Algebraic theory of Colombeauʼs generalized numbers

Journal of Algebra 384 (2013) 194–211

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Journal of Algebra www.elsevier.com/locate/jalgebra

Algebraic theory of Colombeau’s generalized numbers ✩ J. Aragona a , A.R.G. Garcia b,∗ , S.O. Juriaans a a

Instituto de Matemática e Estatística, Universidade de São Paulo, CP 66281, CEP 05311-970, São Paulo, Brazil Universidade Federal Rural do Semi-Árido, Departamento de Ciências Exatas e Naturais, Pós-Graduação em Sistemas de Comunicação e Automação, Pós-Graduação em Matemática, CEP 59.625-900, Mossoró-RN, Brazil

b

a r t i c l e

i n f o

Article history: Received 11 October 2011 Available online 28 March 2013 Communicated by Michel Broué MSC: primary 46F30 secondary 46T20 Keywords: Colombeau’s algebra Generalized numbers Sharp topology Full Algebraic structure

a b s t r a c t Let K denote the commutative ring of Colombeau’s full generalized numbers. Endowed with Scarpalezos’ sharp topology it becomes a topological ring. We study the algebraic and topological properties of this topological ring. In particular, we prove that the group of units of K is dense in the sharp topology, determine its boolean algebra, show that it has minimal primes, describe them completely which results in a complete classification of the maximal ideals. From the description of the prime and maximal ideals, it becomes clear that they should be determined by certain ultra-filters. © 2013 Elsevier Inc. All rights reserved.

Introduction The theory of generalized functions, initially developed by Colombeau, is a relatively young theory with applications to many other fields. Due to an important result of M. Kunzinger and M. Oberguggenberger, it turned out that Colombeau’s theory can be considered as an extension of classical calculus with functions taking values in certain ordered topological algebras. This Colombeau’s calculus theory was developed by J. Aragona, R. Fernandez and S.O. Juriaans (see [3]). Along with several other developments, this makes it clear why it is important to study the algebraic and topological properties of Colombeau’s algebras of generalized numbers. The simplified version of these algebras has already been studied; first by Aragona and Juriaans [7] and more recently by Aragona, Oliveira, ✩

The research of the last author was supported by CNPq-Brazil. Corresponding author. E-mail addresses: [email protected] (J. Aragona), [email protected] (A.R.G. Garcia), [email protected] (S.O. Juriaans).

*

0021-8693/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.03.005

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Scarpalezos, Juriaans [8] and H. Vernaeve [22]. In this paper we study the so-called full algebra of Colombeau’s generalized numbers. Its definition and topology are far more complicated than the simplified version which makes it all the more an interesting object of study. To fix notation, denote by K either R or C and by K the full algebra of Colombeau’s generalized numbers. Aragona and Juriaans developed algebraic and topological methods to study the properties of the ring of the simplified algebra of Colombeau’s generalized numbers Ks (see [7]). Understanding these properties is important because a generalized function can be viewed as a C ∞ function defined on a subset of unit ball of Kn (see [3]). The full algebras depend on more parameters and so, in a certain sense, are more complex to deal with but, in compensation, some of them contain canonically the vector space of all Schwartz distributions. Here we study the algebraic and topological properties of these full algebras. In Section 1, we collect basic definitions, results and notation to be used throughout the paper and, as a rule, most of the proofs are omitted. In Section 2, we present results that are extensions of results obtained by Aragona and Juriaans [7]. In Section 3, we present an accurate look at the structure of the maximal ideals of K based on two basic tools. First, a careful analysis of the set of the representatives of elements of K. Second, the introduction of a set S f consisting of subsets of A0 (K) whose characteristic functions are non-trivial elements of K. The set S f plays the role of the set S in Definition 4.1 of [7]. The family (X A ) A ∈S f of elements of K, where X A is the characteristic function of A, is used to study a number of interesting properties of K, one of them being that the unit group of K is dense. We also derive several characterizations of the units of K as well as a complete description of its maximal ideals. The description of the prime spectrum of K is more complex. Nevertheless we prove the existence of minimal primes and that any prime ideal contains such a minimal prime. Finally, in Section 4, based on the work of Aragona, Juriaans, Scarpalezos and Oliveira [8], we study order relations on R which is used to continue with the study of its algebraic properties. This is used to show that K is not Von Neumann regular and to describe all minimal primes. We refer the reader to [11] for the definition and results of rotationally invariant generalized functions. We have not checked if our results still hold in this case. Some basic references for the theory of generalized functions are [1,9,10,13,14,17] and important new results can be found in [12,16,18–21]. 1. The sharp topology on K In this section we recall basic definitions and results about K with the purpose of fixing terminology. As a rule, proofs are omitted. Notation 1.1. I := ]0, 1], I := [0, 1] and I η := ]0, η[, ∀η ∈ I . A \ B := {a ∈ A | a ∈ / B }. Q denotes the field of rational numbers. K denotes either the field of real or complex numbers, i.e., R or C. K∗ := K \ {0}. N and Z stand respectively for the set of the natural numbers and the set of integers. N∗ := N \ {0} and Z∗ := Z \ {0}. (g) K∗ := K \ {0}. (h) R+ := {x ∈ R | x  0} and R∗+ := {x ∈ R | x > 0}. (a) (b) (c) (d) (e) (f)

(i) Ks denotes the ring of Colombeau’s simplified generalized numbers. ∞ (j) A0 (R) := {ϕ ∈ D (R) | 0 ϕ (x) dx = 12 , ϕ is even and ϕ ≡ const. in V 0 }, where V 0 is a neighborhood of the origin.

∞

j

(l) Aq (R) := {ϕ ∈ A0 (R) | 0 x m ϕ (x) dx = 0, for 1  j , m  q}, where q ∈ N. (m) Γ := {γ : N → R+ | γ (n) < γ (n + 1), ∀n ∈ N and limn→∞ γ (n) = ∞} is the set of the strict increasing sequences diverging to infinity when n → ∞.

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Definition 1.2. Let E (K) be the ring (pointwise operations) of the functions v : A0 (K) → K. Let E M (K) denote the subring of E (K) of those functions v satisfying: (M) ∃ p ∈ N such that ∀ϕ ∈ A p (K), ∃C = C ϕ > 0, ∃η = ηϕ > 0, such that

   v (ϕε )  C ε − p ,

∀0 < ε < η .

These are called moderate functions. Let Γ be as in (m) and define N (K) the ideal of E M (K) of those functions v satisfying: (N) ∃ p ∈ N, ∃γ ∈ Γ such that ∀q  p and ∀ϕ ∈ Aq (K), ∃C = C ϕ > 0, ∃η = ηϕ > 0, such that

   v (ϕε )  C ε γ (q)− p ,

∀0 < ε < η .

These are called null functions. The ring of Colombeau’s full generalized numbers is defined as K = E M (K)/N (K). Unless otherwise stated, if x ∈ K is a generalized number, then xˆ will denote an arbitrary representative of x. There exists a natural embedding of K into K (induced by the map k → (ϕ → k)) and so writing K ⊂ K makes sense. Hence K is a unitary commutative K-algebra. We recall a definition of [2, Definition 6.1.1] and [15]. Definition 1.3. An element v ∈ K is associated to zero, v ≈ 0, if for some (hence for each) representative ( vˆ (ϕ ))ϕ of v we have:

∃ p ∈ N such that

lim vˆ (ϕε ) = 0, ε ↓0

∀ϕ ∈ A p (K).

Two elements v 1 , v 2 ∈ K are associated, v 1 ≈ v 2 , if v 1 − v 2 ≈ 0. If there exists a ∈ K with v ≈ a, then v is said to be associated with a and the latter is called the shadow of v. If we denote by Inv(K) the group of units of K then clearly K∗ ⊂ Inv(K). Another interesting subgroup of Inv(K) is H := {αr• | r ∈ R}, where αˆr• (ϕ ) := (diam(supp(ϕ )))r or



r

αˆr• (ϕ ) := i (ϕ ) , where i (ϕ ) denotes the diameter of the support of

i (ϕε ) = ε i (ϕ ),

ϕ ∈ A0 (K). In particular, we have that

 r ∀ε > 0 and αˆr• (ϕε ) = εr i (ϕ ) .

(1)

ˆ r : I → R+ is defined by αˆ r (ε ) := εr (see [7]). Hence (1) implies that αˆr• (ϕε ) = For r ∈ R, α αˆ r (ε)(i (ϕ ))r . It is convenient to define βr• := α−• log(r ) , r > 0. We now briefly describe the sharp topology on K. The interested reader should consult [4,5].

Definition 1.4. For a given x ∈ K we set A(x) := {r ∈ R | αr• x ≈ 0} and define the valuation of x as V(x) := sup(A(x)). It is easily seen that if x ∈ K, then r ∈ A(x) if, and only if, there exists p ∈ N such that limε↓0 ε −r xˆ (ϕε ) = 0, ∀ϕ ∈ A p (K). From this it easily follows that D : K × K → R+ defined by D (x, y ) := exp(−V(x − y )) is an ultra-metric on K and is invariant under translations. It determines a uniform structure on K called the sharp uniform structure on K and the topology resulting from D is

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called the sharp topology on K and denoted by τsf . Denoting by x := D (x, 0), ∀x ∈ K, we have that the distance between two elements x, y ∈ K is given by D (x, y ) := x − y . Notation 1.5. Let x ∈ K and r ∈ R∗+ . In what follows, B r (x) (resp. B r (x) and S r (x)) denotes the open D-ball (resp. closed D-ball and D-sphere) with center at x and radius r. In case x = 0 we omit it in the notation, writing B r , B r and S r . We have the following easy consequence of what is above and from the definition of D. Corollary 1.6. For given x, y ∈ K, r ∈ R, s ∈ R∗+ and a, b ∈ K, we have: (i) (ii) (iii) (iv) (v) (vi)

x + y  max{ x , y } and xy  x

y ;

x = 0 ⇔ x = 0;

ax = x , if a = 0;

αr• x = e −r x and βs x = s x ;

a = 1, if a = 0;

a − b = 1 − δab (Kronecker’s δ).

Proposition 1.7. (K, τsf ) is a complete topological ring. Proof. It follows from the completeness of the algebras G (Ω) and G (Ω) (see [6]) and from the fact that K is the ring of the constants of such algebras. 2 Proposition 1.8. (K, τsf ) is not a K-topological algebra, is not locally compact and is not separable. We refer the reader to [5] for an equivalent definition of the topological space (K, τsf ) using the concept of generalized semi-norms. 2. Algebraic properties of K In this section we study algebraic properties of K. In what follows, X stands for the topological closure of the set X (except for K). We start with the following: Lemma 2.1. 1. x ∈ B 1 if and only if V(x) > 0. / B 1 , B 1 ∩ B 1 (1) = ∅, B 1 ⊃ B 1 and B 1 =  B1. 2. If x ∈ B 1 , then x ≈ 0, D (1, x) = 1 − x = 1 and hence 1 ∈ The proof of this lemma is easy and a good reference for the arguments to be used is [7]. The next result follows from a well known property of ultra-metric abelian groups. Proposition 2.2. Let (an )n∈N be any sequence in K and let x ∈ B 1 . Then the series In particular



n0 x converges and we have (1 − x)



n

 n

n0 x



n n0 an x

converges in K.

= 1. Therefore (1 − x) ∈ Inv(K).

Corollary 2.3. Inv(K) ⊂ K is open. Proof. Let a ∈ Inv(K), define r := a−1 −1 and let z ∈ B r (a). Then a−1 ( z − a)  a−1

z − a < r −1 r = 1. Therefore a−1 ( z − a) < 1 and hence, by Proposition 2.2, we have that (1 − a−1 (a − z)) ∈ Inv(K). Since z = a(1 − a−1 (a − z)) it follows that z ∈ Inv(K). 2 Theorem 2.4. Let I be a proper ideal of K. Then, for each x ∈ I, we have that 1 − x  1 and D (1, I) = 1. Hence, every maximal ideal of K is closed and rare.

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Proof. If x ∈ I, then x ∈ / Inv(K) and so 1 − x  1. If 1 − x < 1 then, by Proposition 2.2, we have that 1 − (1 − x) = x ∈ Inv(K), a contradiction. Since 1 = 1, the first part is proved. If m is a maximal ideal of K, then, by the first part, we have that 1 ∈ / m and we are done. 2 Lemma 2.5. (i) 0 ∈ Inv(K). (ii) For each x ∈ K, if r = −V(x), then y = αr• x ∈ S 1 , i.e., y = 1. (iii) If x ∈ Inv(K), then V(x) + V(x−1 )  0. Proof. The proof is as in [7].

2

Another result which is analogous to one of [7] is Proposition 2.6. (i) K does not have proper open ideals. (ii) No topological K-module has proper open submodules. (iii) If X is a Hausdorff topological K-modules then for all x ∈ X , x = 0, the set







Inv(K).x := λx  λ ∈ Inv(K)

is unbounded. Whence, Inv(K) is not a bounded subset of K. (iv) A given B ⊂ K is bounded iff B is D-bounded. (v) The only K-topological module which is bounded is {0}. Whence, the only K-topological module which is compact is {0}. Theorem 2.7. Let m be a maximal ideal of K and L := K/m. Then K can be identified with a proper subfield of L, i.e., L is a proper field extension of K. Proof. We follow the proof of [7]. Let π : K → L := K/m be the canonical map. Then, k := π (K)  K. In fact, if L = k, then K = K + m. But K is a discrete subset of K and, hence, from Theorem 2.4 and Proposition 2.6(i), it follows immediately that m ∪ K is a closed set with empty interior. Thus, there exists x ∈ B 1 , such that x ∈ / m ∪ K. Write x = kx − mx , where kx ∈ K and mx ∈ m. Obviously, kx = 0 1 −1 −1 −1 and hence mx = kx − x = kx (1 − k− x x). Indeed, (1 − k x x) = k x m x ∈ m and so (1 − k x x) ∈ m. Since − 1

kx x = x < 1, it follows that mx is a unit, a contradiction. 2 Lemma 2.8. Let R 1 , R 2 be positive real numbers and set r := ln( R 1 ) − ln( R 2 ). Then, αr• . S R 1 = S R 2 . Proof. This is an immediate consequence of Corollary 1.6(iv).

2

Another easily deducible fact is that K has no non-zero nilpotent elements and hence its nilradical is trivial. Consequently, K is contained in a product of integral domains. We will see later that in fact it is contained in a product of ordered fields. 3. Characteristic functions In this section we study the group of units of K, as well as its prime and maximal spectra. The crucial step in this study is to make a careful analysis of the set of zeros of a representative of an element of K. Using this we study some special type of characteristic functions showing that they are related with the prime and maximal ideals of K. One of the main consequences is that the unit group of K is open and dense. The new feature here is the density since we have already proved it

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to be open. The idea behind the proof is closely related to a similar result of [7]. However there is a crucial difference. It is useful for the reader to recall the definition of the sets S and P∗ (S ) defined in [7, Definition 4.1 p. 2217]. We will use this notation in what follows. For A ⊂ A0 (K) let A c be its complement in A0 (K) and denote by Xˆ A the characteristic function of A with domain A0 (K), i.e.,

Xˆ A (ϕ ) =



1, if ϕ ∈ A , 0, if ϕ ∈ / A, i.e., ϕ ∈ A c .

This is clearly a moderate function. Its class in K is denoted by X A and is still called the characteristic function of A ⊂ A0 (K). Definition 3.1. Define

   S f := A ⊂ A0 (K)  ∀ p ∈ N, ∃ϕ ∈ A p (K) such that {ε | ϕε ∈ A } ∈ S , where S := { S ⊂ I | 0 ∈ S ∩ S c }.

 A ∩ A p (K)  A p (K), ∀ p ∈ N. In particular, we have Proposition 3.2. Let A ∈ S f . Then A c ∈ S f and ∅ = A ∩ A p (K) = ∅ and A c ∩ A p (K) = ∅ for all p ∈ N. Proof. (i) Indeed, if A ∈ S f , then for all p ∈ N there exists ϕ ∈ A p (K) such that {ε | ϕε ∈ A } ∈ S . Thus, from Proposition 4.2, item b, p. 2217 of [7], we have that {ε | ϕε ∈ A }c ∈ S ; but {ε | ϕε ∈ A }c = {ε | ϕε ∈/ A } = {ε | ϕε ∈ A c } and therefore {ε | ϕε ∈ A c } ∈ S . Hence for all p ∈ N, there exists ϕ ∈ A p (K) such that {ε | ϕε ∈ A c } ∈ S and, therefore, A c ∈ S f . Reciprocally, if A c ∈ S f , then A = ( A c )c ∈ S f . (ii) If A ∈ S f , then for all p ∈ N, there exists ϕ ∈ A p (K) such that the set A ε := {ε | ϕε ∈ A } ∈ S . As {ϕε | ε ∈ A ε } ⊆ A p (K) and {ϕε | ε ∈ A ε } ⊆ A, it follows that A ∩ A p (K) = ∅. We now prove that A ∩ A p (K)  A p (K). That it is contained is obvious. That A ∩ A p (K) = A p (K) follows from the fact that {ϕε | ε ∈ A cε } ⊂ A p (K) and {ϕε | ε ∈ A cε } ⊂ A c ∈ S f . This implies that A c ∩ A p (K) = ∅. Hence, A ∩ A p (K)  A p (K). 2 We have the following obvious result: Proposition 3.3. Let A ∈ S f . Then, Ann(X A ) = KX A c and K = Ann(X A ) ⊕ Ann(X A c ). Moreover, for each prime ideal p of K, we have that either X A or X A c = 1 − X A belongs to p. Notation 3.4. In what follows the symbol, P (S f ) denotes the set of all subsets of S f . Definition 3.5. Denote by P∗ (S f ) the set of all F ∈ P (S f ) verifying the following conditions: (i) For every A ∈ S f , either A or A c belongs to F but not both. (ii) If A , B ∈ F , then A ∪ B ∈ F . The second condition implies that F is stable under finite union. The following proposition enumerates some properties of the functions X A . Proposition 3.6. (i) If A ∈ S f , then X A ∈ S 1 . (ii) If A , B ∈ F , A = B and X A = X B , then d(X A , X B ) = X A − X B = 1. Hence, the topology of K does not have an enumerable base. (iii) If A ∈ S f , then X A ∈ K \ {0, 1} and X A = X A2 . (iv) (1 − X A )X A = 0.

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• ≈ 0 ⇔ r < 0 (see [5]). Thus, if r ∈ A(X ) (see Proof. (i) To show that X A = 1 observe that α− A r • X ≈ 0 and, therefore, α • ≈ 0 and so A (X ) = ]−∞, 0[. Hence, X = Definition 1.4), then α− A A A r −r e −V(X A ) = e − sup( A (X A )) = e 0 = 1. • (X − X ) ≈ 0. As X = X it follows that α • ≈ 0 and A (X − (ii) Let r ∈ A (X A − X B ). Then, α− A B A B A −r r X B ) = ]−∞, 0[. Hence, X A − X B = e −V(X A −X B ) = e − sup( A (X A −X B )) = e 0 = 1. The items (iii) and (iv) are obvious. 2

Definition 3.7. For F ∈ P∗ (S f ), let g f (F ) := {X A | A ∈ F } be the ideal of K generated by the set of the characteristic functions of A with A ∈ F . Obviously if F ∈ P∗ (S f ) and A , B ∈ F , then X A ∩ B , X A ∪ B ∈ g f (F ). We will use this in our next result. Lemma 3.8. If F ∈ P∗ (S f ), then g f (F ) is a proper ideal of K. Proof. nAssume that g f (F ) = K. Then,

n there exist ai ∈ K and A i ∈ F , i = 1, 2, . . . , n, such that 1= i =1 ai X A i . We define A := i =1 A i . Then, A0 (K) = A ∈ F and X A i · X A = X A i ∩ A = X A i , ∀ both members of the former equation by X A , we obtain X A = i n= 1, 2, . . . , n. Multiplying n ˆ i =1 ai X A i .X A = i =1 ai X A i = 1. Hence, (X A − 1) ∈ N (K). It follows that ∃ p ∈ N, ∃γ ∈ Γ such that

∀q  p and ∀ϕ ∈ Aq (K) (∃C = C ϕ > 0, ∃η = ηϕ > 0) such that |1 − Xˆ A (ϕε )|  C ε γ (q)− p , ∀0 < ε < η . But A ∈ F ⊂ P (S f ) and so, for this ϕ , {ε | ϕε ∈ A } ∈ S ⇒ 0 ∈ {ε | ϕε ∈ A } ∩ {ε | ϕε ∈ A c }, i.e., there exists a sequence {εn }n∈N converging to zero when n → ∞ such that ϕεn ∈ A c . It now follows that 1 = |1 − Xˆ A (ϕεn )|  C (εn )γ (q)− p . As γ is divergent, we can choose q0 ∈ N such that γ (q0 ) − p > 0. Thus, we have 1  C (εn )γ (q0 )− p −−−−→ 0 a contradiction. Therefore, g f (F )  K. 2 n→∞

Theorem 3.9. Let p be a prime ideal of K. Then, there exists a unique Fp ∈ P∗ (S f ) such that g f (Fp ) ⊂ p. In particular, P∗ (S f ) = ∅. Proof. The proof is easy and similar to that given in [7].

2

Theorem 3.9 associates with each prime ideal p of K a set Fp ∈ P∗ (S f ) characterized by g f (Fp ) ⊂ p. Denote by g f (F ), where F ∈ P∗ (S f ), the

τsf -closure of g f (F ) in K.

Definition 3.10. Let x ∈ K and let xˆ be one of its representatives. Define Z (ˆx) := {ϕ ∈ A0 (K) | xˆ (ϕ ) = 0}, the set of zeros of the representative xˆ of x. Lemma 3.11. Let x ∈ K \ {0} and F ∈ P∗ (S f ). The following statements are equivalent: 1. x ∈ g f (F ). 2. There exists A ∈ F such that xX A = x. Proof. A 2 , . . . , A n ∈ F and a1 , a2 , . . . , an ∈ K such that n(i) ⇒ (ii): If x ∈ g f (F ), then there exist A 1 ,

n x = i =1 ai X A i . Definition 3.5(ii) tells us that A := i =1 A i ∈ F and since A i ⊂ A, ∀i = 1, 2, . . . , n it follows that X A i · X A = X A i . Hence,

 xX A =

n

ai X A i X A

i =1

=

n (ai X A i )X A i =1

J. Aragona et al. / Journal of Algebra 384 (2013) 194–211

=

n

201

ai (X A i X A )

i =1

=

n

ai X A i = x.

i =1

Therefore, xX A = x and A ∈ F . (ii) ⇒ (i): If A ∈ F and xX A = x, then x ∈ g f (F ). Therefore, there exist A ∈ F and x ∈ K such that x = xX A , where X A ∈ g f (F ). 2 Theorem 3.12.

/ S f , ∀ representative xˆ of x. (i) x ∈ Inv(K) if and only if Z (ˆx) ∈ / Inv(K) if and only if ∃e ∈ K, e 2 = e such that x · e = 0. In particular, if x ∈ K \ {0} and x ∈ / Inv(K), (ii) x ∈ then x is a zero divisor. Proof. (i) Suppose by contradiction that x ∈ Inv(K) and that Z (ˆx) ∈ S f for some representative xˆ of x. / {0, 1}. Thus, X Z (ˆx) = 0 but xX Z (ˆx) = 0 and, hence, x Then, by Proposition 3.6(iii), we have that X Z (ˆx) ∈ is a zero divisor, a contradiction. To prove the converse, it is enough to show (ii). (ii) We will show that if x ∈ / Inv(K), then x is a zero divisor. In fact, let xˆ be a representative of x. We consider two cases: (a) Z (ˆx) ∈ S f . / Sf . (b) Z (ˆx) ∈ (a) If Z (ˆx) ∈ S f , then, by Proposition 3.6(iii), we have that X Z (ˆx) = 0 and since xX Z (ˆx) = 0 we have that x is a zero divisor. (b) If Z (ˆx) ∈ / S f define

x∗ (ϕ ) =



xˆ (ϕ ), 0,

if ϕ ∈ / Z (ˆx), if ϕ ∈ Z (ˆx),

i.e., x∗ = xˆ X Z (ˆx)c . Then, x∗ (ϕ ) = 0 for some ϕ ∈ A0 (K) and (x∗ − xˆ ) ∈ N (K). So we may substitute xˆ by x∗ and assume that xˆ (ϕ ) = 0. Since x ∈ / Inv(K) it follows that 1x ∈ / E M (K) and, hence, by Definition 1.2, ∀ p ∈ N, ∃ϕ ∈ A p (K) such −p

that ∀C = C ϕ > 0, ∀η = ηϕ > 0, ∃0 < ε0 < η such that | 1xˆ (ϕε0 )| > C ε0 . Taking C = n,

that there exists 0 < εn <

1 , n

η = n1 , we have

such that

  1   (ϕε ) > nεn− p . n   xˆ For this ϕ , we define the set A p (ϕ p ) := {εn | n ∈ N} and let B := {(ϕ p )εn | εn ∈ A p (ϕ p )}. Then, B ∈ S f , i.e., ∀ p ∈ N, ∃ϕ p ∈ A p (K) such that {εn | (ϕ p )εn ∈ B } ∈ S . Indeed, it is enough to show that {εn | (ϕ p )εn ∈ B } ∈ S . For that, it is enough to notice that A p (ϕ p ) = {εn | (ϕ p )εn ∈ B } and, therefore, ∅ = A p (ϕ p ) ∩ I η=1/2 = I η=1/2 because limn→∞ εn  limn→∞ n1 = 0. We now show that xˆ Xˆ B ∈ N (K). Definition 1.2 tells us that we need to prove that ∃ p ∈ N, ∃γ ∈ Γ such that ∀q  p, ∀ϕ ∈ Aq (K), ∃C = C (ϕ ) > 0, ∃η = η(ϕ ) > 0 such that |ˆxXˆ B (ϕε )|  C ε γ (q)− p , ∀0 <

ε < η. In fact, taking p = 0 and γ (q) = q we have that |ˆxXˆB (ϕε )|  C εq < 1 for small ε . Therefore, xˆ Xˆ B ∈ N (K).

2

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Definition 3.13. For x ∈ E M (K) and a ∈ N, define (i) N a (x) := {ϕ ∈ A0 (K) | |x(ϕ )| < αa• (ϕ )};

(ii) Xˆa,x := XˆNa (x) and Xa,x := X Na (x) .

We do not make any attempt to introduce more names and notation than needed. Note that our previous definition (see [7] for the original definition) was used in [22] to define level sets. Our next result is crucial to understand the boolean algebra of K. Lemma 3.14. If A ⊆ A0 (K), then Xˆ A ∈ N (K) if and only if there exist τ : A0 (K) → I and q0 ∈ N such that for all q  q0 and for all ϕ ∈ Aq (K) we have that τ (ϕ ) > 0 and I τ (ϕ ) ⊆ {ε | ϕε ∈ A c }. Proof. Suppose first that Xˆ A ∈ N (K) then there ∃ p ∈ N, ∃γ ∈ Γ such that ∀q  p, ∀ϕ ∈ Aq (K),

∃C = C ϕ > 0, ∃η = ηϕ > 0 such that |Xˆ A (ϕε )|  C ε γ (q)− p , ∀0 < ε < η . As γ is divergent, there exists q0 > p such that γ (q) − p > 2 for all q  q0 . Now choose τ (ϕ ) < η = ηϕ such that 0 < ε < τ (ϕ ) implies that C ε γ (q)− p < 1. We thus have that if 0 < ε < τ (ϕ ) then |Xˆ A (ϕε )| < 1 ⇔ Xˆ A (ϕε ) = 0 ⇔ ϕε ∈ A c ⇔ {ϕε | ε < τ (ϕ )} ⊆ A c , i.e., {ϕε | ε ∈ I τ (ϕ ) } ⊆ A c or equivalently I τ (ϕ ) ⊆ {ε | ϕε ∈ A c }. Conversely if such a function exists, then take p = q0 , ηϕ = τ (ϕ ). It is then easily seen that with this data we have that Xˆ A ∈ N (K).

2

When applying the previous lemma we will work with the restriction of suppose that τ (ϕ ) > 0, for ϕ ∈ Aq with q  q0 .

τ to Aq0 (K), i.e., we will

Lemma 3.15. (i) X Na (x) = 1, ∀a ∈ N iff x ∈ N (K). (ii) X Na (x) = 0, ∀a ∈ N iff x ∈ Inv(K). Proof. (i) Take p = 0 and

γ : N → N, γ (q) = q. We have that XNq (x) = 1 if and only if Xˆ(Nq (x))c =

1 − XˆN q (x) ∈ N (K) and, by Lemma 3.14, there exists τ : A0 (K) → ]0, 1] such that ∀ϕ ∈ Aq (K), I τ (ϕ ) ⊆ {ε | ϕε ∈ N q (x)}, i.e., |x(ϕε )| < αq• (ϕε ) = (i (ϕ ))q εq , ∀0 < ε < τ (ϕ ). So if we set C = C ϕ = (i (ϕ ))q > 0,

η = ηϕ = τ (ϕ ) > 0 we get |x(ϕε )| < (i (ϕ ))q εq = C εγ (q)− p , ∀0 < ε < η(ϕ ). Item (ii) can be proved in a similar way.

2

Proposition 3.16. Let x ∈ K∗ be a non-unit. Then there exists a ∈ N such that S = N a (x) ∈ S f and |xX S | < αa• .

/ Proof. In the proof, x will stand for a representative. Suppose that ∀a ∈ N, we have that S := N a (x) ∈ S f . Then, either X Na (x) ∈ N (K) or X(Na (x))c ∈ N (K). Suppose first that X Na (x) ∈ N (K); then, by Lemma 3.14, there exists τ : A0 (K) → ]0, 1] such that ∀ϕ ∈ Aq (K) we have that I τ (ϕ ) ⊆ {ε | ϕε ∈ ( N a (x))c }. It follows from Definition 3.13 that |x(ϕε )|  αa• (ϕε ) = (i (ϕ ))a εa , ∀0 < ε < τ (ϕ ) and, therefore, | x(ϕ1 ) |  (i (ϕ ))−a ε−a , ∀0 < ε < τ (ϕ ). This implies ε

that 1x ∈ E M (K) and, since x 1x = 1, it follows that x is a unit, a contradiction. On the other hand if X( Na (x))c ∈ N (K) then, since X( Na (x))c = 1 − X Na (x) , it follows that X Na (x) = 1, ∀a ∈ N and, hence, from Lemma 3.15(i), we have that x ∈ N (K), a contradiction. It follows that there must exist an a ∈ N such that N a (x) ∈ S f . The last assertion follows immediately from the definition of S. 2 The following result is frequently used when proving many other results. We shall refer to it as the Approximation Theorem. Theorem 3.17. Let x ∈ K∗ be a non-unit. Then, exactly one of the following conditions holds:

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(a) There exist S ∈ S f and a ∈ N such that (i) xX S = 0; (ii) |xX S c |  αa• X S c (i.e., there exists xˆ a representative of x such that |ˆx(ϕ )|  αˆa• (ϕ ), ∀ϕ ∈ A0 (K)). (b) There exist sequences (an )n∈N ⊂ N and ( S n )n∈N ⊂ S f such that (i) S n ⊃ S n+1 , an < an+1 and xX S n = 0 for all n ∈ N; (ii) The sequence xX S n converges to 0; (iii) |xX S n | < αa•n . Proof. Suppose that x does not satisfy (a). We will show that condition (b) must hold. As x ∈ K \ {0} and x ∈ / Inv(K) it follows, from Proposition 3.16, that there exists a1 ∈ N such that if S 1 := N a1 (x), then S 1 ∈ S f and |xX S 1 | < αa•1 . Let x2 = xX S 1 . If x2 = 0, then xX S 1 = 0 and so (a)(i) holds. From Definition 3.13 we also have that |xX S c |  αa•1 X S c , i.e., (a)(ii) holds and, hence, x satisfies (a), a con1 1 tradiction. So we have that x2 = 0. x2 X S c = 0 with S 1 ∈ S f and, hence, x2 is a non-trivial zero divisor and so it is a non-unit. This 1 allows us to proceed by induction. For completeness, we show how to accomplish the inductive step: There exists an+1 ∈ N such that if S˜ n+1 := N an+1 (xn+1 ), then S˜ n+1 ∈ S f and |xn+1 X S˜ | < αa•n+1 ⇒ n +1

|xX S n ∩ S˜ n+1 | < αa•n+1 . Let S n+1 := S n ∩ S˜ n+1 . Then, S n+1 ⊂ S n ∈ S f and so S n+1 ∈ S f and |xX S n+1 | <

αa•n+1 . Definition 3.13 tells us that

αa•n+1  |xn+1 X S˜ c | = |xX S n ∩ S˜ c | = |xX S˜ c |, n +1

n +1

(2)

n +1

therefore, S˜ nc +1 ⊂ S n . In fact, if ϕ ∈ S˜ nc +1 , then αa•n+1 (ϕ )  |xn+1 (ϕ )| = |x(ϕ )X S n (ϕ )| and since αa•n+1 = (i (ϕ ))an+1 > 0 we have that |x(ϕ )X S n (ϕ )| > 0 ⇒ X S n (ϕ ) = 0, i.e., X S n (ϕ ) = 1 ⇒ ϕ ∈ S n , hence, S˜ nc +1 ⊂ S n and S n ∩ S˜ nc +1 = S˜ nc +1 . Now, ∀ϕ ∈ S˜ nc +1 ⊂ S n we have, from (2), that









αa•n+1 (ϕ )  xn+1 (ϕ ) = x(ϕ ) < αa•n (ϕ ). So we have that

αa•n+1 (ϕ ) < αa•n (ϕ ), ∀ϕ ∈ S˜ nc +1 which implies that 

We may choose S˜ c

an+1

i (ϕ )

 a < i (ϕ ) n ,

∀ϕ ∈ S˜ nc +1 .

ϕ ∈ S˜ nc +1 such that 0 < i (ϕ ) < 1. Hence we obtain that an < an+1 and therefore,

∈ Sf . In this way we construct sequences ( S n )n∈N and {an }n∈N satisfying conditions (i) and (iii). We now show that condition (ii) also holds: from (iii) it follows that for each n ∈ N, |(xX S n )(ϕ )| < αa•n (ϕ ), ∀ϕ ∈ A0 (K) and, hence, xX S n  αa•n = e −an −−−−→ 0. 2 n+1

1

n→∞

Theorem 3.18. x ∈ Inv(K) if and only if there exist r > 0, τ : A0 (K) → I , such that

  xˆ (ϕε )  α • (ϕε ), r

∀0 < ε < τ (ϕ ),

where xˆ is a representative of x. Proof. Suppose that x ∈ Inv(K). Then, from Lemma 3.15, we have that X N q (ˆx) = 0, ∀q ∈ N, i.e.,

XˆNq (ˆx) ∈ N . Lemma 3.14 implies that there exists τ : A0 (K) → ]0, 1] such that ∀ϕ ∈ Aq (K) we have 1 In fact, if ϕ ∈ S˜ nc +1 , then i (ϕε ) = ε i (ϕ ) and {ε | ϕε ∈ S˜ nc +1 } ∈ S . Therefore, there exists i (ϕε0 ) < 1, i.e., ε0 i (ϕ ) < 1. That this happens follows from the fact that i (ϕε ) = ε i (ϕ ) → 0 when

ε0 such that ϕε0 ∈ S˜ nc +1 and ε → 0.

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I τ (ϕ ) ⊂ {ε | ϕε ∈ ( N q (ˆx))c }. Thus, from Definition 3.13, it follows that |ˆx(ϕε )|  αq• (ϕε ), ∀0 < ε < τ (ϕ ), where xˆ is a representative of x. Hence, we may take q = r. Conversely, if there exist r > 0, τ : A0 (K) → ]0, 1], such that

  xˆ (ϕε )  α • (ϕε ), r

∀0 < ε < τ (ϕ ),

then,

    xˆ (ϕε )  i (ϕ ) r εr ,

∀0 < ε < τ (ϕ )

and, hence,

  1   xˆ (ϕ

ε

get

      i (ϕ ) −r ε −r , )

∀0 < ε < τ (ϕ ).

We assert that 1xˆ ∈ E M (K): taking p = r > 0, C = C (ϕ ) = (i (ϕ ))− p > 0 and

  1   xˆ (ϕ

ε

Therefore, x ∈ Inv(K).

      i (ϕ ) −r ε −r = C ε − p , )

η = η(ϕ ) = τ (ϕ ) > 0 to

∀0 < ε < τ (ϕ ).

2

Lemma 3.19. Let x ∈ K, (x = 0) be a non-unit. Then, there exists a ∈ N such that y := x(1 − Xa,ˆx ) + Xa,ˆx ∈ Inv(K).

/ {0, 1}. Define yˆ := Proof. By Proposition 3.16, there exists a ∈ N such that N a (ˆx) ∈ S f , hence, Xa,ˆx ∈ xˆ (1 − Xˆa,ˆx ) + Xˆa,ˆx . Then, we see easily that the class of yˆ defines a unit.

2

Theorem 3.20. Let x ∈ K, (x = 0) be non-unit. Then, there exists a maximal ideal m of K such that x ∈ / m. Hence, the Jacobson radical Rad(K) = {0}.

/ {0, 1}, Proof. By Lemma 3.19, there exists a ∈ N such that y = x(1 − Xa,ˆx ) + Xa,ˆx ∈ Inv(K). As Xa,ˆx ∈ Xa,ˆx ∈ / Inv(K) and K is a ring with unit, there exists a maximal ideal m  K such that Xa,ˆx ∈ m. / m: if x ∈ m, then xX(a,ˆx)c = x(1 − Xa,ˆx ) ∈ m, hence, y = x(1 − Xa,ˆx ) + Xa,ˆx ∈ We shall prove that x ∈ m ∩ Inv(K), a contradiction. 2 We can now state the main result of this section which completely describes the maximal ideals of K and shows that the unit group is dense and open. Theorem 3.21. (1) Let m ⊂ K be an ideal. Then, m is maximal if and only if m = g f (Fm ). (2) Inv(K) is an open and dense subset of K. Proof. (1) (⇒) Suppose that m is a maximal ideal of K. Then, m is a prime ideal of K and hence, by Theorem 3.9, there exists only F = Fm such that g f (F ) ⊂ m. Let x ∈ m \ g f (F ). We now construct a sequence (xn )n∈N in g f (F ) such that xn −−−−→ x in K. For this, we show that x satisfies the n→∞

condition (b) of the Approximation Theorem. Indeed, if x does not satisfy condition (b) of the Approximation Theorem then, it must be that x satisfies condition (a), i.e., there exist S ∈ S f and a ∈ N, such that

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(i) xX S = 0; (ii) |xX S c |  αa• X S c . Recall that S = N a (x) form some a. Suppose that S ∈ F . Then,

X S ∈ g f (F ) ⊂ m

⇒

X S ∈ m.

Thus,

y := x(1 − X S ) + X S ∈ m, a contradiction, because Lemma 3.19 tells us that y is invertible. Now, assume that S c ∈ F . Then,

X S c = (1 − X S ) ∈ g f (F ) and, from item (a) of the Approximation Theorem, it follows that

x = x(1 − X S ) = xX S c ∈ g f (F ) a contradiction because, by assumption, x ∈ / g f (F ). Consequently x does not satisfy condition (a) of the Approximation Theorem and, therefore, must satisfy condition (b), i.e., there exist sequences {an } ⊂ N and ( S n ) ⊂ S f , such that (i) S n ⊃ S n+1 , an < an+1 ; (ii) |xX S n | < αa•n and hence xX S n −→0 as n → ∞. We claim that S nc ∈ F . Indeed, if this were not the case, then, S n ∈ F and, therefore,

X S n ∈ g f (F ) ⊂ m. As done above, we produce a unit y = x(1 − X S n ) + X S n ∈ m, a contradiction. Thus, S nc ∈ F , i.e., X S nc ∈ g f (F ). From this we obtain that xn = xX S nc = x − xX S n is a sequence in g f (F ) which converges

to x. Hence, m = g f (Fm ). (⇐) Let x be a non-zero generalized number not in m. We have to show that the ideal, J say, generated by x and m is non-proper. Obviously, we may suppose that x is not a unit. If the second condition of the Approximation Theorem is satisfied then, as before, we can construct a sequence (xn = xX S nc = x − xX S n ) in m converging to x. Since m is closed we get a contradiction. So x must satisfy the first condition of the Approximation Theorem. With the notation of the same theorem, we have that x + X S is a unit belonging to J . It follows that m is a maximal ideal. (2) From Corollary 2.3 we know that Inv(K) is open in K and hence, it remains to show that Inv(K) is dense. Lemma 2.5(i) tells us that 0 is in the closure of Inv(K).2 So consider a non-zero non-unit x ∈ K. We have that x satisfies either condition (a) or condition (b) of the Approximation Theorem. Assume first that x satisfies condition (a) and define

xn := x(1 − X S ) + αn• X S . From (a) of the Approximation Theorem and Theorem 3.18, we see easily that xn = x + αn• X S are units. As αn converges to 0 as n goes to infinity we have that (xn ) is a sequence of units converging

2

For instance, the sequence (αn• )n∈N ⊂ Inv(K) is

τsf -convergent to zero.

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to x. To finish the proof of the theorem, suppose that x satisfies condition (b) of the Approximation Theorem and let

xn := x(1 − X S n ) + αa•n X S n . Once again, from (b) of the Approximation Theorem and Theorem 3.18, we see easily that xn = x − xX S n + αa•n X S n is a sequence of units converging to x. This completes the proof of the theorem. 2 4. Algebraic properties of R This section relies upon the results obtained in [3,5,8]. We introduce a partial order in R which will induce a total order in every residual class field. The results obtained will be used, in the next sub-section, to show that K contains minimal primes. If F ∈ P∗ (S f ), then g f r (F ) denotes the ideal of R generated by the characteristic functions of elements of F and g f (F ) the ideal of C generated by the same function.

4.1. An order relation on R Lemma 4.1. (See [8].) Let x ∈ R. Then the following statements are equivalent. 1. Every representative xˆ of x satisfies the condition

(∗)

  ∃ N ∈ N such that ∀b > 0, ∀ϕ ∈ A N (K)   ∃η = η(b, ϕ ) ∈ I such that xˆ (ϕε )  −εb , ∀ε ∈ I η .

2. There exists a representative xˆ of x such that xˆ satisfies (∗). 3. There exists a representative x∗ of x such that x∗ (ϕ )  0, ∀ϕ ∈ A0 (K). 4. There exist N ∈ N and a representative x∗ of x such that x∗ (ϕ )  0, ∀ϕ ∈ A N (K). Definition 4.2. An element x ∈ R is non-negative or quasi-positive or q-positive, if it has a representative satisfying one of the conditions of Lemma 4.1. This is denoted by x  0. We say that x is non-positive or quasi-negative or q-negative if −x is q-positive and denote this by x  0. If y ∈ R is another element, then, x  y (resp. x  y) if y − x (resp. x − y) is q-positive. The former definition is not a total order in R since xˆ (ϕ ) = αˆ• 1 (ϕ ) sin(αˆ• −1 (ϕ )), ∀ϕ ∈ A0 (K) defines an element which is neither q-positive nor q-negative. However it defines a partial order such that the positive cone is closed under addition and multiplication. Let x ∈ K and let xˆ be a representative of x. The function |ˆx| : A0 → R+ defined by |ˆx|(ϕ ) = |ˆx(ϕ )| is moderate whose class, |x| := cl[|ˆx|], is independent of the representative. It is called the absolute value of x. Hence, we have a map K  x → |x| ∈ R+ . It now follows from Theorem 3.18 that x ∈ K is a unit if and only if there exists r > 0 such that |x| > αr• . Definition 4.3. Let x ∈ R. Then, x+ := q-negative parts of x.

x+|x| 2

and x− :=

x−|x| 2

are respectively called the q-positive and

Definition 4.4. For u ∈ E M (K), define θu : A0 (K) → K such that θu (ϕ ) = exp(−i Arg(u (ϕ ))) and θu−1 : A0 (K) :→ K such that θu−1 (ϕ ) = exp(i Arg(u (ϕ ))), where Arg(u (ϕ )) denotes the argument of u (ϕ ) ∈ K, with the convention that Arg(0) := 0. In case K = R, the images of θu and θu−1 are subsets of {−1, 1}. If we set Θu = cl[θu ] and Θu−1 = cl[θu−1 ] then, both are invertible generalized numbers and are inverses one of the other. Moreover,

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since u (ϕ )θu (ϕ ) = |u (ϕ )|, ∀ϕ ∈ A0 (K), we have that |u | = u Θu . Note however that Θu and Θu−1 depend on the representative. The following proposition is easily proved. Proposition 4.5. Let x, y ∈ R. Then: 1. 2. 3. 4. 5. 6. 7. 8.

x = x+ if and only if x = |x| iff x is q-positive. x = x− if and only if x = −|x| if and only if x is q-negative. (−x)+ = −(x− ) and (−x)− = −(x+ ). |x|  0  x+ , x−  0  −x+ , |−x| = |x| and |x|  x. |x + y |  |x| + | y |, ||x| − | y ||  |x − y | (triangle inequalities). If x  y and −x  y, then |x|  y.  1+Θ   1−Θ  x+ = x 2 xˆ and x− = x 2 xˆ . If A = {ϕ ∈ A0 (K) | xˆ (ϕ )  0}, then, x+ = xX A and x− = xX A c .

Note that if z ∈ C, then | z| ∈ R, | z|  0 and so we may apply Proposition 4.5 where possible. In particular, the triangle inequalities hold in this context. Proposition 4.6 (Convexity of ideals). Let J be an ideal of K and x, y ∈ K. Then (i) x ∈ J if, and only if, |x| ∈ J. (ii) If x ∈ J and | y |  |x|, then y ∈ J. (iii) If K = R, x ∈ J and 0  y  x, then y ∈ J. Proof. (i) If x ∈ J, then |x| = xΘxˆ ∈ J. Reciprocally, if |x| ∈ J, then x = |x|Θxˆ−1 ∈ J. (ii) If | y | = |x|, then from (i), |x| ∈ J. Therefore x ∈ J hence | y | ∈ J and so y ∈ J. Now, if | y | < |x|, then |x| =  0. Let u = ||xy|| , taking it to be zero on the zero set of a representative of x. Since this is a bounded element it is moderate and so | y | = u |x| ∈ J. (iii) follows from (ii) and Proposition 4.5. 2 Remark 4.7. If z ∈ C then it is√clear that we may write z = x + iy, with x, y ∈ R, where i is the class of the constant function −1.√Define !e ( z) := x and "m( z) := y, the real and imaginary part of z, respectively. Clearly, if zˆ = xˆ + −1 yˆ is a representative of z then xˆ and yˆ are representatives of x and y, respectively. It is also clear that C is an R-module and that the maps !e , "m : C → R are R-epimorphisms. Hence if J
C is an ideal then its image by these epimorphisms are ideals of R which are easily seen to coincide. This ideal is denoted by Jr and called the real part of J. Proposition 4.6 implies that Jr ⊂ J. Note that the involution c : C → C defined by c ( z) = z extends to the involution z ∈ C → z ∈ C, called conjugation. Lemma 4.8. Let J
C be an ideal of C. Then we have that Jr ⊂ J, J = Jr + i Jr , is invariant under conjugation and Jr = J ∩ R. Corollary 4.9. Let F ∈ P∗ (S f ) and z ∈ C. Then g f r (F ) is the real part of g f (F ) and z ∈ g f (F ) if and only if | z| ∈ g f r (F ). Lemma 4.10. Let F ∈ P∗ (S f ) and x, y ∈ R. Then, the following hold: (i) If (x − y ) ∈ g f r (F ) and x− ∈ g f r (F ) then y − ∈ g f r (F ). (ii) Either x+ or x− of x belongs to g f r (F ). Proof. (i) Proposition 4.5 item (v) implies that

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   −    x − y −  =  x − |x| − y − | y |   2 2   1 = x − |x| − y + | y | 2



  |x − y | + | y | − |x|

1 2

 |x − y |, i.e., x− − y −  |x − y |. Since (x − y ) ∈ g f r (F ), it follows from Proposition 4.6 item (ii) that |x− − y − | ∈ g f r (F ) and from the converse of Proposition 4.6 item (i) we have that x− − y − ∈ g f r (F ) and since x− ∈ g f r (F ) it follows that y − ∈ g f r (F ). (ii) We can assume here that x is q-positive and non-q-negative hence, x has a representative xˆ / {±1} which implies that if A := {ϕ ∈ A0 (K) | θxˆ ≡ 1}, then A or A c belongs to F and such that θxˆ ∈ so the conclusion follows from Proposition 4.5 item (viii). 2 Definition 4.11. Let α ∈ K and let negative, if a− ∈ g f r (F ).

α ∈ R/ g f r (F ) be its image in R/ g f r (F ). We say that α is non-

Lemma 4.10 shows that Definition 4.11 gives rise to an intrinsic order of R/ g f r (F ). The following lemma is well known. Lemma 4.12. Let ( A , ) be a commutative unitary partially ordered ring. Then: (i) ( A , ) is totally ordered if and only if for each a ∈ A either a  0 or −a  0. (ii) If  is a total order on A, the nil-radical of A, N ( A ) say, is zero and if a, b  0 implies that ab  0, then A is an integral domain. We can now state our main result of this sub-section. Theorem 4.13. Let F ∈ P∗ (S f ). Then, (R/ g f r (F ), ) is a totally ordered ring. Proof. This follows at once from Lemma 4.12 and the other results of this section. (The reader should also see the proof of Theorem 3.14 of [8].) 2 4.2. The boolean algebra and prime ideals In this sub-section we completely describe the minimal primes and show that K is not Von Neumann regular. If A is a commutative unitary ring, denote by B ( A ) the set of idempotents of A. Theorem 4.14. Let e ∈ B (K) be a non-trivial idempotent. Then, there exists S ∈ S f , such that e = X S . In particular, B (K) is a discrete subset of K. Proof. Let e be a non-trivial idempotent of K. Then e is a zero divisor and thus satisfies one of the conditions of the Approximation Theorem. Suppose that the second condition is satisfied. Then we obtain a sequence (xn = e Xn ) of non-zero elements converging to zero. But note that each xn is a non-trivial idempotent and thus xn = 1. This contradicts the fact that the sequence converges to 0. It follows that the first condition of the Approximation Theorem must hold. Still using the notation of the mentioned theorem we have that e X S c = (e X S c )2 and e X S c + X S = (e X S v )2 + X S is a unit whose inverse is 1e X S c + X S . Note that the last expression makes sense since the Approximation Theorem

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guarantees that if eˆ is any representative of e then 1eˆ is moderate on S c . Multiplying the equation e X S c + X S = (e X S v )2 + X S by the unit 1e X S c + X S we obtain that e X S c + X S = 1. Hence e = X S c . To prove the last part, one only has to notice that the distance between distinct idempotents is 1, proving the discreteness of B (K). 2 Recall that a commutative unitary ring is said to be Von Neumann regular if its principal ideals are generated by an idempotent. The following result is well known. Proposition 4.15. Let A be a commutative unitary ring and let N ( A ) be its nil-radical. Then every prime ideal of A is maximal iff A /N ( A ) is Von Neumann regular. Lemma 4.16. Let γ : A0 (K) → R ∪ {+∞} be defined as follows:



−1 γ (ϕ ) = +∞, if (i (ϕ ))−1 ∈/ N, p, if (i (ϕ )) ∈ N,

where p is the smallest prime dividing (i (ϕ ))−1 . Let x ∈ K be such that xˆ (ϕ ) = αγ•ˆ(ϕ ) (ϕ ) is a representative of x. Then, xK is not generated by an idempotent. Proof. Suppose that J = xK is an idempotent ideal. By Theorem 4.14, there exists S ∈ S f such that J = X S K. From the definition of γ we consider two cases: (a) (b)

γ ( S ) is finite; γ ( S ) is infinite.

(a) Let γ ( S ) be finite, σ := max{ z | z ∈ R ∩ γ ( S )} and let us fix a prime number p > σ . Given ϕ ∈ A0 (K), let εn := (i (ϕ ) pn )−1 . Then εn −−−−→ 0 and γ (ϕεn ) = p ∈/ γ ( S ), ∀n ∈ N hence, ϕεn ∈/ S, i.e., n→∞ ϕεn ∈ S c . From this one gets that xX S c (ϕεn ) = αγ• (ϕεn ) (ϕεn ), ∀n ∈ N and thus, by Lemma 3.14, xX S c = 0.

Writing x = y X S for some y ∈ K we have that 0 = ( y X S )X S c = xX S c = 0, a contradiction. (b) Now, if γ ( S ) is infinite, then there exists a sequence (εn )n∈N ⊂ {ε | ϕε ∈ S } that converges to zero when n → ∞ such that (γ (ϕεn )) ⊂ N is an increasing and strict divergent sequence. Let us assume that X S = yx. Then

XˆS (ϕε ) − yˆ (ϕε )ˆx(ϕε ) = 1 − yˆ (ϕε )ˆx(ϕε ) −−−→ 0. n→∞

But since (γ (ϕεn )) is a divergent increasing sequence it follows easily that y cannot be a moderate function, that is, y ∈ / E M (K), a contradiction, thus proving the result. 2 Theorem 4.17. K is not Von Neumann regular. In particular, there exists F ∈ P∗ (S f ), such that g f (F ) is not closed and K has a prime ideal which is not maximal. Proof. We know from Theorem 3.20 that the nil-radical of K is null, that is, N (K) = {0}. Therefore, from Proposition 4.15 and Lemma 4.16 it follows that K is not Von Neumann regular. 2 Theorem 4.18. For all F ∈ P∗ (S f ), we have g f (F ) is a prime ideal. Proof. Initially, let us assume that K = R. We need to show that condition (ii) of Lemma 4.12 is verified. Thus, there are a, b ∈ R such that a− , b− ∈ g f (F ). Then,

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(ab)− = =

 1 ab − |a||b| 2      1  + a + a− b + + b − − a+ − a− b + − b − 2

= a+ b − + a− b + . Hence, (ab)− ∈ g f (F ). Therefore, a+ b− + a− b+ ∈ g f (F ). This means that the positive cone is invariant under multiplication. Now, let a ∈ R such that a2 ∈ g f (F ). From the definition of the ideal g f (F ), we have that there exists A ∈ F such that a2 = a2 X A . Hence, a2 X A c = 0 and, thus, aX A c ∈ N (R) = 0. Then, a = aX A + aX A c = aX A ∈ g f (F ). Thus, we are done in this case. To complete the proof let us consider the case where K = C. Let x, y ∈ g f (F ) such that xy ∈ g f (F ). Then |xy | = |x|| y | ∈ R ∩ g f (F ) = g f r (F ). From the previous case we have that |x| or | y | belongs to g f r (F ). Convexity of ideals can now be used to complete the proof. 2 Corollary 4.19. { g f r (F ) | F ∈ P∗ (S f )} is the set of minimal prime ideals of R and { g f (F ) | F ∈ P∗ (S f )} is

the set minimal prime ideals of C. Acknowledgment

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