I, J) of ideals is just the unique complete pp-1-type p = p(I, J) such that, for all r ∈ V,

• xr = 0 ∈ p if and only if r ∈ I and

• r | x ∈ p if and only if r ∉ J,

see [1, 3.4] and [6] for more details. Via these indecomposable pp-1-types, pairs of ideals also correspond to indecomposable pure injective V -modules. For every pair (I, J), let PE(I, J) denote the indecomposable pure injective V -module associated to it as the pure injective hull of the indecomposable pp-1-type p(I, J). So PE means pure injective hull. The equivalence relation connecting two pairs (I, J) and (I′, J′) if and only if PE(I, J) and PE(I′, J′) are isomorphic is well characterized, see again [1, 3.4].

Before finishing this section we mention a slight peculiarity. The very attentive reader of [6] and this article might be puzzled by the fact that we never actually need to use condition (3) of the definition of an effectively given Prüfer domain. Combined with 2.1, the following remark implies that if R is a recursive Prüfer domain (i.e. (1) and (2) in the definition of “effectively given” hold) and DPR(R) is recursive then condition (3) in the definition of an effectively given Prüfer domain must also hold.

Remark 2.4. Let a, b ∈ R\{0} and α, s, r ∈ R be such that bα = as and a(α − 1) = br. Then b ∈ aR if and only if (1, α, 1, r) ∈ DPR(R).

Proof. For any domain R, b ∈ aR if and only if b ∈ aR_{𝔭} for all proper prime ideals 𝔭. By [6, 5.5], b ∈ aR_{𝔭} if and only if α ∉ 𝔭 or r ∉ 𝔭. So b ∈ aR if and only if for all prime ideals 𝔭, α ∉ 𝔭 or r ∉ 𝔭.

From the definition of DPR(R), it follows that (1, α, 1, r) ∈ DPR(R) if and only if for all primes 𝔭, 𝔮 such that 𝔭 + 𝔮 ≠ R, α ∉ 𝔭 or r ∉ 𝔮.

If (1, α, 1, r) ∈ DPR(R) then, setting 𝔭 = 𝔮 in the definition of DPR(R), it follows that, for all proper primes 𝔭, α ∉ 𝔭 or r ∉ 𝔭.

Conversely, suppose that for all proper prime ideals 𝔭′, α ∉ 𝔭′ or r ∉ 𝔭′. Suppose that 𝔭, 𝔮 are prime ideals and 𝔭+𝔮 ≠ R. Since R is a Prüfer domain, 𝔭 + 𝔮 ≠ R implies 𝔭 ⊆ 𝔮 or 𝔮 ⊆ 𝔭. Let 𝔭′ = 𝔭 ∪ 𝔮. Then α ∉ 𝔭′ implies α ∉ 𝔭 and r ∉ 𝔭′ implies r ∉ 𝔮. Therefore (1, α, 1, r) ∈ DPR(R). □

3. Recursive sets

Throughout this section R will be a Prüfer domain. However, we will not require that the localizations of R at maximal ideals have dense value groups.

Lemma 3.1. All non-zero finite modules over a Prüfer domain R are of the form ∏_{i=1}^{h} R/𝔪_{i}^{λ_{i}} where 𝔪_{i} is a maximal ideal and λ_{i} ∈ ℕ for every 1 ≤ i ≤ h. If for each maximal ideal 𝔪, R_{𝔪} has dense value group then all non-zero finite modules are of the form ∏_{i=1}^{h} R/𝔪_{i} where each 𝔪_{i} is a maximal ideal.

Proof. Any finite module may be written as a direct sum of indecomposable finite modules, and finite modules are pure injective. Since R is commutative, every indecomposable pure injective is the restriction of a module over R_{𝔪} for some maximal ideal 𝔪. Since R is Prüfer, R_{𝔪} is a valuation domain. If M is a finite module over R_{𝔪} then M is a module over a finite quotient of R_{𝔪}. Thus M is either a module over R_{𝔪}/π^{n}R_{𝔪} where π generates 𝔪R_{𝔪} and R_{𝔪}/𝔪R_{𝔪} is finite, or M is a module over R/𝔪 where 𝔪 is the maximal ideal of R and R/𝔪 is finite.

In order to get the desired result we now just need to note that if π generates the maximal ideal of R_{𝔪} then R_{𝔪}/π^{λ}R_{𝔪} is isomorphic to the Rmodule R/𝔪^{λ} for every positive integer λ.

Finally, when all the value groups are dense, 𝔪 = 𝔪^{2} for every 𝔪. □

Theorem 3.2. Let R be an effectively given Prüfer domain. If T_{R} is decidable, then PP^{⋆}(R) is recursive.

Proof. We claim that, for all p ∈ ℙ, n ∈ ℕ, l ∈ ℕ, (c_{1}, …, c_{l}) ∈ R^{l} and d ∈ R, (p, n, c_{1}, …, c_{l}, d) is in PP^{⋆}(R) (so in PP_{l}(R)) if and only if there is an R-module M such that |M| = p^{n}, for each 1 ≤ j ≤ l, c_{j} ∈ ann_{R}M (the annihilator of M over R) and for all m ∈ M, md = 0 implies m = 0. Note that the condition that M has to satisfy can be expressed as a first order sentence of L_{R} in terms of p, n, d and the c_{j}. Moreover there is an effective procedure which, given a tuple π = (p, n, c_{1}, …, c_{l}, d) with (c_{1}, …, c_{l}) of arbitrary length l, produces this sentence θ_{π}. Since T_{R} is decidable, the set of all sentences of L_{R} true in at least one R-module is recursive. Applying that to the tuples π and the corresponding sentences θ_{π}, we obtain an algorithm deciding, for any given π, whether there exists an R-module M satisfying θ_{π} as required. Hence PP^{⋆}(R) is recursive, provided that we prove our claim. Then let us do that.

First suppose that (p, n, c_{1}, …, c_{l}, d) ∈ PP_{l}(R). Let 𝔪_{1}, …, 𝔪_{s} and k_{1}, …, k_{s} be as asked in the definition of PP_{l}(R), and let λ_{1}, …, λ_{s} ∈ ℕ_{0} satisfy ∑_{i=1}^{s} λ_{i}k_{i} = n. Put M = ∏_{i=1}^{s}(R/𝔪_{i})^{λ_{i}}. So |M| = p^{∑_{i=1}^{s} λ_{i}k_{i}} = p^{n}. Since for 1 ≤ j ≤ l, c_{j} ∈ 𝔪_{i} for all 1 ≤ i ≤ s, c_{j} ∈ ann_{R}M. Since d ∉ 𝔪_{i} for 1 ≤ i ≤ s, if m ∈ M and md = 0 then m = 0.

Now suppose that there exists an R-module M such that |M| = p^{n}, c_{j} ∈ ann_{R}M for all 1 ≤ j ≤ l and the only element of M annihilated by d is 0. By Lemma 3.1, we can assume that M is of the form ∏_{i=1}^{s} R/𝔪_{i}^{λ_{i}} for some suitable maximal ideals 𝔪_{1}, …, 𝔪_{s} and positive integers λ_{1}, …, λ_{s}. So, for each j, c_{j} ∈ ann_{R}M implies that c_{j} ∈ 𝔪_{i} for 1 ≤ i ≤ s. We may assume that if 𝔪_{i}^{2} = 𝔪_{i} then λ_{i} = 1. If λ_{i} = 1 then d ∉ 𝔪_{i} since d ∈ 𝔪_{i} implies (1 + 𝔪_{i})d = 0. Suppose λ_{i} > 1 . Take m ∈ 𝔪_{i}^{λ_{i}−1}\𝔪_{i}^{λ_{i}}. Then d ∈ 𝔪_{i} implies (m + 𝔪_{i}^{λ_{i}})d = 0. So d ∉ 𝔪_{i}. Thus d ∉ 𝔪_{i} for 1 ≤ i ≤ s.

Now |R/𝔪_{i}^{λ_{i}}| = |R/𝔪_{i}|^{λ_{i}}. So p^{n} = |M| = ∏_{i=1}^{s} |R/𝔪_{i}|^{λ_{i}}. Hence |R/𝔪_{i}| = p^{k_{i}} for some k_{i} ∈ ℕ. It follows |M| = p^{∑_{i=1}^{s} λ_{i}k_{i}}. Thus n = ∑_{i=1}^{s} λ_{i}k_{i}. Hence (p, n, (c_{1}, …, c_{l}), d) ∈ PP_{l}(R). □

4. Reducing to divisibility and torsion conditions

When R is an effectively given ring, in order for the theory of R-modules to be decidable it is enough that there is an algorithm which, given a sentence of the following form

(⋆) ⋀_{i=1}^{t} |φ_{1,i}/ψ_{1,i}| = H_{i} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| ≥ E_{k}

(where, for every i = 1, …, t, j = 1, … u and k = 1, … s, φ_{1,i}, ψ_{1,i}, φ_{2,j}, ψ_{2,j} and φ_{3,k}, ψ_{3,k} are pp-1-formulae and H_{i}, E_{k} are integers ≥ 2) answers whether there exists an R-module satisfying it. Moreover, this module can be assumed to be a finite direct sum of indecomposable pure injectives.

As also seen in Facts 2.2 and 2.3, indecomposable pure injective modules N over valuation domains (and more generally Prüfer domains) are ppuniserial, and every pp-1-formula over a valuation domain R is equivalent to a finite sum of formulae of the form a|x ∧ xb = 0. Hence, in order to calculate the size of the Baur-Monk invariants of N, in particular of those occurring in (⋆), it seems enough to handle the problem for pp-pairs φ/ψ where φ, ψ are of the form a|x and xb = 0 with a, b ∈ R. Since every indecomposable pure injective module over a Prüfer domain R is the restriction of an indecomposable pure injective module over R_{𝔪} for some maximal ideal 𝔪, this argument transfers to Prüfer domains. This motivates the following result which this section is dedicated to proving.

Theorem 4.1. Let R be an effectively given Prüfer domain. Suppose that there is an algorithm which, given a sentence

⋀_{i=1}^{t} |φ_{1,i}/ψ_{1,i}| = H_{i} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| ≥ E_{k}

where for i = 1, …, t, j = 1, …, u and k = 1, …, s, H_{i}, E_{k} are integers ≥ 2 and the pp-pairs φ_{1,i} / ψ_{1,i}, φ_{2,j} / ψ_{2,j}, φ_{3,k} / ψ_{3,k} are of the form xb = 0 / c|x or x = x / xd = 0 with b, c, d ∈ R, answers whether there exists an R-module satisfying this sentence. Then the theory T_{R} of R-modules is decidable.

Before starting the proof, we need some preparatory work.

Let Σ be a finite non-empty set of pp-1-formulae. Note that, for every Rmodule M, logical implication (with respect to the theory of M) determines a quasi-order on Σ, which becomes a partial order in the quotient set of Σ with respect to the logical equivalence relation (again with respect to the theory of M). Both the original quasi-order and the quotient order are total if M is an indecomposable pure injective module and R is Prüfer (by Fact 2.3). With this is mind, let us consider all the possible total quasi-orderings on Σ and the corresponding total orderings. To avoid excessively heavy notation, we will identify each total quasi-order on Σ with the corresponding total order, and we will denote by Γ(Σ) the set of these (quasi-)orders. For L ∈ Γ(Σ) (with its relation ≤_{L}) and for φ, ψ ∈ Σ, we write • φ =_{L} ψ to mean that according to L, φ and ψ are equal, that is,

φ ≥_{L} φ and ψ ≥_{L} φ,

• φ >_{L} ψ to mean that according to L, φ is strictly greater than ψ (so φ ≥_{L} ψ holds, but ψ ≥_{L} φ does not).

As an example, if Σ := {φ, ψ} then there are 3 total quasi-orderings, and indeed 3 different related total orders on Σ i.e. those with φ =_{L} ψ, φ >_{L} ψ and ψ >_{L} φ respectively.

For each L ∈ Γ(Σ), write Δ(L) for the following sentence in the language L_{R}:

⋀_{φ=_{L}ψ}(|φ/ψ| = 1 ∧ |ψ/φ| = 1) ∧ ⋀_{φ>_{L}ψ}(|φ/ψ| > 1 ∧ |ψ/φ| = 1).

Note that an R-module M satisfies Δ(L) if and only if the ordering of the pp-formulae in Σ given by L is the same as the inclusion ordering of the sets they define in M.

Recall that, 2.1, when R is a Prüfer domain, for all a, b ∈ R there exist α, r, s ∈ R such that bα = as and a(α − 1) = br (for technical reasons, see the next Lemma, we swap here a with b and r with s). Moreover, if R is effectively given, then, given a, b ∈ R, we can effectively find such α, r, s ∈ R.

Lemma 4.2. Let a, b ∈ R, and let α, r, s ∈ R satisfy aα = br and b(α−1) = as. For all R-modules M,

(i) if |xα = 0 / x = 0 (M)| = 1 then ∃y (ya = x ∧ yb = 0) is equivalent to x = 0 in M,

(ii) if |x(α − 1) = 0 / x = 0 (M)| = 1 then ∃y (ya = x ∧ yb = 0) is

equivalent to a|x ∧ xs = 0 in M, (iii) if |x = x / α|x (M)| = 1 then a|xb is equivalent to r|x + xb = 0 in M

and (iv) if |x = x / (α − 1)|x (M)| = 1 then a|xb is equivalent to x = x in M.

Proof. (i) Suppose that M satisfies |xα = 0 / x = 0| = 1. Let m, m′ ∈ M be such that m′a = m and m′b = 0. Then 0 = m′br = m′aα = mα. So m = 0.

(ii) Suppose that M satisfies |x(α−1) = 0 / x = 0| = 1. Let m, m′ ∈ M be such that m′a = m and m′b = 0. Then a|m and ms = m′as = m′b(α − 1) = 0.

Let m, m′ ∈ M be such that m = m′a and ms = 0. Then m′as = ms = 0. So m′b(α − 1) = 0. Since M satisfies |x(α − 1) = 0 / x = 0| = 1, m′b = 0. So m satisfies ∃y (ya = x ∧ yb = 0).

(iii) Suppose that M satisfies |x = x / α|x| = 1. Let m, m′ ∈ M be such that mb = m′a. Since M satisfies |x = x / α|x| = 1, there exists m′′ ∈ M such that m′ = m′′α. So mb = m′′αa = m′′br. So (m − m′′r)b = 0 and hence m satisfies r|x + xb = 0.

Let m, m′, m′′ ∈ M be such that m′′b = 0 and m = m′r + m′′. Then mb = m′rb = m′aα. So m satisfies a|xb.

(iv) Suppose M satisfies |x = x / (α − 1)|x| = 1. Let m ∈ M. Then there exists m′ ∈ M such that m′(α − 1) = m. So m′as = m′(α − 1)b = mb. Therefore a|mb. □

The next lemma will also be useful later.

Lemma 4.3. Let M be an R-module, a, b, c, d ∈ R. Then (i) |a|x / c|x (M)| = |x = x / c|xa (M)|, (ii) |a|x / xd = 0 (M)| = |x = x / xad = 0 (M)|, (iii) |xb = 0 / xd = 0 (M)| = | ∃y (x = yd ∧ yb = 0) / x = 0 (M)|, (iv) |x = x / xd = 0 (M)| = |d|x / x = 0 (M)|.

Proof. (i) and (ii) follow from considering the abelian group homomorphism from M to Ma/Mc (respectively to Ma/ann_{M}(d)) which sends any m ∈ M to the coset of ma (ann_{M}(d) denotes here the annihilator of d in M, that is, the pp-subgroup of the realizations in M of xd = 0).

(iv) uses the scalar multiplication by d in M.

For (iii) consider the abelian group homomorphism from ann_{M}(b) to the pp-subgroup of M defined by ∃y (x = yd ∧ yb = 0) which sends any m to md. This homomorphism is clearly surjective and m ∈ ann_{M}(b) is in its kernel if and only md = 0. □

Now let X, Y be non-empty finite subsets of R. Let Ω(X) (respectively Ω(Y)) be the set of functions P : X → {1, −1} (respectively Q : Y → {1, −1}). For each (P, Q) ∈ Ω(X) × Ω(Y), write Θ(P, Q) for the following sentence in the language L_{R} (with α ranging over X and β over Y):

⋀_{P (α)=1} |xα = 0 / x = 0| = 1 ∧ ⋀_{P (α)=−1} |x(α − 1) = 0 / x = 0| = 1∧

⋀_{Q(β)=1} |x = x / β|x| = 1 ∧ ⋀_{Q(β)=−1} |x = x / (β − 1)|x| = 1.

Note that |xα = 0 / x = 0| = 1 is satisfied by an R-module N if and only if α ∉ Ass N and |x = x / β|x| = 1 is satisfied by an R-module N if and only if β ∉ Div N.

Let us also point out that the only pairs of pp-formulae occurring in Θ(P, Q) are of the form required by Theorem 4.1.

Lemma 4.4. Let X, Y be non-empty finite subsets of R. If N is an indecomposable pure injective R-module then there exists (P, Q) ∈ Ω(X) × Ω(Y) such that N satisfies Θ(P, Q).

Proof. For N an indecomposable pure injective R-module, Ass N and Div N are proper ideals. Thus for every α ∈ X (respectively β ∈ Y), either α ∉ Ass N (respectively β ∉ Div N) or α − 1 ∉ Ass N (respectively β − 1 ∉ Div N). Let P : X → {1, −1} (respectively Q : Y → {1, −1}) be such that P (α) = 1 (respectively Q(β) = 1) if α ∉ Ass N (respectively β ∉ Div N) and P (α) = −1 (respectively Q(β) = −1) otherwise. Then N satisfies Θ(P, Q). □

Now we are able to prove Theorem 4.1.

Proof. Step 1: Let χ be the sentence labeled (⋆) at the beginning of this section and let Σ be any finite non-empty set of pp-formulae.

Note that there is an R-module satisfying χ if and only if there exists a non-empty subset T of Γ(Σ) and for each L ∈ T, an R-module M_{L} satisfying Δ(L) such that ⨁_{L∈T} M_{L} satisfies χ.

The reverse direction is clear. Conversely, if there exists an R-module satisfying χ, then there exists a finite direct sum of indecomposable pure injective (hence pp-uniserial by Fact 2.3) modules satisfying χ. Take T to be the set of total orderings of Σ determined by the inclusion of pp-subgroups in these direct summands.

Suppose that T ⊆ Γ(Σ) is non-empty. There exist R-modules M_{L} (L ∈ T) satisfying Δ(L) such that ⨁_{L∈T} M_{L} satisfies χ if and only if the following conditions hold:

(1) for each 1 ≤ i ≤ t, ∏_{L∈T} |φ_{1,i}(M_{L})/ψ_{1,i}(M_{L})| = H_{i},

(2) for each 1 ≤ j ≤ u and L ∈ T, |φ_{2,j}(M_{L})/ψ_{2,j}(M_{L})| = 1,

(3) for each 1 ≤ k ≤ s, ∏_{L∈T} |φ_{3,k}(M_{L})/ψ_{3,k}(M_{L})| ≥ E_{k}.

For each 1 ≤ i ≤ t, let F_{i}^{T} be the set of functions f : T → ℕ such that ∏_{L∈T} f(L) = H_{i}. For each 1 ≤ k ≤ s, let G_{k}^{T} be the set of functions g : T → ℕ such that ∏_{L∈T} g(L) ≥ E_{k} and for each L ∈ T, g(L) ≤ E_{k}.

For each pair of tuples f := (f_{1}, …, f_{t}) and g := (g_{1}, …, g_{s}) with f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T} and each L ∈ T, let χ_{(f,g)}^{L} be the sentence

Δ(L) ∧ ⋀_{i=1}^{t} |φ_{1,i}/ψ_{1,i}| = f_{i}(L) ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧

∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| ≥ g_{k}(L).

Now, there exists an R-module M satisfying χ if and only if the following exist

(1) T ⊆ Γ(Σ) non-empty,

(2) a pair of tuples f := (f_{1}, …, f_{l}) and g := (g_{1}, …, g_{s}) with f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T},

(3) for each L ∈ T, an R-module M_{L} satisfying χ_{(f,g)}^{L}.

Since R is a Prüfer domain, we may assume that each φ_{S,i} (with S = 1, 2, 3 and i ranging over the corresponding indices) is of the form ∑_{v=1}^{A_{S,i}} ∃y (ya_{iv}^{S} = x∧yb_{iv}^{S} = 0) and each ψ_{S,i} is of the form ⋀_{g=1}^{B_{S,i}}(c_{iw}^{S}|xd_{iw}^{S}), where the involved scalars are elements of R.

Let Σ be the set of formulae ∃y (ya_{iv}^{S} = x ∧ yb_{iv}^{S} = 0) and c_{iw}^{S}|xd_{iw}^{S} where S ∈ {1, 2, 3}, 1 ≤ i ≤ t if S = 1, 1 ≤ i ≤ u if S = 2, 1 ≤ i ≤ s if S = 3 and 1 ≤ v ≤ A_{S,i}, 1 ≤ w ≤ B_{S,i}.

Let T ⊆ Γ(Σ) and L ∈ T. Let f := (f_{1}, …, f_{l}) and g := (g_{1}, …, g_{s}) be a pair of tuples with f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T}. For each φ_{S,i} there exists σ_{S,i} ∈ Σ such that Δ(L) ⊢ φ_{S,i} ↔ σ_{S,i} and for each ψ_{S,i} there exists τ_{S,i} ∈ Σ such that Δ(L) ⊢ ψ_{i}^{s} ↔ τ_{i}^{s}, moreover each σ_{S,i}, τ_{S,i} can be effectively obtained from the corresponding φ_{S,i}, ψ_{S,i}. Thus χ_{(f,g)}^{L} is equivalent to

Δ(L) ∧ ⋀_{i=1}^{l} |σ_{1,i}/τ_{1,i}| = f_{i}(L) ∧ ⋀_{j=1}^{u} |σ_{2,j}/τ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |σ_{3,k}/τ_{3,k}| ≥ g_{k}(L).

Thus, in order to show that the theory of R-modules is decidable, it is enough that there is an algorithm which given a sentence as in (⋆) with each φ_{S,i} of the form ∃y (ya_{i}^{S} = x ∧ yb_{i}^{S} = 0) and each ψ_{S,i} of the form c_{i}^{S}|xd_{i}^{S}, answers whether there exists an R-module satisfying it.

Step 2: Let χ be the sentence labeled (⋆), as reduced at the end of Step 1. Let X, Y be non-empty finite subsets of R. Note that there is an R-module satisfying χ if and only if there exists a non-empty subset T of Ω(X)×Ω(Y) and for each (P, Q) ∈ T, there exists an R-module M_{(P,Q)} satisfying Θ(P, Q) such that ⨁_{(P,Q)∈T} M_{(P,Q)} satisfies χ.

This follows from Lemma 4.4 since if there exists an R-module satisfying χ then there exists a finite direct sum of indecomposable pure injective Rmodules satisfying χ and if two modules satisfy Θ(P, Q) then so does their direct sum.

Let F_{i}^{T} and G_{k}^{T} be as in Step 1, but adapted to the new setting where the (quasi-)orders L of some subset of Γ(Σ) are replaced by a subset of pairs (P, Q) in Ω(X) × Ω(Y). For each f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T}, let χ_{(f,g)}^{(P,Q)} be the sentence

Θ(P, Q) ∧ ⋀_{i=1}^{t} |φ_{1,i}/ψ_{1,i}| = f_{i}(P, Q) ∧

⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| ≥ g_{k}(P, Q).

Now, there exists an R-module M satisfying χ if and only if the following exist:

(1) T ⊆ Ω(X) × Ω(Y) non-empty,

(2) a pair of tuples f := (f_{1}, …, f_{l}) and g := (g_{1}, …, g_{s}) with f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T},

(3) for each (P, Q) ∈ T, an R-module M_{L} satisfying χ_{(f,g)}^{(P,Q)}.

Using Step 1, we may assume that each φ_{S,i} (with S = 1, 2, 3 and i ranging over the corresponding indices) is of the form ∃y (ya_{i}^{S} = x ∧ xb_{i}^{S} = 0) and each ψ_{S,i} is of the form c_{i}^{S}|xd_{i}^{S}. For each a_{i}^{S}, b_{i}^{S}, let α_{i}^{S}, δ_{i}^{S}, γ_{i}^{S} be such that a_{i}^{S}α_{i}^{S} = b_{i}^{S}δ_{i}^{S} and b(α_{i}^{S} − 1) = a_{i}^{S}γ_{i}^{S}. For each c_{i}^{S}, d_{i}^{s}, let β_{i}^{S}, λ_{i}^{S}, μ_{i}^{S} be such that c_{i}^{S}β_{i}^{S} = d_{i}^{S}λ_{i}^{S} and d_{i}^{S}(β_{i}^{S} − 1) = d_{i}^{S}μ_{i}^{S}. By Fact 2.1, such α_{i}^{S}, δ_{i}^{S}, γ_{i}^{S} and β_{i}^{S}, λ_{i}^{S}, μ_{i}^{S} exist and if R is effectively given then we can find them by searching. Let X, Y be the sets of the α_{i}^{S} and the β_{i}^{S}, respectively, where S = 1, 2, 3 and i ranges over the corresponding indices.

Let T ⊆ Ω(X) × Ω(Y) and (P, Q) ∈ T. Let f := (f_{1}, …, f_{l}) and g := (g_{1}, …, g_{s}) be a pair of tuples with f_{i} ∈ F_{i}^{T} and g_{k} ∈ G_{k}^{T}. By Lemma 4.2, for each φ_{S,i}, there exists a formula σ_{S,i} of the form a|x ∧ xs = 0 such that Θ(P, Q) ⊢ φ_{S,i} ↔ σ_{S,i} and for each ψ_{S,i}, there exists a formula τ_{S,i} of the form r|x + xd = 0 such that Θ(P, Q) ⊢ ψ_{S,i} ↔ τ_{S,i} (and there are algorithms producing these formulae). Thus χ_{(f,g)}^{(P,Q)} is equivalent to

Θ(P, Q) ∧ ⋀_{i=1}^{l} |σ_{1,i}/τ_{1,i}| = f_{i}(L)∧

⋀_{j=1}^{u} |σ_{2,j}/τ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |σ_{3,k}/τ_{3,k}| ≥ g_{k}(L).

Thus, in order to show that the theory of R-modules is decidable, it is enough that there is an algorithm which given a sentence as in (⋆) with each φ_{S,i} of the form a|x ∧ xs = 0 and each ψ_{S,i} of the form r|x + xd = 0, answers whether there exists an R-module satisfying it.

Step 3: Let χ be as in (⋆) with φ_{S,i} equal to a_{i}^{S}|x ∧ xs_{i}^{S} = 0 and ψ_{S,i} equal to r_{i}^{S}|x + xd_{i}^{S} = 0 with a_{i}^{S}, s_{i}^{S}, r_{i}^{S}, d_{i}^{S} ∈ R.

Proceeding as in Step 1 with Σ equal to the set of formulae a_{i}^{S}|x, xs_{i}^{S} = 0, r_{i}^{S}|x and xd_{i}^{S} = 0 one can show that the theory of R-module is decidable if and only if there is an algorithm which, given a sentence

⋀_{i=1}^{t} |φ_{1,i}/ψ_{1,i}| = H_{i} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| ≥ E_{k}

where for i = 1, …, t, j = 1, …, u and k = 1, …, s, H_{i}, E_{k} are integers ≥ 2 and the pp-formulae φ_{1,i}, ψ_{1,i}, φ_{2,j}, ψ_{2,j}, φ_{3,k}, ψ_{3,k} are of the form a|x or xb = 0 with a, b ∈ R, answers whether there exists an R-module satisfying this sentence.

Step 4: Let χ be of the form we reduced to at the end of Step 3. By Lemma 4.3, we can replace in χ

(i) every instance of the form | a|x / c|x | by | x = x / c|xa |, (ii) every instance of the form | a|x / xd = 0 | by | x = x / xad = 0 | (iii) and every instance of the form | xb = 0 / xd = 0| by | ∃y (x = yd ∧

yb = 0) / x = 0 |. Repeating Step 2 and recalling that only pairs of the form xα = 0 / x = 0 and x = x / β|x occur in the sentences Θ(P, Q), we are led to consider a conjunction of invariants sentences involving only pairs of the form x = x / ρ|x + xσ = 0, ρ|x ∧ xσ = 0 / x = 0, x = x / xd = 0 and xb = 0 / c|x. So we can assume that χ is a conjunction of invariants sentences involving only pairs of this form.

Step 5: Suppose that a pair of the form x = x / ρ|x + xσ = 0 or ρ|x ∧ xs = 0 / x = 0 occurs in χ for some ρ, σ ∈ R. Put Σ := {ρ|x, xσ = 0} and take L ∈ Γ(Σ). Then the only pairs that occur in Δ(L) are xσ = 0/ρ|x, which is already of the required final form in the statement of the theorem, and ρ|x / xs = 0, which by Lemma 4.3 can be replaced by x = x / xρσ = 0. Hence all pairs occurring in Δ(L) are of the required form.

Repeating Step 1 of the proof with Σ = {ρ|x, xσ = 0} produces sentences χ_{(f,g)}^{L} where we can replace each instance of x = x / ρ|x + xσ = 0 by x = x / ρ|x or x = x / xσ = 0 as appropriate and each instance of ρ|x ∧ xσ = 0 / x = 0 by ρ|x / x = 0 or xσ = 0 / x = 0 as appropriate. By Lemma 4.3, (iv), we may replace all instances of the pair ρ|x / x = 0 by x = x / xρ = 0. Repeating this process for each ρ, σ ∈ R such that the pair x = x / ρ|x+xσ = 0 or the pair ρ|x ∧ xσ = 0 / x = 0 occurs in χ allows us to reduce to considering sentences of the form required by the statement of the theorem.

□

5. Preparatory lemmas

We assume throughout this section that R is a Prüfer domain such that all the localizations of R at maximal ideals have dense value group.

The focus of this section will be the R-modules

N_{γ}(𝔪) := R_{𝔪}/γ𝔪R_{𝔪}, N′_{β,η}(𝔪) := 𝔪R_{𝔪}/βηR_{𝔪}

where 𝔪 is a maximal ideal of R, γ ∈ R\{0} and β, η ∈ 𝔪\{0}.

It was shown in [13, Proposition 7.8] that, over a valuation domain V with dense value group and finite residue field, the only indecomposable pure injective modules N such that there exists a pp-pair φ/ψ with |φ/ψ(N)| finite but not equal to 1 are those corresponding to the types (βV, ηV) and (𝔭, γ𝔭) where 𝔭 is the maximal ideal of V, γ ∈ V \{0} and β, η ∈ 𝔭\{0}. These types are realized in the uniserial V -modules 𝔭/βηV and V/γ𝔭. Thus all such indecomposable pure injective modules are of the form PE(𝔭/βηV) or PE(V/γ𝔭) (recall that PE means pure injective hull). If the residue field of V is not finite then no such indecomposable pure injective modules exist.

If N is an indecomposable pure injective module over a Prüfer domain R then there exists some maximal ideal 𝔪 such that N is the restriction of an indecomposable pure injective R_{𝔪}-module. Now, if there exists a pp-pair φ/ψ such that |φ/ψ(N)| is finite but not equal to 1 then N is either of the form PE(𝔪R_{𝔪}/βηR_{𝔪}) or of the form PE(R_{𝔪}/γ𝔪R_{𝔪}) where R/𝔪 is finite, β, η ∈ 𝔪R_{𝔪}\{0} and γ ∈ R_{𝔪}\{0}. Since all elements of R_{𝔪} are unit multiples of elements in R, we may assume that β, η ∈ 𝔪\{0} and γ ∈ R\{0}.

Finally, for R any commutative ring and 𝔪 a maximal ideal of R, if M is a module over R_{𝔪}, then taking the pure injective hull of M over R_{𝔪} and then restricting to R is the same as taking the pure injective hull of M as an R-module.

We will need the following result from [13]. Recall that a pp-pair φ/ψ is minimal (in the theory of a given module N over any ring) if φ(N) properly includes its intersection with ψ(N) and there is no intermediate pp-subgroup θ(N) such that φ(N) ⊋ θ(N) ⊋ φ(N) ∩ ψ(N).

Lemma 5.1. ([13, Lemma 7.5 and Corollary 7.6]). Let V be a commutative valuation domain and φ/ψ be a pp-1-pair over V. If N is an indecomposable pure injective V -module and | φ/ψ(N) | is finite and > 1 then φ/ψ is an Nminimal pair. Moreover, if 𝔭 is the maximal ideal of V, then φ(N)/ψ(N) is a 1-dimensional vector space over the residue field V/𝔭, that consequently is finite.

When R is a Prüfer domain, and so every localization at a maximal ideal is a commutative valuation domain, we obtain the following consequence. Let N be an indecomposable pure injective module over R, and let 𝔪 be a maximal ideal of R such that N is a module over R_{𝔪}. Then every pp-1-pair φ/ψ over R with | φ/ψ(N) | finite and greater than 1 is N-minimal and φ/ψ(N) is a 1-dimensional vector space over the residue field R/𝔪, which must therefore be finite.

The minimal pairs of modules, over a valuation domain V with maximal ideal 𝔭 and dense value group, of the form V/γ𝔭 and 𝔭/βηV were described in [13, Section 7] at least for valuation domains with finite residue fields. However, the results in Section 4 focus our interest on pp-pairs of the form xb = 0 / c|x and x = x / xd = 0. We will now prove the results about minimal pairs which we need without the assumption that V has finite residue field.

The following fact can be derived from [5, Theorem 4.3].

Fact 5.2. Let V be a valuation domain and (I, J) be a pair of proper ideals in V. Then PE(I, J) ∈ (xb = 0 ∧ a|x / xd = 0 + c|x) if and only if a ≠ 0, d ≠ 0, c ∈ aJ^{#}, b ∈ dI^{#}, bc ∈ IJ and ad ∉ ann_{V} PE(I, J).

For I an ideal of V, I^{#} denotes ⋃_{r∈V \I}(I : r). Note that Ass PE(I, J) = I^{#} and Div PE(I, J) = J^{#}. For V a valuation domain with maximal ideal 𝔭, γ ∈ V \{0} and β ∈ 𝔭\{0}, (γ𝔭)^{#} = 𝔭 and (βV)^{#} = 𝔭.

For γ ∈ R\{0}, the pure injective hull of R_{𝔪}/γ𝔪R_{𝔪} corresponds to the pair (γ𝔪R_{𝔪}, 𝔪R_{𝔪}) of ideals of R_{𝔪}. The annihilator, as an R_{𝔪}-module, of R_{𝔪}/γ𝔪R_{𝔪}, and hence PE(γ𝔪R_{𝔪}, 𝔪R_{𝔪}), is γ𝔪R_{𝔪}.

For 0 ≠ δ ∈ 𝔪, the pure injective hull of 𝔪R_{𝔪}/δR_{𝔪} corresponds to a pair (βR_{𝔪}, ηR_{𝔪}) of ideals of R_{𝔪} where β, η ∈ 𝔪 and βηR_{𝔪} = δR_{𝔪}. The annihilator, as an R_{𝔪}-module, of 𝔪R_{𝔪}/δR_{𝔪}, and hence of PE(βR_{𝔪}, ηR_{𝔪}), is δR_{𝔪}.

For 0 ≠ δ ∈ 𝔪, the pure injective hull of R_{𝔪}/δR_{𝔪} corresponds to the pair (δR_{𝔪}, 𝔪R_{𝔪}) of R_{𝔪} ideals. The annihilator, as an R_{𝔪}-module, of R_{𝔪}/δR_{𝔪}, and hence of PE(δR_{𝔪}, 𝔪R_{𝔪}), is δR_{𝔪}.

Note that this means that if δ ∈ 𝔪\{0} then PE(δR_{𝔪}, 𝔪) is in the Ziegler closure of both PE(δ𝔪R_{𝔪}, 𝔪R_{𝔪}) and PE(βR_{𝔪}, ηR_{𝔪}) where δR_{𝔪} = βηR_{𝔪} and β, η ∈ 𝔪\{0}. Since Zg_{R_{𝔪}} embeds homeomorphically into Zg_{R} as a closed subset, it doesn’t matter whether we take closures in Zg_{R} or Zg_{R_{𝔪}}.

Lemma 5.3. Let b, c, d ∈ R, 𝔪 be a maximal ideal of R, γ ∈ R\{0} and δ ∈ 𝔪\{0}.

(1) R_{𝔪}/γ𝔪R_{𝔪} opens x = x / xd = 0 if and only if γ ∈ dR_{𝔪}.

(2) R_{𝔪}/γ𝔪R_{𝔪} opens xb = 0 / c|x if and only if c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bc ∈ γ𝔪R_{𝔪}.

(3) 𝔪R_{𝔪}/δR_{𝔪} opens x = x / xd = 0 if and only if δ ∈ d𝔪R_{𝔪}.

(4) 𝔪R_{𝔪}/δR_{𝔪} opens xb = 0 / c|x if and only if c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bc ∈ δR_{𝔪}.

Proof. Each claim can be deduced directly from Fact 5.2. □

Moreover, Fact 5.2 implies that for γ ∈ 𝔪\{0}, R_{𝔪}/γR_{𝔪} opens x = x / xd = 0 if and only if d ∉ γR_{𝔪}, and R_{𝔪}/γR_{𝔪} opens xb = 0 / c|x if and only if b ∈ 𝔪R_{𝔪}, c ∈ 𝔪R_{𝔪} and bc ∈ γ𝔪R_{𝔪}.

Lemma 5.4. Let b, c, d ∈ R, 𝔪 be a maximal ideal of R, γ ∈ R\{0} and δ ∈ 𝔪\{0}.

(1) x = x / xd = 0 is a minimal pair for R_{𝔪}/γ𝔪R_{𝔪} if and only if γR_{𝔪} = dR_{𝔪}.

(2) xb = 0 / c|x is a minimal pair for R_{𝔪}/γ𝔪R_{𝔪} if and only if γ ∉ 𝔪, b ∈ 𝔪 and c ∈ 𝔪.

(3) x = x / xd = 0 is never a minimal pair for 𝔪R_{𝔪}/δR_{𝔪}.

(4) xb = 0 / c|x is a minimal pair for 𝔪R_{𝔪}/δR_{𝔪} if and only if c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bcR_{𝔪} = δR_{𝔪}.

Proof. Recall, [17, Corollary 8.12], that if N is an indecomposable pure injective R-module and φ/ψ is an N-minimal pair then (φ/ψ) isolates N in its Ziegler closure.

(1) If γR_{𝔪} = dR_{𝔪} then, for r ∈ R_{𝔪}, r + γ𝔪R_{𝔪} satisfies xd = 0 if and only if r ∈ 𝔪R_{𝔪}. Thus, as an R-module, x = x/xd = 0 evaluated at R_{𝔪}/γ𝔪R_{𝔪} is isomorphic to the simple R-module R/𝔪. So x = x/xd = 0 is a minimal pair for R_{𝔪}/γ𝔪R_{𝔪}.

For the converse, suppose that R_{𝔪}/γ𝔪R_{𝔪} opens x = x/xd = 0, so γ ∈ dR_{𝔪}. If d ∉ γR_{𝔪} then PE(R_{𝔪}/γR_{𝔪}) opens x = x/xd = 0. Since PE(R_{𝔪}/γR_{𝔪}) is in the Ziegler closure of PE(R_{𝔪}/γ𝔪R_{𝔪}), this implies that x = x/xd = 0 is not a PE(R_{𝔪}/γ𝔪R_{𝔪})-minimal pair and hence also not a R_{𝔪}/γ𝔪R_{𝔪}-minimal pair.

(2) Suppose γ ∉ 𝔪, b, c ∈ 𝔪. Since γ ∉ 𝔪, R_{𝔪}/γ𝔪R_{𝔪} = R/𝔪 is a simple R-module. Therefore xb = 0 / c|x is a R_{𝔪}/γ𝔪R_{𝔪} = R/𝔪-minimal pair if and only if R_{𝔪}/γ𝔪R_{𝔪} opens xb = 0 / c|x. That xb = 0 / c|x is a R_{𝔪}/γ𝔪R_{𝔪} = R/𝔪-minimal pair now follows from (2) in Lemma 5.3.

Suppose that xb = 0 / c|x is an R_{𝔪}/γ𝔪R_{𝔪}-minimal pair. Again from (2) in Lemma 5.3, c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bc ∈ γ𝔪R_{𝔪}. Now, if γ ∈ 𝔪 then PE(R_{𝔪}/γR_{𝔪}) opens xb = 0 / c|x. Hence xb = 0 / c|x is not a R_{𝔪}/γ𝔪R_{𝔪}- minimal pair. So γ ∉ 𝔪.

(3) The module 𝔪R_{𝔪}/δR_{𝔪} opens x = x / xd = 0 if and only if d ∉ δR_{𝔪}, and R_{𝔪}/δR_{𝔪} opens x = x/xd = 0 if and only if d ∉ δR_{𝔪}. Since PE(R_{𝔪}/δR_{𝔪}) is in the closure of PE(𝔪R_{𝔪}/δR_{𝔪}), x = x / xd = 0 is never a 𝔪R_{𝔪}/δR_{𝔪}- minimal pair.

(4) Suppose that c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bcR_{𝔪} = δR_{𝔪}. Then the solution set of xb = 0 in 𝔪R_{𝔪}/δR_{𝔪} is cR_{𝔪}/δR_{𝔪} and the solution set of c|x in 𝔪R_{𝔪}/δR_{𝔪} is c𝔪R_{𝔪}/δR_{𝔪}. So xb = 0 / c|x evaluated at 𝔪R_{𝔪}/δR_{𝔪} is the simple R-module R/𝔪 and hence xb = 0 / c|x is a 𝔪R_{𝔪}/δR_{𝔪}-minimal pair.

Suppose xb = 0 / c|x is a 𝔪R_{𝔪}/δR_{𝔪}-minimal pair. By (4) in Lemma 5.3, c ∈ 𝔪R_{𝔪}, b ∈ 𝔪R_{𝔪} and bc ∈ δR_{𝔪}. Suppose, for a contradiction, that bc ∈ δ𝔪R_{𝔪}. Then R_{𝔪}/δR_{𝔪} opens xb = 0 / c|x. So, we can argue as in (1) and (2) that xb = 0 / c|x is not a 𝔪R_{𝔪}/δR_{𝔪}-minimal pair. □

Now, still for R a Prüfer domain, let us come back to the indecomposable pure injective R-modules of the form PE(𝔪R_{𝔪}/βηR_{𝔪}) and PE(R_{𝔪}/γ𝔪R_{𝔪}) where R/𝔪 is finite, β, η ∈ 𝔪\{0} and γ ∈ R\{0} (those admitting a pp-pair φ/ψ with |φ/ψ(N)| finite but not equal to 1).

The value of |φ/ψ(N)| for a pp-pair φ/ψ when N is one of the above R_{𝔪}-uniserial modules will be determined by conditions of the form a ∈ bR_{𝔪} and a ∈ b𝔪R_{𝔪} with a, b ∈ R. The following lemma, together with Fact 2.1, allows us to convert such conditions into conditions of the form c ∈ 𝔪 (see also the previous Remark 2.4).

Lemma 5.5. Let a, b ∈ R\{0} and let α, r, s ∈ R be such that bα = as and a(α − 1) = br. Then

(1) b ∈ aR_{𝔪} if and only if α ∉ 𝔪 or r ∉ 𝔪;

(2) a ∈ b𝔪R_{𝔪} if and only if α ∈ 𝔪 and r ∈ 𝔪.

Proof. (1) is [6, Lemma 5.5]. (2) follows from (1) since a ∈ b𝔪R_{𝔪} if and only if b ∉ aR_{𝔪}. □

Note that, see [6] just after Lemma 5.5, over a Bézout domain things become even simpler.

This leads us to consider what we call a condition on a maximal ideal (of R), that is, a condition of the form r ∈ M where r ∈ R and M is a variable for a maximal ideal (of R). Let 𝔹 denote the set of Boolean combinations of these conditions. We will say that a maximal ideal 𝔪 of R satisfies such a Boolean combination Δ if when we replace all instances of M by 𝔪, Δ is true in R.

Any Δ ∈ 𝔹 is equivalent to a disjunction of conditions of the form

⋀_{i=1}^{l} a_{i} ∈ M ∧ b ∉ M

for some l ∈ ℕ.

To see this first put Δ into disjunctive normal form and then note that a condition of the form ⋀_{i=1}^{l} b_{i} ∉ M is equivalent to ∏_{i=1}^{l} b_{i} ∉ M.

Note that, when R is Bézout, also a conjunction of the form ⋀_{i=1}^{l} a_{i} ∈ M is equivalent to a single condition a ∈ M where a is the greatest common divisor of a_{1}, …, a_{l}. So, when R is Bézout, each Δ ∈ 𝔹 is equivalent to a disjunction of conditions of the form a ∈ M ∧ b ∉ M.

Now, for every Δ in 𝔹, let PP_{0}(R, Δ) denote the set of all (p, n) ∈ ℙ × ℕ such that there exist s, k_{1}, …, k_{s} ∈ ℕ and maximal ideals 𝔪_{1}, …, 𝔪_{s} of R such that n ∈ Span_{ℕ_{0}}{k_{1}, …, k_{s}} and for all i = 1, …, s,

(1) |R/𝔪_{i}| = p^{k_{i}},

(2) 𝔪_{i} satisfies Δ.

Let PP_{0}(R) be the set of all (p, n, Δ) ∈ ℙ × ℕ × 𝔹 such that (p, n) ∈ PP_{0}(R, Δ).

Lemma 5.6. Suppose that P P^{⋆}(R) is recursive. Then P P_{0}(R) is recursive.

Proof. Let Δ have the form ⋁_{i=1}^{m}(⋀_{h=1}^{l} a_{ih} ∈ M ∧ b_{i} ∉ M) with a_{ih}, b_{i} ∈ R. This can be assumed without loss of generality, adding if necessary 0 for a_{ih} and 1 for b_{i}.

We will now show that (p, n, Δ) ∈ PP_{0}(R) if and only if there exists (δ_{1}, …, δ_{m}) ∈ (ℕ_{0})^{m} such that ∑_{i=1}^{m} δ_{i} = n and for all 1 ≤ i ≤ l, either (p, δ_{i}, a_{i1}, …, a_{il}, b_{i}) ∈ PP^{⋆}(R) or δ_{i} = 0. Since the set of (δ_{1}, …, δ_{m}) ∈ (ℕ_{0})^{m} such that ∑_{i=1}^{m} δ_{i} = n is finite and computable given n, this will imply that if PP^{⋆}(R) is recursive then so is PP_{0}(R).

Suppose that n = ∑_{i=1}^{m} δ_{i}, each δ_{i} ∈ ℕ_{0} and for all 1 ≤ i ≤ m, either (p, δ_{i}, a_{i1}, …, a_{il}, b_{i}) ∈ PP^{⋆}(R) or δ_{i} = 0. So, for each 1 ≤ i ≤ m with δ_{i} ≠ 0, there exist k_{i1}, … k_{is_{i}} ∈ ℕ such that δ_{i} ∈ Span_{ℕ_{0}}{k_{i1}, … k_{is_{i}}} and maximal ideals 𝔪_{i1}, …, 𝔪_{is_{i}} such that |R/𝔪_{ij}| = p^{k_{ij}}, a_{ih} ∈ 𝔪_{ij} for 1 ≤ h ≤ l and b_{i} ∉ 𝔪_{ij}. Thus, for 1 ≤ i ≤ m with δ_{i} ≠ 0 and 1 ≤ j ≤ s_{i}, 𝔪_{ij} satisfies Δ, |R/𝔪_{ij}| = p^{k_{ij}} and n ∈ Span_{ℕ_{0}}{k_{ij} | 1 ≤ i ≤ m and 1 ≤ j ≤ s_{i}}.

Now suppose that (p, n) ∈ ℙ × ℕ and that there exist k_{1}, …, k_{s} ∈ ℕ such that n ∈ Span_{ℕ_{0}}{k_{1}, …, k_{s}} and maximal ideals 𝔪_{1}, …, 𝔪_{s} such that |R/𝔪_{j}| = p^{k_{j}} and 𝔪_{j} satisfies Δ for 1 ≤ j ≤ s. Let λ_{1}, … λ_{s} ∈ ℕ_{0} be such that n = ∑_{j=1}^{s} λ_{j}k_{j}. We may partition {1, …, s} into sets A_{1}, … A_{m} such that, for all 1 ≤ j ≤ s and 1 ≤ i ≤ m, j ∈ A_{i} implies that 𝔪_{j} satisfies a_{ih} ∈ 𝔪_{j} for 1 ≤ h ≤ l and b_{i} ∉ 𝔪_{j}. Let δ_{i} := ∑_{j∈A_{i}} λ_{j}k_{j}. If δ_{i} ≠ 0 then (p, δ_{i}, a_{i1}, … a_{il}, b_{i}) ∈ PP^{⋆}(R) and ∑_{i=1}^{m} δ_{i} = n as required. □

The next definition describes the families of modules we are going to deal with. Indeed the summands of these families were already treated at least implicitly in this section.

Definition 5.7.

(S_{γ}) For γ ∈ R\{0}, let S_{γ} be the set of R-modules of the form ⊕_{i=1}^{m}N_{γ}(𝔪_{𝔦}) where m ∈ ℕ and 𝔪_{1}, …, 𝔪_{m} are maximal ideals of R.

(S′_{β,η}) For β, η ∈ R\{0}, let S′_{β,η} be the set of R-modules of the form ⊕_{i=1}^{m}N′_{β,η}(𝔪_{𝔦}) where m ∈ ℕ and 𝔪_{1}, …, 𝔪_{m} are maximal ideals of R containing both β and η.

(T_{β,η}) For β, η ∈ R\{0}, let T_{β,η} be the set of R-modules of the form ⊕_{i=1}^{m}R/𝔪_{i} where m ∈ ℕ and 𝔪_{1}, …, 𝔪_{m} are maximal ideals of R containing both β and η.

The preparation of the proof of the main theorem culminates in the next proposition.

Proposition 5.8. Let R be a Prüfer domain such that each localization of R at a maximal ideals has dense value group. Suppose that R is effectively given and PP^{⋆}(R) is recursive.

(a) Fix γ ∈ R\{0}. Then there is an algorithm which, given • p ∈ ℙ,

• pp-pairs φ_{i}/ψ_{i} for 1 ≤ i ≤ t + s and φ_{2,j}/ψ_{2,j} for 1 ≤ j ≤ u of the form xb = 0 / c|x or x = x / xd = 0, with b, c, d ∈ R,

• positive integers w, n_{i} for 1 ≤ i ≤ t and l_{i} for t + 1 ≤ i ≤ t + s, answers whether there exists an R-module N ∈ S_{γ} satisfying the sentences | x = x / xγ = 0 | = p^{w} and

⋀_{i=1}^{t} |φ_{i}/ψ_{i}| = p^{n_{i}} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{i=t+1}^{t+s} |φ_{i}/ψ_{i}| ≥ p^{l_{i}}.

(b) Fix β, η ∈ R\{0}. Then there is an algorithm which, given • p ∈ ℙ,

• pp-pairs φ_{i}/ψ_{i} for 1 ≤ i ≤ t + s and φ_{2,j}/ψ_{2,j} for 1 ≤ j ≤ u of the form xb = 0 / c|x or x = x / xd = 0, with b, c, d ∈ R,

• positive integers w, n_{i} for 1 ≤ i ≤ t and l_{i} for t + 1 ≤ i ≤ t + s, answers whether there exists an R-module M ∈ T_{β,η} satisfying the sentences | xη = 0 / β|x | = p^{w} and

⋀_{i=1}^{t} |φ_{i}/ψ_{i}| = p^{n_{i}} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{i=t+1}^{t+s} |φ_{i}/ψ_{i}| ≥ p^{l_{i}}.

(c) Fix β, η ∈ R\{0}. Then there is an algorithm which, given • p ∈ ℙ,

• pp-pairs φ_{i}/ψ_{i} for 1 ≤ i ≤ t + s and φ_{2,j}/ψ_{2,j} for 1 ≤ j ≤ u of the form xb = 0 / c|x or x = x / xd = 0, with b, c, d ∈ R,

• positive integers w, n_{i} for 1 ≤ i ≤ t and l_{i} for t + 1 ≤ i ≤ t + s, answers whether there exists an R-module N′ ∈ S′_{β,γ} satisfying the sentences | xη = 0 / β|x | = p^{w} and

⋀_{i=1}^{t} |φ_{i}/ψ_{i}| = p^{n_{i}} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{i=t+1}^{t+s} |φ_{i}/ψ_{i}| ≥ p^{l_{i}}.

Proof. We provide the proof of (a), and then we explain how it can be adapted to show (b) and (c).

Let Γ be the set of functions f : {1, …, t + s} → {0, 1, ∞}. Let X be the set of pairs (Γ′, δ) where Γ′ ⊆ Γ and δ : Γ′ → ℕ satisfy

(1) ∑_{f∈Γ′} δ(f) = w,

(2) for 1 ≤ i ≤ t and f ∈ Γ′, f(i) ≠ ∞,

(3) for 1 ≤ i ≤ t, ∑_{f∈Γ′, f(i)=1} δ(f) = n_{i},

(4) for t + 1 ≤ i ≤ t + s, either ∑_{f∈Γ′, f(i)=1} δ(f) ≥ l_{i} or there exists f ∈ Γ′ such that f(i) = ∞.

The first condition ensures that the set X is finite and not empty.

Recall that, for each maximal ideal 𝔪, N_{γ}(𝔪) = R_{𝔪}/γ𝔪R_{𝔪}. For each f ∈ Γ we define Δ^{f} ∈ 𝔹, so that a maximal ideal 𝔪 satisfies Δ^{f} if and only if, for i = 1, …, t + s,

(S1) f(i) = 0 implies | φ_{i} / ψ_{i} (N_{γ}(𝔪)) | = 1,

(S2) f(i) = 1 implies φ_{i}/ψ_{i} is an N_{γ}(𝔪)-minimal pair,

(S3) f(i) = ∞ implies | φ_{i} / ψ_{i} (N_{γ}(𝔪)) | > 1 and φ_{i}/ψ_{i} is not an N_{γ}(𝔪)-

minimal pair, (S4) | φ_{2,j} / ψ_{2,j} (N_{γ}(𝔪)) | = 1 for 1 ≤ j ≤ u.

This can be done using Lemmas 5.5, 5.3, (1) and (2), and 5.4, (1) and (2). Note that the conditions in (S4) do not depend of f, and yet are assumed to be part of Δ^{f}.

We claim that there exists N ∈ S_{γ} satisfying the sentences | x = x / xγ = 0 | = p^{w} and

⋀_{i=1}^{t} |φ_{i}/ψ_{i}| = p^{n_{i}} ∧ ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{i=t+1}^{s} |φ_{i}/ψ_{i}| ≥ p^{l_{i}}

if and only if there exists (Γ′, δ) ∈ X such that (p, δ(f), Δ^{f}) ∈ PP_{0}(R) for all f ∈ Γ′ and w = ∑_{f∈Γ′} δ(f). Since, by Lemma 5.6, P P^{⋆}(R) recursive implies P P_{0}(R) recursive, this is enough to prove the proposition.

We first prove the forward direction. Suppose N ∈ S_{γ} satisfies the required sentences. By definition N = ⊕_{h=1}^{m}N_{γ}(𝔪_{h}) for some maximal ideals 𝔪_{1}, …, 𝔪_{m}. Recall that N satisfies | x = x / xγ = 0 | = p^{w}. On the other hand, for each h = 1, …, m, |N_{γ}(𝔪_{h}) / (xγ = 0)(N_{γ}(𝔪_{h}))| = |R_{𝔪_{h}}/𝔪_{h}R_{𝔪_{h}}| = |R/𝔪_{h}|, whence |R/𝔪_{h}| is finite and indeed a power of p. For 1 ≤ h ≤ m, put |R/𝔪_{h}| = p^{k_{h}}. Then ∑_{h=1}^{m} k_{h} = w.

Each 𝔪_{h} satisfies Δ^{f} for exactly one f ∈ Γ because | φ_{2j} / ψ_{2,j}(N_{γ}(𝔪_{h})) | = 1 for 1 ≤ j ≤ u and Δ^{f} simply specifies, for 1 ≤ i ≤ t + s and 𝔪 a maximal ideal, whether | φ_{i} / ψ_{i} (N_{γ}(𝔪)) | = 1, φ_{i}/ψ_{i} is an N_{γ}(𝔪)-minimal pair or neither of these things is true. Let Γ′ be the set of f ∈ Γ such that 𝔪_{h} satisfies Δ^{f} for some 1 ≤ h ≤ m. Since |φ_{i}/ψ_{i} (N)| is finite for 1 ≤ i ≤ t, for all 1 ≤ h ≤ m, |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| is finite. Therefore f ∈ Γ′ implies f(i) ≠ ∞ for 1 ≤ i ≤ t.

For each f ∈ Γ′, let H_{f} be the set of 1 ≤ h ≤ m such that 𝔪_{h} satisfies Δ^{f}. Define δ : Γ′ → ℕ by setting δ(f) := ∑_{h∈H_{f}} k_{h} for every f ∈ Γ′.

We show that (Γ′, δ) ∈ X. We have already seen that w = ∑_{h=1}^{m} k_{h}. Since for each 1 ≤ h ≤ m, h ∈ H_{f} for exactly one f ∈ Γ′, ∑_{f∈Γ′} δ(f) = ∑_{h=1}^{m} k_{h} = w. So δ satisfies condition (1).

We have already proved that Γ′ satisfies condition (2). So let us pass to (3).

Let 1 ≤ i ≤ t. Since N satisfies |φ_{i}(N)/ψ_{i}(N)| = p^{n_{i}}, Lemma 5.1 implies that, for all 1 ≤ h ≤ m, either φ_{i}/ψ_{i} is an N_{γ}(𝔪_{h})-minimal pair or |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| = 1. Let T_{i} be the set of 1 ≤ h ≤ m such that φ_{i}/ψ_{i} is a N_{γ}(𝔪_{h})-minimal pair. So ∑_{h∈T_{i}} k_{h} = n_{i}. Thus

∑_{\begin{matrix}f∈Γ′ \\ f(i)=1 \end{matrix}} δ(f) = ∑_{\begin{matrix}f∈Γ′ \\ f(i)=1 \end{matrix}} ∑_{h∈H_{f}} k_{h} = ∑_{h∈T_{i}} k_{h} = n_{i}.

So δ satisfies condition (3).

Finally let us deal with (4). Let t + 1 ≤ i ≤ t + s. If there exists some 1 ≤ h ≤ m such that φ_{i}/ψ_{i} is not an N_{γ}(𝔪_{h})-minimal pair and |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| ≠ 1 then there is an f ∈ Γ′ such that f(i) = ∞. In this case, (Γ′, δ) satisfies condition (4). So suppose that for all 1 ≤ h ≤ m, φ_{i}/ψ_{i} is an N_{γ}(𝔪_{h})-minimal pair or |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| = 1. Let T_{i} be the set of 1 ≤ h ≤ m such that φ_{i}/ψ_{i} is a N_{γ}(𝔪_{h})-minimal pair. So ∑_{h∈T_{i}} k_{h} ≥ l_{i}. Thus

∑_{\begin{matrix}f∈Γ′ \\ f(i)=1 \end{matrix}} ∑_{h∈H_{f}} k_{h} = ∑_{h∈T_{i}} k_{h} ≥ l_{i}.

So δ satisfies condition (4).

We now just need to confirm that (p, δ(f), Δ^{f}) ∈ PP_{0}(R) for all f ∈ Γ′. By definition δ(f) = ∑_{h∈H_{f}} k_{h}. So δ(f) ∈ Span_{ℕ_{0}}{k_{h} | h ∈ H_{f}}. By definition for each h ∈ H_{f}, 𝔪_{h} satisfies Δ^{f}. So (p, δ(f), Δ^{f}) ∈ PP_{0}(R) for all f ∈ Γ′.

We now prove the reverse direction. Suppose that there exists a pair (Γ′, δ) ∈ X such that (p, δ(f), Δ^{f}) ∈ PP_{0}(R) for all f ∈ Γ′ and w = ∑_{f∈Γ′} δ(f). Using the definition of PP_{0}(R), for each f ∈ Γ′, pick maximal ideals 𝔪_{1}^{f}, … 𝔪_{m_{f}}^{f} such that 𝔪_{h}^{f} satisfies Δ^{f} for 1 ≤ h ≤ m_{f} and

δ(f) ∈ Span_{ℕ_{0}}{k_{h}^{f} | 1 ≤ h ≤ m_{f}} where |R/𝔪_{h}^{f}| = p^{k_{h}^{f}}. For 1 ≤ h ≤ m, let λ_{h} ∈ ℕ_{0} be such that δ(f) = ∑_{h=1}^{m_{f}} λ_{h}k_{h}^{f}.

Let N^{f} := ⊕_{h=1}^{m_{f}}N_{γ}(𝔪_{h}^{f})^{λ_{h}}.

Note that |x = x / xγ = 0 (N^{f})| = p^{δ(f)}.

By definition of Δ^{f}, for all 1 ≤ j ≤ u and 1 ≤ h ≤ m_{f}, |φ_{2,j} /ψ_{2,j} (N_{γ}(𝔪_{h}^{f}))| = 1. Thus |φ_{2,j} / ψ_{2,j} (N^{f})| = 1 for all 1 ≤ j ≤ u. Again by definition of Δ^{f}, if 1 ≤ i ≤ t + s and f(i) = 0, then |φ_{i}/ψ_{i} (N^{f})| = 1. If 1 ≤ i ≤ t + s and f(i) = 1, then φ_{i}/ψ_{i} is an N_{γ}(𝔪_{h})-minimal pair for all 1 ≤ h ≤ m_{f}. Thus |φ_{i}/ψ_{i} (N^{f})| = p^{∑_{h=1}^{m_{f}} λ_{h}k_{h}^{f}} = p^{δ(f)}. Finally, if f(i) = ∞ then |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| ≠ 1 and φ_{i}/ψ_{i} is not a N_{γ}(𝔪_{h})-minimal pair for 1 ≤ h ≤ m_{f}. Thus |φ_{i} / ψ_{i} (N_{γ}(𝔪_{h}))| is infinite. Therefore, if f(i) = ∞ then |φ_{i}/ψ_{i} (N^{f})| is infinite.

Let N := ⊕_{f∈Γ′}N^{f}.

For 1 ≤ j ≤ u, |φ_{2,j}/ψ_{2,j}(N)| = 1 since |φ_{2,j} / ψ_{2,j} (N^{f})| = 1 for each f ∈ Γ′.

Let 1 ≤ i ≤ t. If f ∈ Γ′, f(i) ≠ ∞. Since (Γ′, δ) ∈ X, ∑_{i∈Γ′ f(i)=1} δ(f) = n_{i}. Thus

|φ_{i}/ψ_{i}(N)| = ∏_{f∈Γ′} |φ_{i} / ψ_{i} (N^{f})| =

∏_{\begin{matrix}f∈Γ′ \\ f(i)=1 \end{matrix}} |φ_{i} / ψ_{i} (N^{f})| = ∏_{\begin{matrix}f∈Γ′ \\ f(i)=1 \end{matrix}} p^{δ(f)} = p^{n_{i}}.

Let t + 1 ≤ i ≤ t + s. If f(i) = ∞ for some f ∈ Γ′ then |φ_{i} / ψ_{i} (N^{f})| is infinite and hence |φ_{i}/ψ_{i}(N)| is infinite. So |φ_{i}/ψ_{i}(N)| ≥ l_{i}. So suppose that f(i) ≠ ∞ for all f ∈ Γ′. That |φ_{i}/ψ_{i}(N)| ≥ l_{i} now follows as in the previous paragraph.

Finally

|x = x / xγ = 0 (N)| = ∏_{f∈Γ′} |x = x / xγ = 0 (N^{f})| =

= ∏_{f∈Γ′} p^{δ(f)} = p^{∑_{f∈Γ′} δ(f)} = p^{w}.

Thus we have shown that N satisfies the required sentences.

This concludes the proof of (a).

For (b), when we define Δ^{f} we need to add the conditions β ∈ M and η ∈ M and use Lemmas 5.3, 5.4, (1)-(2), with γ = 1.

For (c), when we define Δ^{f} we need to add the conditions β ∈ M, η ∈ M and use Lemmas 5.3, 5.4, (3)-(4). □

6. The main theorem

Theorem 6.1. Let R be an effectively given Prüfer domain such that each localization of R at a maximal ideal has dense value group. If both DPR^{⋆}(R) and PP^{⋆}(R) are recursive, then T_{R} is decidable.

Proof. By Theorem 4.1, in order to prove that T_{R} is decidable, it is enough that there is an algorithm which, given a conjunction σ of invariants sentences

(1) |φ_{1,i}/ψ_{1,i}| = H_{i}, 1 ≤ i ≤ t,

(2) |φ_{2,j}/ψ_{2,j}| = 1, 1 ≤ j ≤ u,

(3) |φ_{3,k}/ψ_{3,k}| ≥ E_{k}, 1 ≤ k ≤ s, where t, u, s are non negative integers, H_{i} (1 ≤ i ≤ s) and E_{k} (1 ≤ k ≤ s) are integers > 1 and all the involved pp-pairs have the form xη = 0 / β|x or x = x / xγ = 0 with β, η, γ ∈ R, answers whether there is some R-module M satisfying σ.

For σ a sentence as above, define the exponent of σ to be ∏_{i=1}^{t} H_{i} if (1) is non-empty and 1 otherwise.

Our plan is to describe an algorithm for sentences of exponent 1 and then explain how to algorithmically reduce to the exponent 1 case.

Case 1: the exponent of σ is 1.

So (1) is empty. Then there exists an R-module satisfying σ if and only if there exists an R-module satisfying

σ′ := ⋀_{j=1}^{u} |φ_{2,j}/ψ_{2,j}| = 1 ∧ ⋀_{k=1}^{s} |φ_{3,k}/ψ_{3,k}| > 1.

This is because if M satisfies σ′ then M^{ℵ_{0}} satisfies σ. We may now proceed as in [6, Theorem 7.1].

Case 2: the exponent of σ is strictly greater than 1.

We now describe an algorithm which given a sentence σ with exponent strictly greater than 1 produces finitely many sentences σ_{1}, …, σ_{l} such that their exponents are strictly smaller than that of σ and there exists an Rmodule satisfying σ if and only if there exists an R-module satisfying one of the sentences σ_{1}, …, σ_{l}.

Given a sentence σ, we can apply this algorithm finitely many times to produce sentences σ_{1}, …, σ_{l} with exponent 1 such that σ is true in some R-module if and only if one of the sentences σ_{1}, …, σ_{l} is true in some Rmodule. So we are done.

Let p ∈ ℙ divide H_{1} and h_{1} ∈ ℕ be maximal such that p^{h_{1}}|H_{1}. We will deal with the cases when φ_{11}/ψ_{11} is of the form xη = 0 / β|x and x = x / xγ = 0 separately.

Subcase 2.1: φ_{11}/ψ_{11} is xη = 0 / β|x.

Let Ω_{σ} be the set of pairs (f, g) of functions f, g : {1, …, t+s} → ℕ_{0}∪{∞} such that

• f(1) = 0 (respectively g(1) = 0) implies f(i) = 0 (respectively g(i) = 0) for 1 ≤ i ≤ t + s,

• f(1) + g(1) = h_{1},

• p^{f(i)+g(i)}|H_{i} for 2 ≤ i ≤ t,

• either p^{f(t+k)} ≤ E_{k} or f(t + k) = ∞ for 1 ≤ k ≤ s, and

• either p^{g(t+k)} ≤ E_{k} or g(t + k) = ∞ for 1 ≤ k ≤ s.

For each (f, g) ∈ Ω_{σ}, let σ_{f} (respectively σ_{g}) be the conjunction of invariants sentences

(1) |φ_{1,i}/ψ_{1,i}| = p^{f(i)} (respectively = p^{g(i)}) for 1 ≤ i ≤ t,

(2) |φ_{2,j}/ψ_{2,j}| = 1 for 1 ≤ j ≤ u, (3) |φ_{3,k}/ψ_{3,k}| = p^{f(t+k)} (respectively = p^{g(t+k)}) for 1 ≤ k ≤ s and

f(t + k) ≠ ∞ (respectively g(t + k) ≠ ∞),

(4) |φ_{3,k}/ψ_{3,k}| ≥ p^{⌈^{log_{p} E_{k}}⌉} for 1 ≤ k ≤ s and f(t + k) = ∞ (respectively g(t + k) = ∞).

Here, for r a real number, ⌈r⌉ denotes the minimal integer greater than or equal to r . For each (f, g) ∈ Ω_{σ}, let σ′_{(f,g)} be the conjunction of invariants sentences

(1) |φ_{1,i}/ψ_{1,i}| = H_{i}/p^{f(i)+g(i)} for 1 ≤ i ≤ t,

(2) |φ_{2,j}/ψ_{2,j}| = 1 for 1 ≤ j ≤ u,

(3) |φ_{3,k}/ψ_{3,k}| ≥ ⌈(E_{k}/p^{f(t+k)+g(t+k)})⌉ for 1 ≤ k ≤ s, f(t + k) ≠ ∞ and g(t + k) ≠ ∞.

Note that, for each (f, g) ∈ Ω_{σ}, the exponent of σ′_{(f,g)} is strictly less that the exponent of σ since p^{f(1)+g(1)} = p^{h_{1}} > 1.

If p^{f(t+k)+g(t+k)} ≥ E_{k} then ⌈(E_{k}/p^{f(t+k)+g(t+k)})⌉ = 1. For that value of k, |φ_{3,k}/ψ_{3,k}| ≥ ⌈(E_{k}/p^{f(t+k)+g(t+k)})⌉ is satisfied by all R-modules and so this condition may be removed.

Now we claim that there exists an R-module M satisfying σ if and only if there exists (f, g) ∈ Ω_{σ}, N_{f} ∈ S′_{β,η} satisfying σ_{f}, N_{g} ∈ T_{β,η} satisfying σ_{g} and an R-module M′ satisfying σ′_{(f,g)}.

The reverse direction follows directly from the definitions of σ_{f}, σ_{g} and σ′_{(f,g)}.

So assume that there is an R-module M satisfying σ. We may suppose that M is a finite direct sum of indecomposable pure injective modules ⨁_{h=1}^{m} N_{h}. So ∏_{h=1}^{m} |xη = 0 / β|x (N_{h})| = H_{1}. In particular, |xη = 0 / β|x (N_{h})| is finite for each 1 ≤ h ≤ m. For each 1 ≤ h ≤ m, |xη = 0 / β|x (N_{h})| is either 1 or q^{l} for some prime q. If |xη = 0 / β|x (N_{h})| = q^{l} for some l ∈ ℕ then, by Lemma 5.1, xη = 0 / β|x is an N_{h}-minimal pair. So N_{h} is either the pure injective hull of R/𝔪_{h} or 𝔪_{h}R_{𝔪_{h}}/βηR_{𝔪_{h}} for some maximal ideal 𝔪_{h} with β, η ∈ 𝔪_{h}\{0} and such that |R/𝔪_{h}| = q^{l}.

Let M′ be the direct sum of the modules N_{h} for 1 ≤ h ≤ m such that p does not divide |xη = 0 / β|x (N_{h})|. Let L (respectively N) be the direct sum of the modules N_{h} for 1 ≤ h ≤ m such that p divides |xη = 0 / β|x (N_{h})| and N_{h} the pure injective hull of R/𝔪_{h} with β, η ∈ 𝔪_{h} (respectively 𝔪_{h}R_{𝔪_{h}}/βηR_{𝔪_{h}} with β, η ∈ 𝔪_{h}\{0}).

Note that |xη = 0 / β|x(L⊕N)| = p^{h_{1}} and for any pp-pair φ/ψ, |φ/ψ(L⊕ N)| is either 1, a power of p or infinite. Moreover, if |xη = 0 / β|x (L)| = 1 (respectively |xη = 0 / β|x (N)| = 1) then L = 0 (respectively N = 0).

Define (f, g) ∈ Ω_{σ} by setting

f(i) = log_{p} |φ_{1i}/ψ_{1i}(N)| and g(i) = log_{p} |φ_{1i}/ψ_{1i}(L)|

for 1 ≤ i ≤ t and, for 1 ≤ k ≤ s,

f(t + k) = \{ \begin{matrix}log_{p} |φ_{3,k}/ψ_{3,k}(N)|, & if |φ_{3,k}/ψ_{3,k}(N)| ≤ E_{k}, \\ ∞, & otherwise \end{matrix}

and

g(t + k) = \{ \begin{matrix}log_{p} |φ_{3,k}/ψ_{3,k}(L)|, & if |φ_{3,k}/ψ_{3,k}(L)| ≤ E_{k}, \\ ∞, & otherwise. \end{matrix}

Then N satisfies σ_{f} and L satisfies σ_{g}. That (f, g) ∈ Ω_{σ} follows from the definition of N and L and the above discussion. We now just need to check that M′ satisfies σ′_{(f,g)}. For 1 ≤ i ≤ t,

|φ_{1,i}/ψ_{1,i}(N)| · |φ_{1,i}/ψ_{1,i}(L)| · |φ_{1,i}/ψ_{1,i}(M′)| =
= p^{f(i)+g(i)} · |φ_{1i}/ψ_{1i}(M′)| = H_{i}.

So M′ satisfies the invariants sentences in (1) of the definition of σ′_{(f,g)}. Finally, suppose 1 ≤ k ≤ s, f(t + k) ≠ ∞ and g(t + k) ≠ ∞. Then

p^{f(t+k)+g(t+k)} · |φ_{3,k}/ψ_{3,k}(M′)| =
= |φ_{3,k}/ψ_{3,k}(N)| · |φ_{3,k}/ψ_{3,k}(L)| · |φ_{3,k}/ψ_{3,k}(M′)| ≥ E_{k}.

Therefore |φ_{3,k}/ψ_{3,k}(M′)| ≥ ⌈E_{k}/p^{f(t+k)+g(t+k)}⌉. So M′ satisfies the invariants sentences in (2) of the definition of σ′_{(f,g)}.

Thus we have proved the claim.

By definition, N is elementary equivalent to a module in S′_{β,η} and L is elementary equivalent to a module in T_{β,η}. If f(1) = 0 (respectively g(1) = 0) then σ_{f} (respectively σ_{g}) holds for the zero module. If f(1) ≠ 0 then, by Proposition 5.8(c), there is an algorithm which given the sentence σ_{f} answers whether there exists N ∈ S′_{β,η} satisfying σ_{f}. Similarly, if g(1) ≠ 0 then, by Proposition 5.8(b), there is an algorithm which given the sentence σ_{g} answers whether there exists L ∈ T_{β,η} satisfying σ_{g}.

Now σ is true in some R-module if and only if there exists (f, g) ∈ Ω_{σ} such that σ_{f} is true in some module in S′_{β,η} and σ_{g} is true in some module in T_{β,γ} and σ′_{(f,g)} is true in some R-module.

Subcase 2.2: φ_{11}/ψ_{11} is x = x / xγ = 0.

Let Ω_{σ} be the set of functions f : {1, …, t + s} → ℕ_{0} ∪ {∞} such that • f(1) = h_{1},

• p^{f(i)}|H_{i} for 2 ≤ i ≤ t,

• either p^{f(t+k)} ≤ E_{k} or f(t + k) = ∞ for 1 ≤ k ≤ s. For each f ∈ Ω_{σ}, let σ_{f} be the conjunction of invariants sentences

(1) |φ_{1,i}/ψ_{1,i}| = p^{f(i)}, for 1 ≤ i ≤ t,

(2) |φ_{2,j}/ψ_{2,j}| = 1 for 1 ≤ j ≤ u,

(3) |φ_{3,k}/ψ_{3,k}| = p^{f(t+k)}, if 1 ≤ k ≤ s and f(t + k) ≠ ∞,

(4) |φ_{3,k}/ψ_{3,k}| ≥ p^{⌈^{log_{p} E_{k}}⌉}, if 1 ≤ k ≤ s and f(t + k) = ∞.

For each f ∈ Ω_{σ}, let σ′_{f} be the conjunction of invariants sentences

(1) |φ_{1,i}/ψ_{1,i}| = H_{i}/p^{f(i)},

(2) |φ_{2,j}/ψ_{2,j}| = 1 for 1 ≤ j ≤ u,

(3) |φ_{3,k}/ψ_{3,k}| ≥ ⌈(E_{k}/p^{f(t+k)})⌉ for f(t + k) ≠ ∞.

Note that if p^{f(t+k)} ≥ E_{k} then ⌈(E_{k}/p^{f(t+k)})⌉ = 1 and so the last condition for that value of k is satisfied by all R-modules.

We claim that there exists an R-module M satisfying σ if and only if there exist f ∈ Ω_{σ}, N_{f} ∈ S_{γ} satisfying σ_{f}, and an R-module M′ satisfying σ′_{f}.

The proof of this claim is as the previous one except we use the fact that the only indecomposable pure injective R-modules N such that x = x / xγ = 0 is an N-minimal pair are the pure injective hulls of N_{γ}(𝔪).

We can now use Proposition 5.8(a) instead of 5.8(b) and (c) to get the required algorithm. □

Recall that, over a Bézout domain R, if PP(R) is recursive then PP^{⋆}(R) is recursive and if DPR(R) is recursive then DPR^{⋆}(R) is recursive (see §1). Moreover, [6, 6.4], if T_{R} is decidable then DPR(R) is recursive.

When R is a Bézout domain, we obtain the following theorem as a corollary to Theorem 6.1 and Theorem 3.2.

Theorem 6.2. Let R be an effectively given Bézout domain such that each localization of R at a maximal ideal has dense value group. Then T_{R} is decidable if and only if both DPR(R) and PP(R) are recursive.

The following remark follows directly from the definition of PP(R).

Remark 6.3. Let q ∈ ℙ and t ∈ ℕ. Let R be a Prüfer domain such that for all maximal ideals 𝔪, |R/𝔪| = q^{t}. Then (p, n, c, d) ∈ PP(R) if and only if p = q, t divides n and there exists some maximal ideal 𝔪 such that c ∈ 𝔪 and d ∉ 𝔪.

When the Krull dimension of R is 1, we can say a bit more.

Proposition 6.4. Let R be an effectively given Bézout domain of Krull dimension 1 such that, when 𝔪 ranges over the maximal ideals of R, either all the residue fields R/𝔪 are infinite, or all the residue fields R/𝔪 are of the same finite cardinality and the localizations R_{𝔪} have dense value group. Then T_{R} is decidable.

Proof. Note that the case of infinite residue fields is already treated in [6, Corollary 6.7]. So we focus on the second case, when the residue fields of R have a common finite size, q^{t} say, where q is a prime and t is a positive integer. It suffices to prove that then both PP(R) and DPR(R) are recursive. It is shown in [8, Lemma 3.3] that, when R is an effectively given Bézout domain with Krull dimension 1, the prime radical relation is recursive.

The argument for DPR(R) is just the same as in the proof of [6, Corollary 6.7]. So we just need to show that PP(R) is recursive.

First we show that Jac(R), the Jacobson radical of R, is a recursive subset of R. If Jac(R) = 0 then it is finite and hence recursive. Suppose Jac(R) ≠ 0. Fix a ∈ Jac(R) non-zero. Then a is contained in all maximal ideals and all non-zero prime ideals are maximal. Therefore rad(aR) = Jac(R). Since the prime radical relation is recursive, Jac(R) is a recursive subset of R.

By 6.3, if all residue fields of R are of the same finite size then (p, n, c, d) ∈ PP(R) if and only if t divides n and there exists some maximal ideal 𝔪 such that c ∈ 𝔪 and d ∉ 𝔪. Therefore, it is enough to show that the set of pairs (c, d) ∈ R^{2} such that there exists some maximal ideal 𝔪 with c ∈ 𝔪 and d ∉ 𝔪 is recursive.

Suppose c ≠ 0. Then c ∈ 𝔪 and d ∉ 𝔪 implies that d ∉ rad(cR). Conversely, if d ∉ rad(cR) then there exists some prime ideal 𝔪 such that c ∈ 𝔪 and d ∉ 𝔪. Since c ≠ 0, 𝔪 is non-zero and hence, since R is a domain with Krull dimension 1, 𝔪 is maximal.

Now, if c = 0 then c is a member of all maximal ideals. So there exists a maximal ideal 𝔪 such that c ∈ 𝔪 and d ∉ 𝔪 if and only if d ∉ J(R).

We have shown that the set of (p, n, c, d) ∈ PP(R) with c ≠ 0 is recursive and that the set of (p, n, 0, d) ∈ PP(R) is recursive. So PP(R) is recursive.

□

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(L. Gregory) Dipartimento di Matematica e Fisica, Università degli Studi della Campania “Luigi Vanvitelli”, Viale Lincoln 5, 81100 Caserta, Italy

E-mail address: lorna.gregory@gmail.com

(S. L’Innocente) University of Camerino, School of Science and Technologies, Division of Mathematics, Via Madonna delle Carceri 9, 62032 Camerino, Italy

E-mail address: sonia.linnocente@unicam.it

(C. Toffalori) University of Camerino, School of Science and Technologies, Division of Mathematics, Via Madonna delle Carceri 9, 62032 Camerino, Italy

E-mail address: carlo.toffalori@unicam.it