Fields Institute Communications Volume 00, 0000

Preorders, rings, lattice-ordered groups and formal power series.

Gérard Leloup

UPRESA 7056 (équipe de logique, Paris VII) and Département de Mathématiques, Faculté des Sciences Avenue Olivier Messiaen B.P. 535 72085 LE MANS CEDEX FRANCE

                              leloup@logique.jussieu.fr

Abstract. We give examples of the links between partially ordered abelian groups and integral domains, by means of groups of divisibility. The definition of valuations can be generalized to any integral domain. Jaffard called those functions “semi-valuations”. We focus here on the case where the group of divisibility is lattice-ordered, i.e., the domain is a Gauss domain. In this case, there is a one-to-one correspondence between the prime ideals which are closed under gcds, and the prime subgroups of the group of divisibility. We examine the case of Bezout domains, unique factorization domains, Prüfer rings, and we give a definition of boolean rings which is based on their groups of divisibility. Finally, we use Ribenboim’s generalized formal power series to construct Gauss domains with given groups of divisibility.

Introduction.

There are several ways of approaching the study of valuations. One way is based on general properties of integral domains. We will point out some aspects of this approach. For some logicians, the language of valued fields is the language of fields together with a one place predicate for the valuation ring. Indeed, if R is a valuation ring with quotient field K and with group of units U, then the value group is isomorphic to (K^{∗}/U, ·), and its positive cone is (R\{0})/U. In this case, the valuation is the quotient map from K^{∗} onto K^{∗}/U. More generally, if R is any integral domain, then (R\{0})/U is the positive cone of a compatible order relation on K^{∗}/U, and the canonical mapping from K^{∗} onto K^{∗}/U is a generalization of a valuation. The ordered group K^{∗}/U is called the group of divisibility of R. Some properties of integral domains can be formulated in terms of properties of their groups of divisibility. For example, this occurs for “being a valuation ring”, “being

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society

2000 Mathematics Subject Classification. 6, 13.

a unique factorization domain”, and, in case of Prüfer rings, “to be a boolean ring”. Many authors have used groups of divisibility to solve ring-theoretic problems. In particular, unique factorization domains, Bezout domains and their overrings have been studied. Such results were often written for experts in ordered-group theory, and their statements require a background in ordered-groups larger than the one assumed here. Among open problems is the characterization of those ordered groups which occur as groups of divisibility. Only partial results are known; the reader can find more properties of groups of divisibility and bibliographies, for example, in [1],

[2] and [17]. The aim of the present work is only to give examples of the kind of information that the group of divisibility gives about an integral domain. So, we only point out the following three aspects.

The correspondence between those prime ideals of an integral domain which are closed under gcds and the prime subgroups of its divisibility group, in the case where the divisibility group is a lattice-ordered group (section 4).

Boolean rings (section 5).

In the lattice-ordered case, a construction of integral domains with a given group of divisibility, similar to the construction of formal power series (section 6).

We begin by recalling Jaffard’s definition of semi-valuations and demi-valuations (section 1). In section 2, we recall some properties of lattice-ordered groups. Then we define a topology on the set of all prime subgroups of a lattice ordered group. Some properties of boolean families of valuation rings, and their proofs, are translated into the language of groups (see [9], [10] and [7]). Section 3 gives complements about “Gauss domains” and “Bezout domains”.

The sources of this work are the following references: [4] ch. 1 to 4, [24], [25], [18], [8], and [9], [10].

1 Semi-valuations.

An ordered abelian group is an abelian group together with a compatible order. Let (Γ, ·) be an abelian group. We know that there is a one-to-one mapping between the family of all compatible orders ≤ on Γ, and the family of all subsets P of Γ closed under · and such that P ∩P^{−1} = {1}. In this case, P = {x ∈ Γ | x ≥ 1}, and x ≤ y if and only if yx^{−1} ∈ P. P is called the positive cone of the ordered group. If (Γ, ≤) is an ordered group, its positive cone will be denoted by Γ_{+}.

Let (Γ, ·) be an abelian group and let Γ_{+} be a subset of Γ closed under · such that Γ_{+} contains 1. Define x ≤ y to mean that yx^{−1} ∈ Γ_{+}. Then ≤ is a preorder relation and it is compatible with ·.

Conversely, if Γ is an abelian group with a compatible preorder relation ≤, let Γ_{+} denote the subset of all elements ≥ 1. Then for all x and y in Γ, x ≤ y iff yx^{−1} ∈ Γ_{+}; furthermore, Γ_{+} contains {1} and is closed under ·.

A preorder relation is an order relation if and only if Γ_{+} ∩ (Γ_{+})^{−1} = 1.

Set U = Γ_{+} ∩ Γ_{+}^{−1}. Then U is a subgroup of Γ.

If · is commutative, then ≤ is compatible with · , hence Γ/U is a preordered group. But (x ≤ y and y ≤ x) ⇔ x · y^{−1} ∈ U, hence Γ/U is an ordered group. In this case, the quotient map is a homomorphism from Γ onto the ordered group Γ/U.

Conversely, let Γ be an abelian group, G be an ordered abelian group, and let v be a homomorphism from Γ onto G, and set Γ_{+} = {x ∈ Γ | v(x) ≥ 1}. Then Γ_{+} is closed under · and contains 1. If v′ is a homomorphism from Γ onto an ordered group G′ such that {x ∈ Γ | v′(x) ≥ 1} = Γ, then (G′, G′_{+}) ≃ (Γ/U, Γ_{+}/U) ≃ (G, G_{+}).

Now, assume that the ordered group Γ is the multiplicative group of a field K, and that Γ_{+} = R ∩ Γ = R\{0}, where R is a subring of K such that 1 ∈ R. Then ≤ satisfies: ∀x_{1}, ∀x_{2}, ∀y, y ≤ x_{1} and y ≤ x_{2} ⇒ y ≤ x_{1}−x_{2}. Indeed, y ≤ x_{1} and y ≤ x_{2} is equivalent to 1 ≤ x_{1}y^{−1} and 1 ≤ x_{2}y^{−1}, that is x_{1}y^{−1} ∈ R and x_{2}y^{−1} ∈ R. Since R is a ring, it follows that x_{1}y^{−1} − x_{2}y^{−1} ∈ R, i.e., 1 ≤ x_{1}y^{−1} − x_{2}y^{−1}. Hence: y ≤ x_{1} − x_{2} (in case x_{1} − x_{2} ≠ 0).

Conversely, if ≤, defined on Γ = K\{0}, satisfies ∀x_{1}, ∀x_{2}, ∀y, y ≤ x_{1} and y ≤ x_{2} ⇒ y ≤ x_{1} − x_{2}, then Γ_{+} ∪ {0} is a ring. Hence:

There is a one-to-one mapping from the set of all subrings of K containing 1 onto the set of all preorder relations ≤ such that ∀x_{1}, ∀x_{2}, ∀y, y ≤ x_{1} et y ≤ x_{2} ⇒ y ≤ x_{1} − x_{2}.

Let U be the multiplicative subgroup Γ_{+} ∩ Γ_{+}^{−1}. U is the group of all units of R. Then the canonical mapping v from K onto (K^{∗}/U) ∪ {∞} enjoys ∀x, ∀y, ∀z, (v(z) ≤ v(x) and v(z) ≤ v(y)) ⇒ v(z) ≤ v(x − y) (we extend the mapping v to 0 by setting v(0) = ∞, where ∞ is an element bigger than all the elements of Γ, and such that ∀g ∈ Γ, ∞ · g = ∞). Conversely, assume that f is a homomorphism from K onto an ordered group that satisfies these conditions. Then the inverse image of the positive cone is a ring containing 1.

Definition 1.1 Let R be an integral domain, U be its group of units and K be its quotient field. The ordered group (K^{∗}/U, (R\{0})/U) is called the group of divisibility of R.

Example 1.2 Let P be the set of all prime integers. Then the group of divisibility of ℤ is the group ∑_{p∈P} ℤ, equipped with the product order.

Remark 1.3 In [14], Jaffard gives an example of a group that cannot be considered as a group of divisibility (see also example 4.4.2 in [17]).

Definition 1.4 An ordered group Γ is a lattice-ordered group (or a l-o group) if any two elements x and y of Γ have a supremum, denoted by sup(x, y) or x ∨ y, and an infimum, denoted by inf(x, y) or x ∧ y.

Example 1.5 Let T_{i}, i ∈ I, be a totally ordered group, and Γ be the product ∏_{i∈I} T_{i}, ordered by (x_{i})_{i∈I} ≤ (y_{i})_{i∈I} ⇔ ∀i ∈ I, x_{i} ≤ y_{i}. Then Γ is a lattice-ordered group.

Definitions 1.6 Let R be an integral domain. R is a Gauss domain (or a pseudo-Bezout domain, or a GCD domain) if its group of divisibility is a lattice ordered group. R is a valuation ring if its group of divisibility is a totally ordered group.

Definitions 1.7 ([13]) Let K be a field, Γ be an ordered group and v be a group homomorphism from K^{∗} onto Γ. We extend the definition of v to 0 by setting v(0) = ∞, where ∞ is an element bigger than all the elements of Γ and such that ∀g ∈ Γ, ∞ · g = g · ∞ = ∞. The mapping v will be called a semi-valuation if and only if ∀x, ∀y, ∀z, (v(z) ≤ v(x) and v(z) ≤ v(y)) ⇒ v(z) ≤ v(x − y). We have Γ ≃ K^{∗}/ ker v.

If Γ is lattice ordered, v will be called a demi-valuation. In this case, the condition “∀x, ∀y, ∀z, (v(z) ≤ v(x) and v(z) ≤ v(y)) ⇒ v(z) ≤ v(x − y)” is equivalent to “∀x, ∀y, v(x) ∧ v(y) ≤ v(x − y)”, and the ring R_{v} = {x ∈ K | v(x) ≥ v(1)} is a Gauss domain.

If Γ is totally ordered, v is a valuation. In this case, the condition “∀x, ∀y, ∀z, (v(z) ≤ v(x) and v(z) ≤ v(y)) ⇒ v(z) ≤ v(x − y)” is equivalent to “∀x, ∀y, min(v(x), v(y)) ≤ v(x − y)”, and the ring R_{v} = {x ∈ K | v(x) ≥ v(1)} is a valuation ring.

Note that the word semi-valuation has been used by many authors with different meanings.

Lemma 1.8 ([13]) Let R be an integral domain, and let K be its field of fractions. Assume that v is a homomorphism from R to an ordered group Γ such that ∀x, ∀y, ∀z, (v(z) ≤ v(x) and v(z) ≤ v(y)) ⇒ v(z) ≤ v(x − y). Then v extends in a natural way to a unique semi-valuation from K onto Γ.

Example 1.9 ([13]) Let k be a field and define v from k[X_{1}, …, X_{n}] onto the lattice-ordered group ℤ^{n} by setting:

v(X_{1}) = (1, 0, …, 0), …, v(X_{n}) = (0, …, 0, 1).

More generally, let I be a finite set of n-tuples of integers, and, for any ī = (i_{1}, …, i_{n}) in I, aī ∈ k\{0}, set v(∑ aīX^{i_{1}} ⋯ X^{i_{n}}) = inf ī. Extend v in the obvious way to the field of fractions k(X_{1}, …, X_{n}). The mapping v is a demi-valuation. It is a valuation if n = 1.

2 Complements about lattice-ordered groups.

In this section, we will use additive notations for groups. We give some definitions and properties that we will use in later sections.

2.1 Positive part. For more details and for the proofs, see one of the following: [2], [4], [6], [12], [15].

Definitions 2.1 Let (Γ, +, ≤) be a lattice-ordered group. For all g in Γ we set g_{+} = g ∨ 0 and g_{−} = (−g) ∨ 0 = −(g ∧ 0). g_{+} is called the positive part of g, and g_{−} is called the negative part of g.

Lemma 2.2 Let (Γ, +, ≤) be a lattice-ordered group. Then for all g ∈ Γ, we have g_{+} ≥ 0, g_{−} ≥ 0, g = g_{+} − g_{−}, g_{+} + g_{−} = g_{+} ∨ g_{−}, g_{+} ∧ g_{−} = 0.

Lemma 2.3 Let (Γ, +, ≤) be a lattice-ordered group. Take a, b in Γ_{+} and c in Γ. If c = a − b and a ∧ b = 0, then a = c_{+} and b = c_{−}.

2.2 Embedding theorem.

Theorem 2.4 ([20], or any of [2], [4], [6], [12], [15]). Any lattice-ordered abelian group Γ is embedded in a direct product ∏_{i∈I} T_{i} of totally ordered abelian groups T_{i} in such a way that, for every j ∈ I, p_{j}(Γ) = T_{j}, where p_{j} : ∏_{i∈I} T_{i} → T_{j} is the projection map. We say that Γ is a sub-direct product of ∏_{i∈I} T_{i}.

2.3 l-subgroups, prime subgroups.

Definition 2.5 Let (Γ, +, ≤) be an ordered group, and H be a subgroup of Γ. H is a convex subgroup of Γ if ∀a ∈ H, ∀b ∈ H, ∀c ∈ Γ, a < c < b ⇒ c ∈ H.

Definitions 2.6 Let Γ be a lattice-ordered group, and H be a subgroup of Γ. H is a l-subgroup of Γ if ∀g ∈ H, ∀h ∈ H, g ∧ h ∈ H. (It follows g ∨ h ∈ H.) H is a prime subgroup of Γ if H is a convex l-subgroup and enjoys ∀g ∈ Γ,

∀h ∈ Γ, 0 ≤ g ∧ h ∈ H ⇒ (g ∈ H or h ∈ H).

Proposition 2.7 ([25]). Let (Γ, +, ≤) be an abelian lattice-ordered group, and H be a non-empty subset of Γ. Then H is a convex l-subgroup of Γ if and only if the conditions (1) to (3) below hold.

(1) H = H_{+} − H_{+},

(2) (g ∈ H_{+}, 0 ≤ h ≤ g) ⇒ h ∈ H_{+}

(3) H_{+} is closed under +.

Lemma 2.8 ([11]). Let (Γ, +, ≤) be an abelian lattice-ordered group. Then the sum of any two convex subgroups is a convex subgroup.

Lemma 2.8 can also be deduced from [6], theorem 7.4.

It is possible to derive characterizations of l-subgroups and prime subgroups, from the following properties.

Lemma 2.9 ([25]). Let (Γ, +, ≤) be an ordered abelian group and G be a subgroup of Γ. For all h ∈ Γ, we set h + G ≥ 0 + G ⇔ ∃g ∈ G, h ≥ g. Then G is convex if and only if (Γ/G, ≤) is an ordered group.

Proposition 2.10 (Any of [2], [4], [6], [12], [15]). Let (Γ, +, ≤) be a latticeordered abelian group and G be a subgroup of Γ. Then G is a convex l-subgroup of Γ if and only if Γ/G is a lattice-ordered group (with the same order as in lemma 2.9). If G is a convex l-subgroup, then the following properties (i), (ii), (iii) are equivalent.

(i) G is a prime subgroup. (ii) Γ_{+}\G_{+} is closed under inf. (iii) Γ/G is totally ordered.

2.4 Final sets. We recall some trivial properties of final subsets of latticeordered groups.

Definitions 2.11 Let (E, ≤) be an ordered set. A subset F of the positive cone E_{+} of E will be called a final subset of E_{+} if (a ∈ F and b ≥ a) ⇒ b ∈ F. A subset Δ of E+ will be called an initial subset of E_{+} if (a ∈ Δ and 0 ≤ b ≤ a) ⇒ b ∈ Δ.

Fact 2.12 Let Γ be an abelian lattice-ordered group, and F ⊂ Γ_{+}. F is a final subset of Γ_{+} if and only if Γ_{+}\F is an initial subset of Γ_{+}.

Definition 2.13 Let Γ be an abelian lattice-ordered group. A subset C of Γ will be called a convex centered set if C is convex, non-empty and a ∈ C ⇒ (−a ∈ C and a_{+} ∈ C) (from which follows that a_{−} = (−a)_{+} ∈ C and 0 ∈ C).

Proposition 2.14 Let (Γ, +, ≤) be an abelian lattice-ordered group. A subset C of G is a convex centered set if and only if there exists a non-empty initial set Δ of Γ_{+} such that C = Δ − Δ.

Corollary 2.15 Let Γ be a lattice-ordered group. F ↦ (Γ_{+}\F) − (Γ_{+}\F) is a one to one correspondence from the set of all final subsets of Γ_{+} onto the set of all convex centered sets of Γ.

Notation 2.16 Let Γ be an abelian lattice-ordered group. In the following, if F is a final subset of Γ_{+}, cvx(F) will denote the convex centered set (Γ_{+}\F)−(Γ_{+}\F).

2.5 Weak units.

Definition 2.17 Let Γ be a lattice-ordered group and g ∈ Γ. g is called a weak unit if g > 0 and ∀h ∈ Γ_{+}, g ∧ h = 0 ⇒ h = 0.

Example 2.18 In the lattice-ordered group ℝ × ⋯ × ℝ, (1, …, 1) is a weak unit.

Lemma 2.19 ([4]) Let Γ be an abelian lattice-ordered group, F = {F | F is a final subset of Γ_{+} closed under inf }, ordered by ⊂. We have the following. For all prime subgroup G, Γ_{+}\G_{+} ∈ F. For all F ∈ F, Γ_{+}\F is a maximal element different from Γ_{+} if and only if Γ_{+}\F is the positive cone of a minimal prime subgroup of Γ.

Proof This lemma can be deduced from [4]. We give a proof based on the definitions and properties of subsections 2.3 and 2.4. Let G be a prime subgroup. By using definition 2.6, it is easy to check that Γ_{+}\G_{+} belongs to F.

Let F be a maximal element of F, and g, h in Γ_{+} such that g + h ∈ F, and assume g ∉ F. By maximality of F, the minimal element of F containing F and g is Γ_{+}. Hence there is a ∈ F such that g ∧ a = 0. Then h = (g + h) ∧ (a + h) ∈ F. Accordingly, Γ_{+}\F is closed under +. By proposition 2.7, it follows that Γ_{+}\F is the positive cone of a convex l-subgroup G. Moreover, G is prime because G_{+} satisfies g ∧ h ∈ G_{+} ⇒ (g ∈ G_{+} or h ∈ G_{+}). Now, G is minimal because F is maximal.

Let G be a minimal prime subgroup, and F ≠ Γ_{+} be a maximal element of F containing Γ_{+}\G_{+}. Then Γ_{+}\G_{+} is the positive cone of a minimal prime subgroup, so F = Γ_{+}\G_{+}.

Lemma 2.20 ([4]) In an abelian lattice-ordered group, the set of all weak units is equal to the complement in the positive cone of the union of all the minimal prime subgroups.

Proof This lemma can be deduced from [4]. We give a proof based on the definitions and properties of subsections 2.3 and 2.4. Let (Γ, +) be an abelian lattice-ordered group, and g ∈ Γ_{+}, g > 0.

Assume that g is not a weak unit, so there exists h > 0 such that g ∧ h = 0. Let F_{0} be a maximal element of {F | F is a final subset of Γ_{+} closed under inf and containing h}. By lemma 2.19, Γ_{+}\F_{0} is the positive cone of a minimal prime subgroup G. But g ∉ F_{0}, because otherwise, 0 = g ∧ h ∈ F_{0}, and g ∈ G.

Assume that g is contained in a minimal prime subgroup G, and set F_{0} = Γ_{+}\G_{+}. Then, by lemma 2.19, F_{0} is a maximal element of F, and doesn’t contain g. Therefore the final subset closed under inf generated by F_{0} and g is equal to Γ_{+}. Consequently, there is h ∈ F_{0} (h > 0) such that g ∧ h = 0.

2.6 Topology on the set of all prime subgroups. In [9], [10], Ershov defines a Zariski topology on a family of valuation rings. From the bijection between valuation rings and minimal prime subgroups of a lattice-ordered group (cf. section 5), we deduce a topology on the set G of all minimal prime subgroups of a latticeordered group. In this section, the author gives a definition of this topology on G and he transposes some results to this case. In [4] and [6] there is the definition of the spectral topology on the set of all prime subgroups. The Hausdorff topology induced on G is generated by the complements of the sets V_{g} (with notations of definition 2.21).

Definitions 2.21 If Γ is an abelian lattice-ordered group, let G be the set of all minimal prime subgroups of Γ, and for all g ∈ Γ, set V_{g} := {G ∈ G | g_{+} ∈ G}. The sets V_{g}, g ∈ Γ, form a basis of open subsets of a topology on G, (we see that it is sufficient to consider the V_{g}’s for g ∈ Γ_{+}).

Let Γ be an abelian lattice-ordered group. We know that {0} = ⋂{G | G ∈ G} (cf. [4]). Hence for all g ∈ Γ_{+}, V_{g} = G iff g = 0.

It is not difficult to check that, if g and h are elements of the positive cone of Γ, then V_{g} ∪ V_{h} = V_{g∧h}, V_{g} ∩ V_{h} = V_{g∨h} = V_{g+h}.

Proposition 2.22 The topology defined on the set of all minimal prime subgroups of an abelian lattice-ordered group is quasi-compact.

Proof Let Γ be an abelian lattice-ordered group, let I be a set of Γ_{+}, and assume that G = ⋃{V_{g} | g ∈ I} is a covering of G. Let F be the final subset of Γ_{+} closed under inf and generated by the elements of I. (By definition, F is the set of all elements that are greater or equal to the inf of some finite subset of I.) If F ≠ Γ_{+}, then F is contained in a maximal final subset closed under inf, and the complement in Γ_{+} is the positive cone of a minimal prime subgroup G (cf. lemma 2.19). But in this case, G doesn’t contain any element of I, a contradiction with the hypothesis (covering). Consequently, F = Γ_{+}, and there exists g_{1}, …, g_{n} in I such that g_{1} ∧ ⋯ ∧ g_{n} = 0. Then V_{g_{1}} ∪ ⋯ ∪ V_{g_{n}} = V_{g_{1}∧···∧g_{n}} = V_{0} = G. We have constructed a finite subcovering.

Definition 2.23 A topological space is said to be boolean if it is compact and totally disconnected.

Let Γ be an abelian lattice-ordered group, and g, h in Γ_{+}. By lemma 2.20, V_{g∨h} = Ø if and only if g ∨ h is a weak unit, and V_{g∧h} = G if and only if g ∧ h = 0 (by above discussion). Hence G\V_{g} = V_{h} if and only if g ∧ h = 0 and g ∨ h is a weak unit. Now, assume that for every g ∈ Γ_{+} there exists h ∈ Γ_{+} such that g ∧ h = 0 and g ∨ h is a weak unit. Hence the topology defined in 2.21 is the same as the topology defined in [4] and [6], and this topology is Hausdorff. Note that in this case, every open set V_{g} is a closed set. It follows from proposition 2.22 that G is a boolean space. So, we have the following proposition.

Proposition 2.24 Let Γ be an abelian lattice-ordered group, then the following (i) and (ii) are equivalent.

(i) ∀g ∈ Γ, ∃h ∈ Γ, G\V_{g} = V_{h}. (ii) ∀g ∈ Γ_{+}, ∃h ∈ Γ_{+}, g ∧ h = 0, and g ∨ h is a weak unit. If (i), (ii) are satisfied, then G is a boolean space.

The following definitions are motivated by the properties of families of valuation rings, see [9], [10].

Definitions 2.25 Let Γ be a lattice-ordered group. We will say that Γ is independent if for all minimal prime subgroups G_{1} ≠ G_{2}, we have G_{1} + G_{2} = Γ. We will say that Γ is locally complemented if the topological space G is locally complemented, in other words, if for all g and h in Γ_{+} there exists c ∈ Γ_{+} such that c ≤ h and V_{c} = V_{h} ∪ (G\V_{g}).

The following propositions are trivial consequences of the definitions. We give the proofs for completeness.

Proposition 2.26 Let Γ be a lattice-ordered group, then (1), (2), (3) are equivalent.

(1) Γ is independent.

(2) Any proper convex l-subgroup contains at most one minimal prime subgroup.

(3) Any proper prime subgroup contains exactly one minimal prime subgroup.

Proof (2) ⇔ (3)is well known. Indeed, from proposition 2.10, we deduce that any convex l-subgroup that contains a prime subgroup is a prime subgroup.

Let G_{1} and G_{2} be minimal prime subgroups. By lemma 2.8, G_{1}+G_{2} is a convex l-subgroup. If G_{1} + G_{2} ≠ Γ, then there exists a proper convex l-subgroup that contains two minimal prime subgroups. Reciprocally, if there exists a proper convex l-subgroup containing two minimal prime subgroups G_{1} and G_{2}, then G_{1} + G_{2} is contained in this convex l-subgroup, and is different from Γ. Consequently (1) ⇔ (2).

Proposition 2.27 Any lattice-ordered group such that, for all g, there exists h with V_{h} = G\V_{g} is locally complemented.

Proof Indeed, this condition can be called “complemented”, and for any h′, we set c = h ∧ h′, then V_{c} = V_{h} ∪ V_{h′} = V_{h′} ∪ (G\V_{g}).

3 Gauss domains - Bezout domains.

Gauss domains are defined in definition 1.6.

Let K be a field and R be a subring of K containing 1. The preorder relation defined by R is, in fact, the relation “divides” (x|y ⇔ ∃z ∈ R, y = zx). If v is the semi-valuation corresponding to R, then v(x) and v(y) have a sup if and only if x and y have a lcm (defined up to a unit), v(x) and v(y) have an inf if and only if x and y have a gcd (defined up to a unit). Furthermore, it is well known that if two elements x and y of R have a lcm z in R, then they have a gcd \frac{xy}{z}. Hence, if R is an integral domain, then R is a Gauss domain if and only if any two elements of R have a lcm in R.

It is also well known that if R is an integral domain, and a, b, c are elements of R, then c is a lcm of a and b if and only if cR = aR ∩ bR.

Lemma 3.1 ([5], exercises § 1, n^{∘} 21, or [18]). Any Gauss domain is integrally closed in its quotient field.

From section 1, we deduce the following characterization.

Proposition 3.2 Let K be a commutative field, R be a subring of K containing 1 and U = R ∩ (R\{0})^{−1} be the group of all units of R. Then:

(K^{∗}/U) is an ordered group

(K^{∗}/U) is totally ordered  iff R is a valuation ring of K
iff ∀x ∈ K, x ∉ R ⇒ x^{−1} ∈ R
iff K = R ∪ (R\{0})^{−1}
(K^{∗}/U) is lattice-ordered  iff the intersection of any two principal

ideals is a principal ideal

iff R is a Gauss domain.

Now we give an example of a property of integral domains that depends on their groups of divisibility.

Theorem 3.3 ([17]). Let R be an integral domain. Then the following (a), (b), (c) are equivalent.

(a) R is factorial (or a unique factorization domain).

(b) The group of divisibility of R is a lattice-ordered group Γ and any strictly decreasing sequence of elements of the positive cone Γ_{+} is finite.

(c) The group of divisibility of R is isomorphic to a direct sum ∑_{i∈I} ℤ, ordered by (x_{i})_{i∈I} ≥ 0 ⇔ ∀i ∈ I, x_{i} ≥ 0.

Proof (a) ⇔ (b) is trivial. Indeed, we know that a factorial ring is a Gauss domain such that any strictly increasing sequence of principal ideals is finite (see, for example, [5], exercises § 3, n^{∘} 2, or [18]). Now, let v be the demi-valuation defined by this Gauss domain, I be a well-ordered set and for i ∈ I, let (a_{i})_{i∈I} be a sequence of elements of R. Then the sequence of principal ideals (a_{i}R)_{i∈I} is increasing if and only if the sequence (v(a_{i}))_{i∈I} is decreasing.

(b) ⇔ (c) is a theorem of Ward and Birkhoff, see, for example, [23], [3] or [15] pp. 181-182.

Note that in [17], theorem 3.3 is proved in another way, and implies the theorem of Ward and Birkhoff. Hence Mott uses ring theory to provide the proof of a theorem of group theory.

Definition 3.4 Let R be a Gauss domain. We say that R is a Bezout domain if for all a and b in R there exist a′ and b′ in R such that the gcd of a and b is equal to aa′ + bb′ (this is equivalent to “any finitely generated ideal is principal”).

We know some simple examples of Bezout domains: ℤ and the polynomial ring k[X], with k a field. As an example of an integral domain that is not a Bezout domain, take the ring R[X, Y] with R a commutative integral domain with unit element. The gcd of X and Y is 1, but for any a and b, 1 ≠ aX + bY.

We recall an example of an integrally closed domain that is not a Gauss domain. From general properties of quadratic fields, we know that the ring ℤ[\sqrt{−5}] is integrally closed because −5 ≢ 1 modulo 4. We also know that the gcd of 3 and 2 + \sqrt{−5} is 1. Indeed, N(3) = 3^{2} + 0^{2} = 9, N(2 + \sqrt{−5}) = 2^{2} + 5 · 1^{2} = 9, (the norm N is defined as follows: N(a + b\sqrt{−5}) = a^{2} + 5b^{2}) and if x divides y in ℤ[\sqrt{−5}], then N(x) divides N(y) in ℕ. But there is no element x with N(x) = 3, and the only elements with norm 1 are 1 and −1, that is, the units of ℤ[\sqrt{−5}]. The elements 3 and 2+\sqrt{−5} divide 9 and 6+3\sqrt{−5} (because (2+\sqrt{−5})(2−\sqrt{−5}) = 9). N(9) = N(6 + 3\sqrt{−5}) = 81, therefore, if 3 and 2 + \sqrt{−5} have a lcm in ℤ[\sqrt{−5}] then its norm divides 81 and is a multiple of 9, hence it is equal to 27. Since 3 is not the norm of any element of ℤ[\sqrt{−5}], we deduce that 27 can’t be the norm of a multiple of 3. It follows that 3 and 2 + \sqrt{−5} don’t have a lcm.

Theorem 3.5 ([25], or [5], exercises § 1, n^{∘} 20 b). Let K be a commutative field, and R be a subring of K. Assume that R is the intersection of a finite number of non-comparable valuation rings. Then R is a Bezout domain.

Proposition 3.6 Let R be a Gauss domain, K be the quotient field of R and v be the associated demi-valuation. For g ∈ vK, set V_{g} := {x ∈ K | v(x) ≥ g}. Then there exists a topology such that the collection of all the V_{g}, g ∈ vK, is a basis of open neighbourhoods of 0. With this topology, K is a topological ring.

Proof In the valuation field case, this is well known. The proof can be generalized to the demi-valuation case.

Remark 3.7 The property “x ↦ x^{−1} is continuous” is a first order property in the language of fields together with a predicate for the ring R.

Proof It is equivalent to ∀x_{0}, ∀β ∈ vK, ∃α ∈ vK, ∀x, v(x − x_{0}) ≥ α ⇒ v(\frac{1}{x} − \frac{1}{x_{0}} ) ≥ β.

Now, in the language (+, ·, 1, 0, −,^{−1} , R), with R(x) iff x belongs to the demivaluation ring: ∀x_{0}, ∀y, ∃z, ∀x, R((x − x_{0})z^{−1}) ⇒ R((x^{−1} − x_{0}^{−1})y^{−1}).

Proposition 3.8 Let R be a Gauss domain with quotient field K such that vK contains weak units. Then K is a topological field.

Proof Let x_{0} ∈ K^{∗}, g ∈ vK^{∗}, u be a weak unit, g′ = (u+v(x_{0}))∨(g +2v(x_{0})). First we prove that, for all x ∈ x_{0} +V_{g′}, we have v(x) = v(x_{0}). Let x ∈ x_{0} +V_{g′},

v(x) = v(x − x_{0} + x_{0}) ≥ v(x − x_{0}) ∧ v(x_{0}) ≥ (u + v(x_{0})) ∧ v(x_{0}) = v(x_{0}). Assume that v(x) > v(x_{0}), say v(x) = v(x_{0}) + h, with h > 0. Then u ∧ h > 0, because u is a weak unit, and v(x_{0}) = v(x_{0} −x+x) ≥ v(x_{0} −x)∧v(x) ≥ (u+v(x_{0}))∧(h+v(x_{0})) = (u ∧ h) + v(x_{0}) > v(x_{0}), a contradiction.

Now, v(x^{−1}−x_{0}^{−1}) = v(\frac{x_{0}−x}{xx_{0}}) = v(x_{0}−x)−v(x)−v(x_{0}) = v(x−x_{0})−2v(x_{0}) ≥ g. Hence x ↦ x^{−1} is continuous.

In section 6 we will see examples of demi-valued fields that are not topological fields. We will see also that in a Gauss domain, the assertion ∀x, ∀y, v(x) ≠ v(y) ⇒ v(x + y) = v(x) ∧ v(y) is false.

4 Ideals closed under gcds.

In this section, R is a Gauss domain, K is its quotient field, U is its group of units and Γ is the multiplicative group K^{∗}/U. We will abuse notation and write the group Γ additively.

If F is a final subset of Γ_{+}, the subset v^{−1}(F) = {x ∈ R | v(x) ∈ F } satisfies v^{−1}(F)R ⊂ v^{−1}(F). If E ⊂ R satisfies ER ⊂ E, then v(E) is a final subset of Γ_{+}. Furthermore, ∀x ∈ R, v(x) ∈ E ⇒ x ∈ v(E). Therefore v^{−1}(v(E)) = E. It follows that the mapping E ↦ v(E) is a one to one mapping from the set of all subsets E of R such that ER ⊂ E (including ideals) onto the set of all final subsets of Γ_{+}. Note that R doesn’t need to be a Gauss domain, this fact remains valid with R an integral domain.

From corollary 2.15, we deduce that the mapping E ↦ cvx(v(E)) is a one to one mapping from the set of all subsets E of R such that ER ⊂ E (including ideals) onto the set of all convex centered sets of Γ (with notations of 2.16).

If v(E) satisfies ∀(a, b) ∈ Γ_{+} × Γ_{+}, a + b ∈ v(E) ⇒ a ∈ v(E) or b ∈ v(E), then Γ_{+}\v(E) satisfies ∀(a, b) ∈ Γ_{+} × Γ_{+}, (a ∈ (Γ_{+}\v(E)) and b ∈ (Γ_{+}\v(E))) ⇒ a + b ∈ Γ_{+}\v(E) and by proposition 2.7, the corresponding convex centered set is a convex l-subgroup of Γ. The converse is trivial. So, the mapping E ↦ cvx(v(E)) induces a one to one mapping from the set of all subsets E of R such that ER ⊂ E and ∀(a, b) ∈ Γ_{+} × Γ_{+}, a + b ∈ v(E) ⇒ (a ∈ v(E) or b ∈ v(E)) (including prime ideals), onto the set of all convex l-subgroups of Γ.

Proposition 4.1 (Yakabe, [24]) v induces a one to one mapping from the set of all ideals closed under gcds onto the set of all final subsets closed under inf.

Proof Let F be a final subset of Γ_{+}, E = v^{−1}(F), x, y in E and z be a gcd of x and y in R. Then v(x)∧v(y) = v(z), hence v(x)∧v(y) ∈ v(E) ⇔ z ∈ E : F is closed under inf if and only if v^{−1}(F) is closed under gcds. Now v(x + y) ≥ v(x) ∧ v(y), hence if F is closed under inf, then v(x + y) ∈ F, and x + y ∈ E : E is an ideal.

Proposition 4.2 (Yakabe, [25]) The mapping I ↦ cvx(v(I)) induces a one to one mapping from the set of all prime ideals closed under gcds onto the set of all prime subgroups of Γ. Its restriction is a one to one mapping from the set of all maximal ideals closed under gcds onto the set of all minimal prime subgroups of Γ.

Proof If I is a prime ideal, then, by above discussion, C = cvx(v(I)) is a convex l-subgroup of Γ. If I is closed under gcd, then, by proposition 4.1, Γ_{+}\C_{+} is closed under inf. It follows that x ∧ y ∈ C_{+} ⇒ x ∈ C_{+} or y ∈ C_{+}. This is equivalent to saying that Γ is a prime subgroup of Γ.

If C is a prime subgroup of Γ, then by definition 2.6, Γ_{+}\C_{+} enjoys hypothesis of proposition 4.1, and the subset I = v^{−1}(Γ_{+}\C_{+}) is an ideal closed under gcds. Let x and y in R such that xy ∈ I. Then v(x)+v(y) ∈ v(I) and since C is a convex l-subgroup, by proposition 2.7, v(x) ∈ v(I) or v(y) ∈ v(I), and x ∈ I or y ∈ I. Hence I is prime.

Now, recall that in a Bezout domain, all the ideals are closed under gcds. It follows:

Corollary 4.3 If R is a Bezout domain, then I ↦ cvx(v(I)) induces a one to one mapping from the set of all prime ideals of R onto the set of all prime subgroups of Γ. Its restriction is a one to one mapping from the set of all maximal ideals of R onto the set of all minimal prime subgroups of Γ.

Lemma 4.4 We have the following properties. Any non-unit of R belongs to a prime ideal closed under gcds. The domain R is equal to the intersection of the family of all the R_{P}, with P ideal which is closed under gcds.

Proposition 4.5 Let P be a prime ideal of R. Then the local ring R_{P} is a valuation ring of the fraction field K of R if and only if P is closed under gcds.

From lemma 4.4 and proposition 4.5 we deduce the following classical result (see [2]).

Corollary 4.6 If R is a Bezout domain, then R is the intersection of a family of valuation rings of its field of fractions.

Definition 4.7 If R′ is an integral domain such that for any prime ideal P, R′_{P} is a valuation ring, then R′ is a Prüfer ring. To prove that R′ is a Prüfer ring, it is sufficient to prove that for any maximal ideal M, R′_{M} is a valuation ring.

Remark 4.8 Let R′ be a commutative ring. Then R′ is a Bezout domain if and only if R′ is a Gauss domain and a Prüfer ring.

5 Boolean rings.

Properties of boolean rings have already been studied (see for example [9], [10] and [7]). Here, the author points out some definitions and properties that depend on the group of divisibility of the ring.

Definition 5.1 A boolean ring is a Prüfer ring such that the group of divisibility is a lattice-ordered group and satisfies “for any x in the positive cone, there exists y in the positive cone such that the inf of x and y is the unit element and the sup of x and y is a weak unit”. Equivalently, a boolean ring is a Bezout domain such that the group of divisibility satisfies “for any x in the positive cone, there exists y in the positive cone such that the inf of x and y is the unit element and the sup of x and y is a weak unit”.

We deduce that the group of divisibility of a boolean ring contains weak units, and that its fraction field is a topological field (cf. Proposition 3.8).

It follows that for any abelian lattice-ordered group Γ, there exists a boolean ring with divisibility group Γ if and only if Γ enjoys “for any x in the positive cone, there exists y in the positive cone such that the inf of x and y is the unit element and, the sup of x and y is a weak unit”. The construction is based on polynomials but we will see another construction in section 6.

First, we recall a characterization of the Jacobson radical of a Bezout domain.

Lemma 5.2 Let R be a Bezout domain, J be the Jacobson radical of R (J is the intersection of all maximal ideals of R) and x ∈ R. Then x ∈ J ⇔ (x^{−1} ∉ R and ∀y, (x, y) = 1 ⇒ y^{−1} ∈ R). In other words, a non unit element x belongs to the Jacobson radical if and only if the gcd of x and of any non unit element y is a non unit element.

Hence, we have the following.

Lemma 5.3 Let R be an integral domain with quotient field K and group of units U. R is a boolean ring if and only if R is a Bezout domain such that for any x there exists y such that the gcd of x and y is 1 and the lcm of x and y belongs to the Jacobson radical.

Now, from proposition we deduce:

Proposition 5.4 Let Γ be the group of divisibility of a boolean ring, and G be the set of all prime subgroups of Γ. Then G is a boolean space, and for any g ∈ Γ there exists h ∈ Γ such that G\V_{g} = V_{h}.

Notations 5.5 Let R be an integral domain. We will denote by specm(R) the maximal spectrum of R, that is, the set of all maximal ideals of R. For R_{v} a valuation ring, M(R_{v}) will denote the maximal ideal of R_{v}. W_{R} will denote the set of all valuation rings R_{v} of the quotient field of R such that R ∩ M(R_{v}) ∈ specm(R).

Definitions 5.6 Let K be a field and W be a family of valuation rings of K. Let R_{W} be the intersection of all the elements of W. R_{W} is the holomorphy ring of W. The Zariski topology on W is the topology that admits the sets V_{x} := {R_{v} ∈ W | x ∉ M(R_{v})}, with x ∈ K, as elements of a basis of open sets.

We have also V_{x} = {R_{v} ∈ W | x^{−1} ∈ R_{v}}.

Let R be a Prüfer domain, then {R_{M} | M ∈ specm(R)} ⊂ W_{R}. Now, by a theorem of Chevalley, for all prime ideals P of R there exists a valuation ring R_{v} of the quotient field of R such that M(R_{v}) ∩ R = P. Therefore W_{R} = {R_{M} | M ∈ specm(R)}. But R is an integrally closed domain, hence R = ⋂{R_{M} | M ∈ specm(R)} = R_{W_{R}}.

Lemma 5.7 Let R be a Bezout domain with quotient field K, G be the family of all minimal prime subgroups of the group of divisibility of R, and let v be the demivaluation of K with ring R. For all M ∈ Specm(R), let G_{M} be the corresponding minimal prime subgroup of the group of divisibility (cf proposition 4.2 and corollary 4.3). Then for all x ∈ K, x^{−1} ∈ R_{M} if and only if v(x)_{+} ∈ G_{M}. It follows that the topological spaces W_{R} and G are homeomorphic.

Proof Let x ∈ K and y, z be elements of R such that v(y) = v(x)_{+} and v(z) = v(x)_{−}. There is a unit u such that x = \frac{yu}{z}. For all y′ and z′ in R such that x = \frac{y′}{z′}, we have v(x) = v(y′) − v(z′) with v(y′) ≥ 0 and v(z′) ≥ 0. Hence we deduce from lemma 2.3, that there exist t ∈ R and a unit e such that y′ = ty and z′ = tze.

Therefore x^{−1} ∈ R_{M} ⇔ (∃y′ ∈ R\M and z′ ∈ R, x^{−1} = \frac{z′}{y′}) ⇔ y ∉ M ⇔ v(x)_{+} ∈ G_{M}.

Definition 5.8 Let K be a field and W be a family of valuation rings of K. W is said to be boolean if the canonical topology is compact and ∀x ∈ K, ∀y ∈ K, ∃z ∈ K, V_{x} ∩ V_{y} = V_{z}, ∀x ∈ K, ∃x′ ∈ K, W \V_{x} = V_{x′}.

It follows that the family V_{x}, x ∈ K is closed under boolean operations.

Proposition 5.9 Let R be a Bezout domain, and G be the family of all minimal prime subgroups of the group of divisibility of R. Then the following (1), (2), (3) are equivalent.

(1) R is a boolean ring,

(2) the family W_{R} is boolean,

(3) the family G enjoys the hypothesis of proposition 2.24. ((1) ⇔ (2) is a theorem of Ershov, cf. [9], [10]).

Proof This follows from definition 5.1, proposition 5.4, proposition 2.24, proposition 2.22 and from the one-to-one mapping of lemma 5.7.

Definitions 5.10 A family of valuation rings is said to be independent if any two rings of the family are independent. An integrally closed domain R is said to be independent if the family W_{R} is independent. An integrally closed ring R is said to be locally complemented if the family W_{R} is locally complimented, in other words, if for all non zero x and y in R, there exists z ∈ R such that V_{z} = V_{x} ∪ (W_{R}\V_{y}).

Proposition 5.11 Let R be a Bezout domain.

(1) R is independent if and only if its group of divisibility is independent.

(2) R is locally complemented if and only if its group of divisibility is locally complemented.

Proof (1) Let M_{1} and M_{2} be maximal ideals of R. The rings R_{M_{1}} and R_{M_{2}} are independent if and only if R_{M_{1}}R_{M_{2}} = K^{∗} (with K the fraction field of R), if and only if (R\M_{1})(R\M_{2}) = R, if and only if (cvx(M_{1}))_{+} + (cvx(M_{2}))_{+} = v(K^{∗})_{+}, if and only if cvx(M_{1})+ cvx(M_{2}) = v(K^{∗}).

(2) Follows from the definitions and from the one-to-one mapping of lemma 5.7.

A Bezout domain is independent if and only if every prime ideal is contained in exactly one maximal ideal. This can also be deduced from proposition 2.26.

6 Formal power series.

Definition 6.1 Let R be an integral domain, Γ a torsion free abelian group. R^{Γ} denotes the group of all mappings from Γ to R. If σ ∈ R^{Γ}, then the support of σ will be the subset of all elements g of Γ such that σ(g) = 0. We will denote by Supp(σ) the support of σ, and elements of R^{Γ} will be denoted by σ = ∑_{g∈S} σ_{g}X^{g}, with S a subset of Γ containing Supp(σ).

Definitions 6.2 A subset S of an ordered set is said to be narrow if any subset of S consisting of elements pairwise incomparable is finite, artinian if any strictly decreasing sequence of S is finite. In an abelian ordered group, AN will denote the family of all artinian and narrow subsets.

Notation 6.3 Let R be an integral domain, Γ an ordered abelian group. R[[Γ, AN]] will denote the subset of all elements of R^{Γ} with support in AN, R[Γ] will denote the subset of all elements of R^{Γ} with finite support.

Theorem 6.4 (Priess-Crampe & Ribenboim, [19]). Let R be an integral domain, Γ an ordered abelian group. One can define a product on R[[Γ, AN]] in the following way. Let σ = ∑_{g∈S} σ_{g}X^{g} and τ = ∑_{g∈S′} σ_{g}X^{g} in R[[Γ, AN]] then

στ = ∑_{g∈S+S′}( ∑_{h∈S,h′∈S′,h+h′=g} σ_{h}τ_{h′})X^{g}.

Then R[[Γ, AN]] is a ring, and R[Γ] is a subring of R[[Γ, AN]].

Proposition 6.5 (Ribenboim, [22]) Let R be an integral domain, Γ an ordered abelian group. Then R[Γ] and R[[Γ, AN]] are integral domains.

Remark 6.6 In [22], Γ needs not be a lattice-ordered group . Note that if Γ is lattice-ordered, we can prove proposition 6.5 in the following way. We denote by ≤ the partial order on Γ. Let k be the quotient field of R and ≤′ be a total order on Γ, finer than ≤. We know that any ≤-artinian and narrow subset of Γ is ≤′ artinian and narrow, i.e., well-ordered (cf. [8]). Hence R[[Γ, AN]] is a subring of the ring of formal power series k[[Γ]] (with Γ ordered by ≤′). It follows that R[Γ] and R[[Γ, AN]] are integral domains.

Notation 6.7 Let R be an integral domain, Γ be an ordered abelian group. The quotient field of R[[Γ, AN]] will be denoted by R((Γ, AN)), R(Γ) will be the quotient field R[Γ].

Proposition 6.8 Let R be an integral domain, Γ be an abelian lattice-ordered group. For σ = ∑_{g∈S} σ_{g}X^{g}, σ_{g} ∈ R\{0}, S ∈ AN, we set v(x) = inf S. We extend the definition to the quotient field R((Γ, AN)) of R[[Γ, AN]], v is a demi-valuation. Its restriction to R(Γ) is also a demi-valuation.

Proof Jaffard already proved that the restriction of v to R(Γ) is a demivaluation (see [13] and [14]). Let x and y in R[[Γ, AN]], we prove v(xy) = v(x)+v(y) (the proof of v(x − y) ≥ v(x) ∧ v(y) is left to the reader). By embedding theorem 2.4, Γ is embedded in a direct product ∏_{i∈I} T_{i} of totally ordered abelian groups T_{i}; in such a way that, for any j ∈ I, p_{j}(Γ) = T_{j}, where p_{j} : ∏_{i∈I} T_{i} → T_{j} is the projection map. Now, it is sufficient to prove that, for all i ∈ I, p_{i} ∘ v(xy) = p_{i} ∘ v(x) + p_{i} ∘ v(y). Let i_{0} ∈ I and consider a well-order on I such that i_{0} is the lowest element. Let T be the totally ordered group ∏_{i∈I} T_{i} lexicographically ordered. This total order is finer than the product order on ∏_{i∈I} T_{i}, hence any artinian and narrow subset of Γ remains artinian and narrow in T, i.e., well-ordered. It follows that R[[Γ, AN]] ⊂ R[[T]]. Let v_{0} be the valuation of R[[T]], then v_{0}(xy) = v_{0}(x) + v_{0}(y). But, by construction, p_{i_{0}} ∘ v(x) = min p_{i_{0}}(Supp(x)) = p_{i_{0}} ∘ v_{0}(x). Hence p_{i_{0}} ∘v(xy) = p_{i_{0}}(v_{0}(x)+v_{0}(y)) = p_{i_{0}} ∘v_{0}(x)+p_{i_{0}} ∘v_{0}(y) = p_{i_{0}} ∘v(x)+p_{i_{0}} ∘v(y). Now, we extend v to R((Γ, AN)) by lemma 1.8.

Proposition 6.9 Let R be an infinite integral domain and Γ be an abelian lattice-ordered group, then the demi-valuation rings of R(Γ) and R((Γ, AN)) are Bezout domains.

Proof The demi-valuation ring of R(Γ) has already been studied by Jaffard (see [13] and [14]). We focus on R((Γ, AN)). Two elements of the quotient field can be expressed with a common denominator, therefore, we can restrict to formal power series x = ∑ a_{i}X^{g_{i}} and y = ∑ b_{j}X^{g_{j}}. Let I (resp. J) be the set of all minimal elements of the support of x (resp. y), hence v(x) = inf I and v(y) = inf J. By hypothesis, I and J are finite and R is infinite, hence there exists c ∈ R such that ∀(i, j) ∈ I ×J, cb_{j} ≠ −a_{i}. Consequently, I ∪J ⊂ Supp(x+cy) ⊂ Supp(x)∪Supp(y). Hence v(x + cy) = inf(Supp(x + cy)) ≤ inf(I ∪ J). Let l ∈ Supp(x + cy), say l ∈ Supp(x), then there is i ∈ I such that i ≤ l. It follows that inf(Supp(x + cy)) ≥ inf(I ∪ J), and v(x + cy) = inf (I ∪ J) = inf I∧ inf J = v(x) ∧ v(y). Let z be a gcd of x and y (it exists because the ring of a demi-valuation is a Gauss domain). We have v(\frac{z}{x+cy}) = v(\frac{cz}{x+cy}) = 0 and z = \frac{z}{x+cy} · x + \frac{cz}{x+cy} · y.

As an immediate consequence of definition 5.1, we have the following.

Proposition 6.10 Let k be an infinite field and Γ be an abelian lattice-ordered group. Then the demi-valuation ring of k((Γ, AN)) is a boolean ring if and only if for every x in the positive cone of Γ, there exists y in the positive cone of Γ such that the inf of x and y is the unit element and the sup of x and y is a weak unit.

By theorem 3.3, we can state:

Proposition 6.11 Let Γ be an abelian lattice ordered group, k be a field, and K be any subfield of k((Γ, AN)) containing k(Γ). Let v be the natural demi-valuation of k((Γ, AN)). Then the following (a), (b), (c) are equivalent.

(a) The demi-valuation ring of (K, v) is a factorial ring.

(b) The positive cone of Γ is artinian.

(c) There is a set I such that Γ is isomorphic to the lattice-ordered group ∑_{i∈I} ℤ.

Now, we can give examples to illustrate some remarks of section 3.

Lemma 6.12 Let k be a commutative field and Γ be a lattice-ordered group without weak unit. Then K = k(Γ) is not a topological field.

Proof We use notations of proposition 3.6. Let g > 0. For any h ∈ vK, we pick a h′ ≥ g ∨ h > 0, and t > 0 such that h′ ∧ t = 0, set y = \frac{X^{t+h′} +X^{t}}{X^{t}−X^{h′}}.

v(y−1) = v(\frac{X^{t+h′} +X^{t}−X^{t}+X^{h′}}{X^{t}−X^{h′}}) = v(X^{h′})+v(X^{t}+1)−v(X^{t}−X^{h′}) = h′+0−0 = h′ : y ∈ 1 + V_{h′} ⊂ 1 + V_{h}.

v(y^{−1}−1) = v(\frac{X^{t}−X^{h′} −X^{t+h′} −X^{t}}{X^{t}(X^{h′} +1)} ) = v(−X^{h′})+v(1+X^{t})−v(X^{t})−v(1+X^{h′}) = h′ + 0 − t − 0 = h′ − t.

If h′ − t ≥ g, then h′ ≥ g + t, hence h′ ∧ t = t, a contradiction. Consequently, y^{−1} ∉ 1 + V_{g} : the mapping x ↦ x^{−1} is not continuous in 1.

We can also see that, in a Gauss domain, it is not true, in general, that ∀x, ∀y, v(x) ≠ v(y) ⇒ v(x + y) = v(x) ∧ v(y). Indeed, in the demi-valuation ring of k(X_{1}, X_{2}, X_{3}, X_{4}), let x = X_{1}X_{2}X_{4} + X_{2}X_{3} and y = X_{1}X_{3}X_{4} − X_{2}X_{3}, then v(x) = (0, 1, 0, 0), v(y) = (0, 0, 1, 0), v(x)∧v(y) = (0, 0, 0, 0), but x+y = X_{1}X_{2}X_{4}+ X_{1}X_{3}X_{4}, v(x + y) = (1, 0, 0, 1).

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