AATOMICITY AND FACTORIZATION OF PUISEUX MONOIDSByMARLY GOTTIA DISSERTATION PRESENTED TO THE GRADUATE SCHOOLOF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYUNIVERSITY OF FLORIDA2019 a r X i v : . [ m a t h . A C ] J un (cid:13) p -adic Puiseux Monoids . . . . . . . . . . . . . . . . . . . . . . . . . 485 FACTORIZATION INVARIANTS . . . . . . . . . . . . . . . . . . . . . . . . . 535.1 Sets of Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.1.1 Full Systems of Sets of Lengths for BFMs . . . . . . . . . . . . . . . 535.1.2 Puiseux Monoids with Arithmetic Sets of Lengths . . . . . . . . . . 545.2 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.2.1 Union of Sets of Lengths and Local Elasticity . . . . . . . . . . . . . 605.2.2 Prime Reciprocal Puiseux Monoids . . . . . . . . . . . . . . . . . . 675.2.3 Multiplicatively Cyclic Puiseux Monoids . . . . . . . . . . . . . . . 705.3 Tame Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.3.1 Omega Primality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.3.2 Tameness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786 FACTORIAL ELEMENTS OF PUISEUX MONOIDS . . . . . . . . . . . . . . 816.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.2 Molecules of Numerical Monoids . . . . . . . . . . . . . . . . . . . . . . . . 826.2.1 Atoms and Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . 826.2.2 Betti Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.2.3 On the Sizes of the Sets of Molecules . . . . . . . . . . . . . . . . . 846.3 Molecules of Generic Puiseux Monoids . . . . . . . . . . . . . . . . . . . . 866.4 Molecules of Prime Reciprocal Monoids . . . . . . . . . . . . . . . . . . . . 907 PUISEUX ALGEBRAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.1 Monoid Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.2 Factorial Elements of Puiseux Algebras . . . . . . . . . . . . . . . . . . . . 98REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103BIOGRAPHICAL SKETCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086IST OF TABLESTable page5-1 U k for k ∈ { , . . . , } . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637IST OF FIGURESFigure page6-1 Atoms, molecules, and non-molecules of the numerical monoids N = (cid:104) , (cid:105) , N = (cid:104) , , (cid:105) , N = (cid:104) , , , , (cid:105) , and N = (cid:104) , (cid:105) . . . . . . . . . . . . . . . . . . 826-2 The factorization graphs of 90 ∈ Betti( N ) and 84 / ∈ Betti( N ), where N is thenumerical monoid (cid:104) , , , , (cid:105) . . . . . . . . . . . . . . . . . . . . . . . . . 848bstract of Dissertation Presented to the Graduate Schoolof the University of Florida in Partial Fulfillment of theRequirements for the Degree of Doctor of PhilosophyATOMICITY AND FACTORIZATION OF PUISEUX MONOIDSByMarly GottiDecember 2019Chair: Peter SinMajor: MathematicsA commutative and cancellative monoid (or an integral domain) is called atomicif each non-invertible element can be expressed as a product of irreducibles. Manyalgebraic properties of monoids/domains are determined by their atomic structure. Forinstance, a ring of algebraic integers has class group of size at most two if and only ifit is half-factorial (i.e., the lengths of any two irreducible factorizations of an elementare equal). Most atomic monoids are not unique factorization monoids (UFMs). Infactorization theory one studies how far is an atomic monoid from being a UFM. Duringthe last four decades, the factorization theory of many classes of atomic monoids/domains,including numerical/affine monoids, Krull monoids, Dedekind domains, and Noetheriandomains, have been systematically investigated.A Puiseux monoid is an additive submonoid of the nonnegative cone of rationalnumbers. Although Puiseux monoids are torsion-free rank-one monoids, their atomicstructure is rich and highly complex. For this reason, they have been important objectsto construct crucial examples in commutative algebra and factorization theory. In 1974Anne Grams used a Puiseux monoid to construct the first example of an atomic domainnot satisfying the ACCP, disproving Cohn’s conjecture that every atomic domain satisfiesthe ACCP. Even recently, Jim Coykendall and Felix Gotti have used Puiseux monoids toconstruct the first atomic monoids with monoid algebras (over a field) that are not atomic,answering a question posed by Robert Gilmer back in the 1980s.9his dissertation is focused on the investigation of the atomic structure andfactorization theory of Puiseux monoids. Here we established various sufficient conditionsfor a Puiseux monoid to be atomic (or satisfy the ACCP). We do the same for twoof the most important atomic properties: the finite-factorization property and thebounded-factorization property. Then we compare these four atomic properties in thecontext of Puiseux monoids. This leads us to construct and study several classes ofPuiseux monoids with distinct atomic structure. Our investigation provides sufficientevidence to believe that the class of Puiseux monoids is the simplest class with enoughcomplexity to find monoids satisfying almost every fundamental atomic behavior.10HAPTER 1INTRODUCTIONFactorization theory studies the phenomenon of non-unique factorizations intoirreducibles in commutative cancellative monoids and integral domains. Factorizationtheory originated from commutative algebra and algebraic number theory, most of itsinitial motivation was the study of factorization into primes in ring of integers, Dedekinddomains, and Krull domains. During the last four decades factorization theory has becomean autonomous field and has been actively investigated in connection with other areas,including number theory, combinatorics, and convex geometry. The primary goal offactorization theory is to measure how far an atomic monoid or an integral domain isfrom being factorial or half-factorial (i.e., the number of irreducible factors in any twofactorizations of a given element are the same).The origin of factorization theory lies in algebraic number theory, and one of theprimary motivations was the fact that the ring of integers O K of an algebraic number field K usually fails to be a UFD. Consider for example the ring of integers Z [ √− Z [ √−
5] as 6 = 2 · − √− √− , and the elements 2 , , − √− √− Z [ √−
5] (this example is surveyed in one of my recent papers, [19]). Throughouthistory some theories have been developed to understand the phenomenon of non-uniquefactorizations, including C. F. Gauss’s theory of binary quadratic forms for quadraticfields, L. Kronecker’s divisor theory, and R. Dedekind’s ideal theory. In the mid-twentiethcentury, L. Carlitz characterized the half-factorial rings of integers in terms of their classnumber [12], and W. Narkiewicz began a systematic study of non-unique factorizations onring of integers (see [62] and references therein). On the other hand, R. Gilmer studiedfactorization properties of more general integral domains [5]. During the last four decadesand motivated by the work of Gilmer, many authors have influenced the development11f factorization theory in general integral domains (see [4] and references therein). Sincea large number of factorization properties of integral domains do not depend on thedomains’s additive structure, to investigate such properties it often suffices to focus onthe corresponding multiplicative monoids. As a result, several techniques to measurethe non-uniqueness of factorizations have been systematically developed and abstractedto the context of commutative cancellative monoids. Many factorization invariants andarithmetic statistics have been introduced, including the set of lengths, the union of sets oflengths, the elasticity, the catenary degree, and the tame degree.In this thesis, we present results on the factorization theory and atomic structureof a class of commutative and cancellative monoids called Puiseux monoids (i.e, additivesubmonoids of Q ≥ ). Although the atomicity of Puiseux monoids has earned attentiononly in the last few years (see [20, 46] and references therein), since the 1970s Puiseuxmonoids have been crucial in the construction of numerous examples in commutativering theory. Back in 1974, A. Grams [56] used an atomic Puiseux monoid as the mainingredient to construct the first example of an atomic integral domain that does notsatisfy the ACCP, and thus she refuted P. Cohn’s assumption that every atomic integraldomain satisfies the ACCP. In addition, in [2], A. Anderson et al. appealed to Puiseuxmonoids to construct various needed examples of integral domains satisfying certainprescribed properties. More recently, Puiseux monoids have played an important rolein [25], where J. Coykendall and F. Gotti partially answered a question on the atomicity ofmonoid rings posed by R. Gilmer back in the 1980s (see [42, page 189]).Puiseux monoids have also been important in factorization theory. For instance,the class of Puiseux monoids comprises the first (and only) example known so far ofprimary atomic monoids with irrational elasticity (this class was found in [36, Section 4]via [54, Theorem 3.2]). A Puiseux monoid is a suitable additive structure containingsimultaneously several copies of numerical monoids independently generated. This fact hasbeen harnessed by A. Geroldinger and W. Schmid to achieve a nice realization theorem12or the sets of lengths of numerical monoids [41, Theorem 3.3]. In [45] Puiseux monoidswere studied in connection with Krull monoids and transfer homomorphisms. In addition,Puiseux monoids have been recently studied in [8] in connection to factorizations of uppertriangular matrices. Finally, some connections between Puiseux monoids and music theoryhave been recently highlighted by M. Bras-Amoros in the Monthly article [11]. A briefsurvey on Puiseux monoids can be found in [21].This thesis is organized as follows. Chapter 1 provides the background informationand sets up the notation needed to study the atomic structure of Puiseux monoids givenin Chapter 2 and Chapter 3. In these two chapters, the atomic structure of some familiesof Puiseux monoids is fully characterized (Proposition 3.1.4, Corollary 3.2.3, Proposition4.2.7). In addition, the chain of implications 3-1 is shown not to be reversible by providingresults and examples in the realm of Puiseux monoids (Corollary 3.1.5, Theorem 3.2.2,Theorem 3.3.1).Furthermore, Chapters 4, 5, and 6 explore the factorization invariants of Puiseuxmonoids and Puiseux algebras. Sets of lengths, the union of set of lengths, the elasticity, k -elasticities, among other factorization invariants are analyzed (Theorem 5.1.5,Proposition 5.2.4, Theorem 5.2.11). In addition, the connection between moleculesand atomicity in Puiseux monoids and Puiseux algebras is investigated (Theorem 6.3.4,Theorem 7.2.7). 13HAPTER 2ALGEBRAIC BACKGROUND OF PUISEUX MONOIDS In this section we introduce most of the relevant concepts on commutative monoidsand factorization theory required to follow our exposition. General references forbackground information can be found in [57] for commutative monoids and in [37] foratomic monoids and factorization theory.
We let N := { , , . . . } denote the set of positive integers and set N := N ∪ { } . Inaddition, we let P denote the set of all prime numbers. For X ⊆ R and r ∈ R , we set X ≥ r := { x ∈ X : x ≥ r } and we use the notations X >r , X ≤ r , and X
A monoid M is atomic if each element of M \ U ( M ) can be expressedas a sum of atoms, and M is antimatter if A ( M ) = ∅ .Every finitely generated monoid is atomic [37, Proposition 2.7.8(4)], while itimmediately follows that every abelian group is an antimatter monoid.A subset I of M is an ideal of M if I + M = I (or, equivalently, I + M ⊆ I ). An ideal I is principal if I = x + M for some x ∈ M , and M satisfies the ascending chain conditionon principal ideals (or ACCP ) provided that every increasing sequence of principal idealsof M eventually stabilizes. It is well known that every monoid satisfying the ACCP mustbe atomic [37, Proposition 1.1.4].An equivalence relation ρ ⊆ M × M is a congruence if it is compatible with theoperation of the monoid M , i.e., for all x, y, z ∈ M with ( x, y ) ∈ ρ it follows that( z + x, z + y ) ∈ ρ . It can be readily verified that the set M/ρ consisting of the equivalenceclasses of a congruence ρ is a commutative semigroup with identity. For x, y ∈ M , we saythat x divides y in M and write x | M y provided that x + x (cid:48) = y for some x (cid:48) ∈ M . Twoelements x, y ∈ M are associates if y = u + x for some u ∈ U ( M ). Being associates definesa congruence on M whose semigroup of classes is a reduced monoid, which we denote by M red . Observe that M is reduced if and only if M red = M .The Grothendieck group gp ( M ) of M is the abelian group (unique up to isomorphism)satisfying that any abelian group containing a homomorphic image of M also containsa homomorphic image of gp ( M ). The rank of a monoid M is the rank of the Z -module gp ( M ) or, equivalently, the dimension of the Q -vector space Q ⊗ Z gp ( M ). The monoid M is torsion-free if for all x, y ∈ M and n ∈ N , the equality nx = ny implies that x = y .Clearly, M is a torsion-free monoid if and only if gp ( M ) is a torsion-free group.A numerical monoid is a submonoid N of ( N , +) satisfying that | N \ N | < ∞ . If N (cid:54) = N , then max( N \ N ) is the Frobenius number of N . Numerical monoids are finitely15enerated and, therefore, atomic with finitely many atoms. The embedding dimension of N is the cardinality of A ( N ). For an introduction to numerical monoids, see [34], and forsome of their many applications, see [7]. A multiplicative monoid F is called free with basis P if every element x ∈ F can bewritten uniquely in the form x = (cid:89) p ∈ P p v p ( x ) , where v p ( x ) ∈ N and v p ( x ) > p ∈ P . The monoid F is determined by P up to isomorphism. By the Fundamental Theorem of Arithmetic, themultiplicative monoid N is free on the set of prime numbers. In this case, we can extend v p to Q ≥ as follows. For r ∈ Q > let v p ( r ) := v p ( n ( r )) − v p ( d ( r )) and set v p (0) = ∞ . The map v p is called the p - adic valuation on Q . It is not hard to verify that v p is semi-additive , i.e., v p ( r + s ) ≥ min { v p ( r ) , v p ( s ) } for all r, s ∈ Q ≥ . Definition 2.1.2.
Let M be a reduced monoid. The factorization monoid of M , denotedby Z ( M ), is the free commutative monoid on A ( M ). The elements of Z ( M ) are called factorizations .If z = a · · · a n ∈ Z ( M ), where a , . . . , a n ∈ A ( M ), then | z | := n is the length of z . Theunique monoid homomorphism π : Z ( M ) → M satisfying that π ( a ) = a for all a ∈ A ( M ) isthe factorization homomorphism of M . For each x ∈ M , Z ( x ) := π − ( x ) ⊆ Z ( M ) and L ( x ) := {| z | : z ∈ Z ( x ) } are the set of factorizations and the set of lengths of x , respectively. Factorizationinvariants stemming from the sets of lengths have been studied for several classes ofatomic monoids and domains; see, for instance, [15, 18, 22, 23]. In particular, the sets oflengths of numerical monoids have been studied in [1, 14, 41]. In [41] the sets of lengths16f numerical monoids were studied using techniques involving Puiseux monoids. Anoverview of sets of lengths and the role they play in factorization theory can be found inthe Monthly article [35].By restricting the size of the sets of factorizations/lengths, one obtains subclasses ofatomic monoids that have been systematically studied by many authors. We say that areduced atomic monoid M is1. a UFM (or a factorial monoid ) if | Z ( x ) | = 1 for all x ∈ M ,2. an HFM (or a half-factorial monoid ) if | L ( x ) | = 1 for all x ∈ M ,3. an FFM (or a finite-factorization monoid ) if | Z ( x ) | < ∞ for all x ∈ M , and4. a BFM (or a bounded-factorization monoid ) if | L ( x ) | < ∞ for all x ∈ M . In this section we study some algebraic aspects of Puiseux monoids.
Definition 2.2.1. A Puiseux monoid is an additive submonoid of Q ≥ .Note that Puiseux monoids are natural generalizations of numerical monoids. Asnumerical monoids, it is clear that Puiseux monoids are reduced. However, as we shall seelater, Puiseux monoids are not, in general, finitely generated or atomic. Recall that a monoid M is torsion-free if for all x, y ∈ M and n ∈ N , the equality nx = ny implies that x = y . Each Puiseux monoid M is obviously torsion-free and,therefore, gp ( M ) is a torsion-free group. Moreover, for a Puiseux monoid M , one can takethe Grothendieck group gp ( M ) to be an additive subgroup of Q , specifically, gp ( M ) = { x − y : x, y ∈ M } . (2-1)Puiseux monoids can be characterized as follows. Proposition 2.2.2.
For a nontrivial monoid M the following statements are equivalent.1. M is a rank- torsion-free monoid that is not a group.2. M is isomorphic to a Puiseux monoid. roof. To argue (1) ⇒ (2), first note that gp ( M ) is a rank-1 torsion-free abelian group.Therefore it follows from [32, Section 85] that gp ( M ) is isomorphic to a subgroup of( Q , +), and one can assume that M is a submonoid of ( Q , +). Since M is not a group,[42, Theorem 2.9] ensures that either M ⊆ Q ≤ or M ⊆ Q ≥ . So M is isomorphic to aPuiseux monoid. To verify (2) ⇒ (1), let us assume that M ⊆ gp ( M ) ⊆ Q . As gp ( M ) isa subgroup of ( Q , +), it is a rank-1 torsion-free abelian group. This implies that M is arank-1 torsion-free monoid. Since M is nontrivial and reduced, it cannot be a group, whichcompletes our proof.Puiseux monoids are abundant, as the next proposition illustrates. Proposition 2.2.3.
There are uncountably many non-isomorphic Puiseux monoids.Proof.
Consider the assignment G (cid:55)→ M G := G ∩ Q ≥ sending each subgroup G of( Q , +) to a Puiseux monoid. Clearly, gp ( M G ) ∼ = G . In addition, for all subgroups G and G (cid:48) of ( Q , +), each monoid isomorphism between M G and M G (cid:48) naturally extends toa group isomorphism between G and G (cid:48) . Hence our assignment sends non-isomorphicgroups to non-isomorphic monoids. It follows from [32, Corollary 85.2] that there areuncountably many non-isomorphic rank-1 torsion-free abelian groups. As a result, thereare uncountably many non-isomorphic Puiseux monoids. Given a monoid M with Grothendieck group gp ( M ), the sets • M (cid:48) := (cid:8) x ∈ gp ( M ) : there exists N ∈ N such that nx ∈ M for all n ≥ N (cid:9) , • (cid:102) M := (cid:8) x ∈ gp ( M ) : there exists n ∈ N such that nx ∈ M (cid:9) , and • (cid:99) M := (cid:8) x ∈ gp ( M ) : there exists c ∈ M such that c + nx ∈ M for all n ∈ N (cid:9) are called the seminormal closure , root closure , and complete integral closure of M ,respectively. It is not hard to verify that M ⊆ M (cid:48) ⊆ (cid:102) M ⊆ (cid:99) M ⊆ gp ( M ) for any monoid M . It has been recently proved by Geroldinger et al. that for Puiseux monoids the threeclosures coincide. 18 roposition 2.2.4. [36, Proposition 3.1] Let M be a Puiseux monoid, and let n =gcd( n ( M • )) . Then M (cid:48) = (cid:102) M = (cid:99) M = gp ( M ) ∩ Q ≥ = n (cid:104) /d : d ∈ d ( M • ) (cid:105) . (2-2)A monoid M is said to be root-closed provided that (cid:102) M = M . In addition, M is calleda Pr¨ufer monoid if M is the union of an ascending sequence of cyclic submonoids. Corollary 2.2.5.
For a Puiseux monoid M , the following statements are equivalent.1. M is root-closed.2. M = n (cid:104) /d : d ∈ d ( M • ) (cid:105) , where n = gcd n ( M • ) .3. gp( M ) = M ∪ − M .4. M is a Pr¨ufer monoid.Proof. The equivalences (1) ⇔ (2) ⇔ (3) follow from Proposition 2.2.4, while theequivalence (1) ⇔ (4) follows from [42, Theorem 13.5].We now characterize finitely generated Puiseux monoids in terms of its root closures. Proposition 2.2.6.
For a Puiseux monoid M the following statements are equivalent.1. (cid:102) M ∼ = ( N , +) .2. M is finitely generated.3. d ( M • ) is finite.4. M is isomorphic to a numerical monoid.Proof. To prove (1) ⇒ (2), suppose that (cid:102) M ∼ = ( N , +). Proposition 2.2.4 ensuresthat d ( M • ) is finite. Now if (cid:96) := lcm d ( M • ), then (cid:96)M is submonoid of ( N , +) that isisomorphic to M . Hence M is finitely generated. To argue (2) ⇒ (3), it suffices to noticethat if S is a finite generating set of M , then every element of d ( M • ) divides lcm d ( S • ).For (3) ⇒ (4), let (cid:96) := lcm d ( M • ). Then note that (cid:96)M is a submonoid of ( N , +) that isisomorphic to M . As a result, M is isomorphic to a numerical monoid. To prove (4) ⇒ (1), assume that M is a numerical monoid and that gp ( M ) is a subgroup of ( Z , +). By19efinition of (cid:102) M , it follows that (cid:102) M ⊆ N . On the other hand, the fact that N \ M is finiteimmediately implies that N ⊆ (cid:102) M . Thus, (cid:102) M = ( N , +). Corollary 2.2.7.
A Puiseux monoid M is not finitely generated if and only if (cid:102) M isantimatter.Proof. Suppose first that M is not finitely generated. Set n = gcd( n ( M • )). It followsfrom Proposition 2.2.4 that (cid:102) M = (cid:104) n/d : d ∈ d ( M • ) (cid:105) . Fix d ∈ d ( M • ). Since d ( M • ) is aninfinite set that is closed under taking least common multiples, there exists d (cid:48) ∈ d ( M • )such that d (cid:48) properly divides d . As a consequence, n/d (cid:48) properly divides n/d in (cid:102) M and so n/d / ∈ A ( (cid:102) M ). As none of the elements in the generating set { n/d : d ∈ d ( M • ) } of (cid:102) M isan atom, (cid:102) M must be antimatter. The reverse implication is an immediate consequence ofProposition 2.2.6.The conductor of a monoid M , denoted by ( M : (cid:99) M ), is defined to be( M : (cid:99) M ) = { x ∈ gp ( M ) : x + (cid:99) M ⊆ M } . (2-3)For a numerical monoid N , the term ‘conductor’ also refers to the number f ( N ) + 1,where f ( N ) denotes the Frobenius number of N . This does not generate ambiguity as thefollowing example illustrates. Example 2.2.8.
Let N be a numerical monoid, and let f ( N ) be the Frobenius numberof N . It follows from (2-1) that gp ( N ) = Z . Therefore Proposition 2.2.4 guarantees that (cid:98) N = N . For each n ∈ N with n ≥ f ( N ) + 1, it is clear that n + (cid:98) N = n + N ⊆ N . Onthe other hand, for each n ∈ Z with n ≤ f ( N ) the fact that f ( N ) ∈ n + (cid:98) N implies that n + (cid:98) N (cid:42) N . As a result, ( N : (cid:98) N ) = { n ∈ Z : n ≥ f ( N ) + 1 } . (2-4)As the equality of sets (2-4) shows, the minimum of ( N : (cid:98) N ) is f ( N ) + 1, namely, theconductor number of N , as defined in the context of numerical monoids.20he conductor of a Puiseux monoid was first considered in [36], where the followingresult was established. Proposition 2.2.9.
Let M be a Puiseux monoid. Then the following statements hold.1. If M is root-closed, then ( M : (cid:99) M ) = (cid:99) M = M .2. If M is not root-closed and σ = sup (cid:99) M \ M .(a) If σ = ∞ , then ( M : (cid:99) M ) = ∅ .(b) If σ < ∞ , then ( M : (cid:99) M ) = M ≥ σ . Remark 2.2.10.
With notation as in Proposition 2.2.9.2, although (cid:99) M >σ = M >σ holds, itcan happen that (cid:99) M ≥ σ (cid:54) = M ≥ σ . For instance, consider the Puiseux monoid { } ∪ Q > . As we are about to show, homomorphisms between Puiseux monoids are given byrational multiplication.
Proposition 2.2.11.
The homomorphisms between Puiseux monoids are given by rationalmultiplication.Proof.
Every rational-multiplication map is clearly a homomorphism. Suppose, onthe other hand, that ϕ : M → M (cid:48) is a homomorphism between Puiseux monoids.As the trivial homomorphism is multiplication by 0, one can assume without loss ofgenerality that ϕ is nontrivial. Let { n , . . . , n k } be the minimal generating set of theadditive monoid N := M ∩ N . Since ϕ is nontrivial, k ≥ ϕ ( n j ) (cid:54) = 0 for some j ∈ { , . . . , k } . Set q = ϕ ( n j ) /n j and then take r ∈ M • and c , . . . , c k ∈ N satisfying that n ( r ) = c n + · · · + c k n k . As n i ϕ ( n j ) = ϕ ( n i n j ) = n j ϕ ( n i ) for every i ∈ { , . . . , k } , ϕ ( r ) = 1 d ( r ) ϕ ( n ( r )) = 1 d ( r ) k (cid:88) i =1 c i ϕ ( n i ) = 1 d ( r ) k (cid:88) i =1 c i n i ϕ ( n j ) n j = rq. Thus, the homomorphism ϕ is multiplication by q ∈ Q > .21HAPTER 3ATOMICITY OF PUISEUX MONOIDSIt is well known that in the class consisting of all monoids, the following chain ofimplications holds. UFM ⇒ HFM reduced ⇒ FFM ⇒ BFM ⇒ ACCP ⇒ atomic monoid (3-1)It is also known that, in general, none of the implications in (3-1) is reversible (even in theclass of integral domains [2]). In this chapter, we provide various examples to illustratethat none of the above implications, except the first one, is reversible in the class ofPuiseux monoids. We characterize the Puiseux monoids belonging to the first two classesof the chain of implications (3-1). For each of the last four classes, we find a family ofPuiseux monoids belonging to such a class but not to the class right before. We begin this section with an easy characterization of finitely generated Puiseuxmonoids in terms of their atomicity.
Proposition 3.1.1.
A Puiseux monoid M is finitely generated if and only if M is atomicand A ( M ) is finite.Proof. The direct implication follows immediately from the fact that finitely generatedPuiseux monoids are isomorphic to numerical monoids via Proposition 2.2.6. The reverseimplication is also obvious because the atomicity of M means that M is generated by A ( M ), which is finite.Corollary 2.2.7 yields, however, instances of non-finitely generated Puiseux monoidscontaining no atoms. As the next example shows, for every n ∈ N there exists anon-finitely generated Puiseux monoid containing exactly n atoms. Example 3.1.2.
Let m ∈ N , and take distinct prime numbers p and q with q > m .Consider the Puiseux monoid M = (cid:10) { m, . . . , m − } ∪ { qp − m − i : i ∈ N } (cid:11) . To verify that22 ( M ) = { m, . . . , m − } , write a ∈ { m, . . . , m − } as a = a (cid:48) + N (cid:88) n =1 c n qp m + n , (3-2)where a (cid:48) ∈ { } ∪ { m, . . . , m − } and c n ∈ N for every n ∈ { , . . . , N } . After cleaningdenominators in (3-2), one finds that q | a − a (cid:48) . So a = a (cid:48) and c = · · · = c N = 0, whichimplies that a ∈ A ( M ). Thus, { m, . . . , m − } ⊆ A ( M ). Clearly, qp − m − i / ∈ A ( M ) for any i ∈ N . Hence A ( M ) = { m, . . . , m − } , and so |A ( M ) | = m . As d ( M • ) is not finite, itfollows from Proposition 2.2.6 that M is not finitely generated.Perhaps the class of non-finitely generated Puiseux monoids that has been mostthoroughly studied is the class of multiplicatively cyclic Puiseux monoids [20]. Definition 3.1.3.
For r ∈ Q > , we call M r = (cid:104) r n : n ∈ N (cid:105) a multiplicatively cyclic Puiseux monoid.Although M r is, indeed, a rational cyclic semiring, we shall only be concerned herewith its additive structure. The atomicity of multiplicatively cyclic Puiseux monoids wasfirst studied in [51, Section 6] while several factorization aspects were recently investigatedin [20]. Proposition 3.1.4.
For r ∈ Q > , consider the multiplicatively cyclic Puiseux monoid M r .Then the following statements hold.1. If r ≥ , then M r is atomic and • either r ∈ N and so M r = N , • or r / ∈ N and so A ( M r ) = { r n : n ∈ N } .2. If r < , then • either n ( r ) = 1 and so M r is antimatter, • or n ( r ) (cid:54) = 1 and M r is atomic with A ( M r ) = { r n : n ∈ N } .Proof. To argue (1), suppose that r ≥
1. If r ∈ N , then it easily follows that M r = N .Then we assume that r / ∈ N . Clearly, A ( M r ) ⊆ { r n : n ∈ N } . To check the reverse23nequality, fix j ∈ N and write r j = (cid:80) Ni =0 α i r i for some N ∈ N and α i ∈ N for every i ∈ { , . . . , N } . As ( r n ) n ∈ N is an increasing sequence, one can assume that N ≤ j . Then,after cleaning denominators in r j = (cid:80) Ni =0 α i r i we obtain N = j as well as α j = 1 and α i = 0 for every i (cid:54) = j . Hence r j ∈ A ( M r ) for every j ∈ N , which yields the secondstatement of (1).Now suppose that r <
1. If n ( r ) = 1 then r n = d ( r ) r n +1 for every n ∈ N , and so M r is antimatter, which is the first statement of (2). Finally, suppose that n ( r ) >
1. Fix j ∈ N , and notice that r i (cid:45) M r r j for any i < j . Then write r j = (cid:80) j + ki = j β i r i , for some k ∈ N and β i ∈ N for every i ∈ { j, . . . , j + k } . Notice that β j ∈ { , } . Suppose by acontradiction that β j = 0. In this case, k ≥
1. Let p be a prime dividing n ( r ), and let α bethe maximum power of p dividing n ( r ). From r j = (cid:80) j + ki = j β i r i one obtains αj = v p (cid:0) r j (cid:1) = v p (cid:18) k (cid:88) i =1 β j + i r j + i (cid:19) ≥ min i ∈{ ,...,k } (cid:8) v p (cid:0) β j + i r j + i (cid:1)(cid:9) ≥ α ( j + m ) , (3-3)where m = min { i ∈ { , . . . , k } : β j + i (cid:54) = 0 } . The inequality (3-3) yields the desiredcontradiction. Hence r j ∈ A ( M r ) for every j ∈ N , which implies the second statementof (2). Corollary 3.1.5.
For each r ∈ Q ∩ (0 , with n ( r ) (cid:54) = 1 , the monoid M r is an atomicmonoid that does not satisfy the ACCP.Proof. Proposition 3.1.4 guarantees the atomicity of M r . To verify that M r does notsatisfy the ACCP, consider the sequence of principal ideals ( n ( r ) r n + M r ) n ∈ N . Since n ( r ) r n = d ( r ) r n +1 = ( d ( r ) − n ( r )) r n +1 + n ( r ) r n +1 , n ( r ) r n +1 | M r n ( r ) r n for every n ∈ N . Therefore ( n ( r ) r n + M r ) n ∈ N is an ascending chain ofprincipal ideals. In addition, it is clear that such a chain of ideals does not stabilize. Hence M r does not satisfy the ACCP, which completes the proof.24 .2 A Class of ACCP Puiseux Monoids We proceed to present a class of ACCP Puiseux monoids containing a subclass ofmonoids that are not BFMs.
Definition 3.2.1.
A Puiseux monoid M is called prime reciprocal provided that thereexists a generating set S of M satisfying the following two conditions. • d ( S ) ⊆ P . • d ( a ) = d ( a (cid:48) ) implies that a = a (cid:48) for all a, a (cid:48) ∈ S .Let us prove that all prime reciprocal Puiseux monoids satisfy the ACCP. Theorem 3.2.2.
Every prime reciprocal Puiseux monoid satisfies the ACCP.Proof.
Let ( p n ) n ∈ N be a strictly increasing sequence of prime numbers, and let ( a n ) n ∈ N bea sequence of positive integers such that p n (cid:45) a n . Now set M = (cid:104) a n /p n : n ∈ N (cid:105) . It is nothard to verify that for all x ∈ M there exist k, n ∈ N and α i ∈ { , . . . , p i } such that x = n + k (cid:88) i =1 α i a i p i . (3-4)Let us now check that the sum decomposition in (3-4) is unique. To do this, take k (cid:48) , m ∈ N and β i ∈ { , . . . , p i } such that n + k (cid:88) i =1 α i a i p i = m + k (cid:48) (cid:88) i =1 β i a i p i . (3-5)Suppose, without loss of generality, that k = k (cid:48) . After isolating ( α i − β i ) a i /p i in (3-5) andapplying the p i -adic valuation, we obtain that p i | α i − β i , which implies that α i = β i foreach i ∈ { , . . . , k } . As a consequence, n = m and we can conclude that the uniqueness ofthe decomposition in (3-4) holds.Now suppose, by way of contradiction, that M does not satisfy the ACCP. Then thereexists a strictly decreasing sequence ( q n ) n ∈ N of elements in M such that q + M (cid:40) q + M (cid:40) · · · . (3-6)25n the unique sum decomposition of q as in (3-4), let N be the integer and let p n , . . . , p n k be the distinct prime denominators of the atoms with nonzero coefficients. By (3-6), forevery n ∈ N there exists r n +1 ∈ M such that q n = q n +1 + r n +1 and, therefore, n (cid:88) i =1 r i +1 ≤ q n +1 + n (cid:88) i =1 r i +1 = q for every n ∈ N . Thus, lim n →∞ r n = 0. Then, for any finite subset A of A ( M ) there exist alarge enough (cid:96) ∈ N and a ∈ A ( M ) such that a | M r (cid:96) and a / ∈ A . Hence we can find t ∈ N and (possibly repeated) p m , . . . , p m t ∈ P satisfying that |{ p m , . . . , p m t }| > N + k and a m i /p m i ∈ A ( r i +1 ) for each i ∈ { , . . . , t } . Take z i ∈ Z ( r i ) containing the atom a m i /p m i ,and take z t +1 ∈ Z ( q t +1 ). As q = q t +1 + (cid:80) ti =1 r i +1 , we have z = z t +1 + (cid:80) ti =1 z i ∈ Z ( q ).By the uniqueness of the sum decomposition in (3-4), the factorization z contains at least d ( a ) copies of each atom a for which d ( a ) ∈ P := { p m , . . . , p m t } \ { p n , . . . , p n k } . Since |{ p m , . . . , p m t }| > N + k , it follows that | P | > N . Thus, N ≥ (cid:88) a ∈A ( M ) | d ( a ) ∈ P d ( a ) a ≥ | P | > N, which is a contradiction. Hence M satisfies the ACCP. Corollary 3.2.3.
Prime reciprocal Puiseux monoids are atomic.Proof.
It follows from Theorem 3.2.2, along with the chain of implications we exhibited atthe beginning of this chapter.
Corollary 3.2.4.
There are Puiseux monoids satisfying the ACCP that are not BFMs.Proof.
Consider the Puiseux monoid M = (cid:104) /p : p ∈ P (cid:105) . We have seen in Theorem 3.2.2that M satisfies the ACCP. However, it is clear that p ∈ L (1) for every p ∈ P . As | L (1) | = ∞ , the monoid M is not a BFM. Our next main goal is to find a large class of Puiseux monoids that are BFMs. Thisamounts to prove the following result. 26 heorem 3.3.1.
Let M be a Puiseux monoid. If is not a limit point of M • , then M isa BFM.Proof. It is clear that A ( M ) consists of those elements of M • that cannot be written asthe sum of two positive elements of M . Since 0 is not a limit point of M there exists (cid:15) > (cid:15) < x for all x ∈ M • . Now we show that M = (cid:104)A ( M ) (cid:105) . Take x ∈ M • . Since (cid:15) isa lower bound for M • , the element x can be written as the sum of at most (cid:98) x/(cid:15) (cid:99) elementsof M • . Take the maximum natural m such that x = a + · · · + a m for some a , . . . , a m ∈ M • .By the maximality of m , it follows that a i ∈ A ( M ) for every i ∈ { , . . . , m } , which meansthat x ∈ (cid:104)A ( M ) (cid:105) . Hence M is atomic. We have already noticed that every element x in M • can be written as the sum of at most (cid:98) x/(cid:15) (cid:99) atoms, i.e., | L ( x ) | ≤ (cid:98) x/(cid:15) (cid:99) for all x ∈ M .Thus, M is a BFM.The converse of Theorem 3.3.1 does not hold. The following example sheds some lightupon this observation. Example 3.3.2.
Let ( p n ) n ∈ N and ( q n ) n ∈ N be two strictly increasing sequence of primenumbers such that q n > p n for every n ∈ N . Set M := (cid:10) p n q n : n ∈ N (cid:11) . It follows from [51,Corollary 5.6] that M is atomic, and it is easy to verify that A ( M ) = { p n /q n : n ∈ N } . Toargue that M is indeed a BFM, take x ∈ M • and note that since both sequences ( p n ) n ∈ N and ( q n ) n ∈ N are strictly increasing, there exists N ∈ N such that q n (cid:45) d ( x ) and p n > x for every n ≥ N . As a result, if z ∈ Z ( x ), then none of the atoms in { p n /q n : n > N } can appear in z . From this, one can deduce that Z ( x ) is finite. Then L ( x ) is finite for any x ∈ M , and so M is a BFM. However, q n > p n for every n ∈ N implies that 0 is a limitpoint of M • .As we have seen in Corollary 3.2.4, not every ACCP Puiseux monoid is a BFM.However, under a mild assumption on conductors, each of these atomic conditions isequivalent to having 0 as a limit point. The next theorem, along with the next twoexamples, has been recently established in [36].27 heorem 3.3.3. [36, Theorem 3.4] If M is a nontrivial Puiseux monoid with nonemptyconductor, then the following conditions are equivalent.1. is not a limit point of M • .2. M is a BFM.3. M satisfies the ACCP. As Corollary 3.2.4 and Example 3.3.2 indicate, without the nonempty-conductorcondition, none of the last two statements in Theorem 3.3.3 implies its predecessor. Inaddition, even inside the class of Puiseux monoids with nonempty conductor neither beingatomic nor being an FFM is equivalent to being a BFM (or satisfying the ACCP).
Example 3.3.4.
Consider the Puiseux monoid M := { } ∪ Q ≥ . It is clear that theconductor of M is nonempty. In addition, it follows from Theorem 3.3.3 that M is a BFM.Note that A ( M ) = [1 , M is far from being an FFM; for instance, the formalsum (1 + 1 /n ) + ( x − − /n ) is a length-2 factorization in Z ( x ) for all x ∈ (2 , ∩ Q and n ≥ (cid:6) x − (cid:7) , which implies that | Z ( x ) | = ∞ for all x ∈ M > . Example 3.3.5.
Now consider the Puiseux monoid M = (cid:104) /p : p ∈ P (cid:105) ∪ Q ≥ . Since themonoid (cid:104) /p : p ∈ P (cid:105) is atomic by Theorem 3.2.2, it is not hard to check that M is alsoatomic. It follows from Proposition 2.2.9 that M has nonempty conductor. Since 0 is alimit point of M • , Theorem 3.3.3 ensures that M does not satisfy the ACCP. We are in a position now to begin our study of the atomic structure of Puiseuxmonoids generated by monotone sequences.
Definition 3.4.1.
A Puiseux monoid M is said to be increasing (resp., decreasing ) ifit can be generated by an increasing (resp., decreasing) sequence. A Puiseux monoid is monotone if it is either increasing or decreasing.Not every Puiseux monoid is monotone, as the next example illustrates.28 xample 3.4.2. Let p , p , . . . be an increasing enumeration of the set of prime numbers.Consider the Puiseux monoid M = (cid:104) A ∪ B (cid:105) , where A = (cid:26) p n : n ∈ N (cid:27) and B = (cid:26) p n − − p n − : n ∈ N (cid:27) . It follows immediately that both A and B belong to A ( M ). So M is atomic, and A ( M ) = A ∪ B . Every generating set of M must contain A ∪ B and so will have atleast two limit points, namely, 0 and 1. Since every monotone sequence of rationals canhave at most one limit point in the real line, we conclude that M is not monotone.The following proposition describes the atomic structure of the family of increasingPuiseux monoids. Proposition 3.4.3.
Every increasing Puiseux monoid is atomic. Moreover, if ( r n ) n ∈ N is an increasing sequence of positive rationals generating a Puiseux monoid M , then A ( M ) = { r n : r n / ∈ (cid:104) r , . . . , r n − (cid:105)} .Proof. The fact that M is atomic follows from observing that r is a lower bound for M • and so 0 is not a limit point of M . To prove the second statement, set A = { r n : r n / ∈ (cid:104) r , . . . , r n − (cid:105)} , and rename the elements of A in a strictly increasing sequence (possibly finite), namely,( a n ) n ∈ N . Note that M = (cid:104) A (cid:105) and a n / ∈ (cid:104) a , . . . , a n − (cid:105) for any n ∈ N . Since a is thesmallest nonzero element of M , we obtain that a ∈ A ( M ). Suppose now that n is anatural such that 2 ≤ n ≤ | A | . Because ( a n ) n ∈ N is a strictly increasing sequence and a n / ∈ (cid:104) a , . . . , a n − (cid:105) , one finds that a n cannot be written as a sum of elements in M in anon-trivial manner. Hence a n is an atom for every n ∈ N and, therefore, we can concludethat A ( M ) = A .Now we use Proposition 3.4.3 to show that every Puiseux monoid that is notisomorphic to a numerical monoid has an atomic submonoid with infinitely many atoms.Let us first prove the next lemma. 29 emma 3.4.4. Let M be a nontrivial Puiseux monoid. Then d ( M • ) is finite if and only if M is isomorphic to a numerical monoid.Proof. Suppose first that d ( M • ) is finite. Since M is not trivial, M • is not empty. Take a ∈ N to be the least common multiple of d ( M • ). Since aM is a submonoid of N , itis isomorphic to a numerical monoid. Furthermore, the map ϕ : M → aM defined by ϕ ( x ) = ax is a monoid isomorphism. Thus, M is isomorphic to a numerical monoid.Conversely, suppose that M is isomorphic to a numerical monoid. Because every numericalmonoid is finitely generated, so is M . Hence d ( M • ) is finite. Proposition 3.4.5. If M is a nontrivial Puiseux monoid, then it satisfies exactly one ofthe following conditions:1. M is isomorphic to a numerical monoid;2. M contains an atomic submonoid with infinitely many atoms.Proof. Suppose that M is not isomorphic to any numerical monoid. Take r ∈ M • . ByLemma 3.4.4, the set d ( M • ) is not finite. Therefore d ( (cid:104) r (cid:105) • ) is strictly contained in d ( M • ).Take r (cid:48) ∈ M • such that d ( r (cid:48) ) / ∈ d ( (cid:104) r (cid:105) • ). Let r be the sum of m copies of r (cid:48) , where m isa natural number so that gcd( m , d ( r (cid:48) )) = 1 and m r (cid:48) > r . Setting r = m r (cid:48) , we noticethat r > r and r / ∈ (cid:104) r (cid:105) . Now suppose that r , . . . , r n ∈ M have been already chosen sothat r i +1 > r i and r i +1 / ∈ (cid:104) r , . . . , r i (cid:105) for i = 1 , . . . , n −
1. Once again, by using Lemma 3.4.4we can guarantee that d ( M • ) \ d ( (cid:104) r , . . . , r n (cid:105) • ) is not empty. Take r (cid:48) n +1 ∈ M • such that d ( r (cid:48) n +1 ) / ∈ d ( (cid:104) r , . . . , r n (cid:105) • ), and choose m n +1 ∈ N so that gcd( m n +1 , d ( r (cid:48) n +1 )) = 1 and m n +1 r (cid:48) n +1 > r n . Taking r n +1 = m n +1 r (cid:48) n +1 , one finds that r n +1 > r n and r n +1 / ∈ (cid:104) r , . . . , r n (cid:105) .Using the method just described, we obtain an infinite sequence ( r n ) n ∈ N of elements in M satisfying that r n +1 > r n and r n +1 / ∈ (cid:104) r , . . . , r n (cid:105) for every n ∈ N . By Proposition 3.4.3, thesubmonoid N = (cid:104) r n : n ∈ N (cid:105) is atomic and A ( N ) = { r n : n ∈ N } . Hence M has an atomicsubmonoid with infinitely many atoms, namely, N .30inally, note that conditions (1) and (2) exclude each other; this is because asubmonoid of a numerical monoid is either trivial or isomorphic to a numerical monoidand so it must contain only finitely many atoms.Now we split the family of increasing Puiseux monoids into two fundamentalsubfamilies. We will see that these two subfamilies have different behavior. We saythat a sequence of rationals is strongly increasing if it increases to infinity. On the otherhand, a bounded increasing sequence of rationals is called weakly increasing . Definition 3.4.6.
A Puiseux monoid is said to be strongly (resp., weakly ) increasing if itcan be generated by a strongly (resp., weakly) increasing sequence. Proposition 3.4.7.
Every increasing Puiseux monoid is either strongly increasing orweakly increasing. A Puiseux monoid is both strongly and weakly increasing if and only ifit is isomorphic to a numerical monoid.Proof.
The first statement follows straightforwardly. For the second statement, supposethat M is a Puiseux monoid that is both strongly and weakly increasing. By Proposition 3.4.3,the monoid M is atomic, and its set of atoms can be listed increasingly. Let ( a n ) n ∈ N bean increasing sequence with underlying set A ( M ). Suppose, by way of contradiction,that A ( M ) is not finite. Since M is strongly increasing, ( a n ) n ∈ N must be unbounded.However, the fact that M is weakly decreasing forces ( a n ) n ∈ N to be bounded, which is acontradiction. Hence A ( M ) is finite, which implies that M is isomorphic to a numericalmonoid.To prove the converse implication, take M to be a Puiseux monoid isomorphic to anumerical monoid. So M is finitely generated, namely, M = (cid:104) r , . . . , r n (cid:105) for some n ∈ N and r < · · · < r n . The sequence ( a n ) n ∈ N defined by a k = r k if k ≤ n and a k = kr n if k > n is an unbounded increasing sequence generating M . Similarly, the sequence ( b n ) n ∈ N definedby b k = r k if k ≤ n and b k = r n if k > n is a bounded increasing sequence generating M .Consequently, M is both strongly and weakly increasing.31e will show that the strongly increasing property is hereditary on the class ofstrongly increasing Puiseux monoids. We will require the following lemma. Lemma 3.4.8.
Let R be an infinite subset of Q ≥ . If R does not contain any limit points,then it is the underlying set of a strongly increasing sequence.Proof. For every r ∈ R and every subset S of R , the interval [0 , r ] must contain onlyfinitely many elements of S ; otherwise there would be a limit point of S in [0 , r ]. Thereforeevery nonempty subset of R has a minimum element. So the sequence ( r n ) n ∈ N recurrentlydefined by r = min R and r n = min R \ { r , . . . , r n − } is strictly increasing and has R asits underlying set. Since R is infinite and contains no limit points, the increasing sequence( r n ) n ∈ N must be unbounded. Hence R is the underlying set of the strongly increasingsequence ( r n ) n ∈ N . Theorem 3.4.9.
A nontrivial Puiseux monoid M is strongly increasing if and only ifevery submonoid of M is increasing.Proof. If M is finitely generated, then it is isomorphic to a numerical monoid, and thestatement of the theorem follows immediately. So we will assume for the rest of this proofthat M is not finitely generated. Suppose that M is strongly increasing. Let us start byverifying that M does not have any real limit points. By Proposition 3.4.3, the monoid M is atomic. As M is atomic and non-finitely generated, |A ( M ) | = ∞ . Let ( a n ) n ∈ N bean increasing sequence with underlying set A ( M ). Since M is strongly increasing and A ( M ) is an infinite subset contained in every generating set of M , the sequence ( a n ) n ∈ N is unbounded. Therefore, for every r ∈ R , the interval [0 , r ] contains only finitely manyelements of ( a n ) n ∈ N , say a , . . . , a k for k ∈ N . Since (cid:104) a , . . . , a k (cid:105) ∩ [0 , r ] is a finite set, itfollows that M ∩ [0 , r ] is finite as well. Because | [0 , r ] ∩ M | < ∞ for all r ∈ R , it followsthat M does not have any limit points in R .Now suppose that N is a nontrivial submonoid of M . Notice that, being a subset of M , the monoid N cannot have any limit points in R . Thus, by Lemma 3.4.8, the set N
32s the underlying set of a strongly increasing sequence of rationals. Hence N is a stronglyincreasing Puiseux monoid, and the direct implication follows.For the converse implication, suppose that M is not strongly increasing. We willcheck that, in this case, M contains a submonoid that is not increasing. If M is notincreasing, then M is a submonoid of itself that is not increasing. Suppose, therefore, that M is increasing. By Proposition 3.4.3, the monoid M is atomic, and we can list its atomsincreasingly. Let ( a n ) n ∈ N be an increasing sequence with underlying set A ( M ). Because M is not strongly increasing, there exists a positive real (cid:96) that is the limit of the sequence( a n ) n ∈ N . Since (cid:96) is a limit point of M , which is closed under addition, it follows that 2 (cid:96) and 3 (cid:96) are both limit points of M . Let ( b n ) n ∈ N and ( c n ) n ∈ N be sequences in M havinginfinite underlying sets such that lim b n = 2 (cid:96) and lim c n = 3 (cid:96) . Furthermore, assume thatfor each n ∈ N , | b n − (cid:96) | < (cid:96) | c n − (cid:96) | < (cid:96) . (3-7)Take N to be the submonoid of M generated by the set A := { b n , c n : n ∈ N } . Note that A contains at least two limit points. Let us verify that N is atomic with A ( N ) = A . Theinequalities (3-7) immediately imply that A is bounded from above by 3 (cid:96) + (cid:96)/
4. On theother hand, proving that A ( N ) = A amounts to showing that the sets A and A + A aredisjoint. To verify this, it suffices to note thatinf( A + A ) = inf (cid:8) b m + b n , b m + c n , c m + c n : m, n ∈ N (cid:9) ≥ min (cid:26) (cid:96) − (cid:96) , (cid:96) − (cid:96) , (cid:96) − (cid:96) (cid:27) > (cid:96) + (cid:96) ≥ sup A. Thus, A ( N ) = A . Since every increasing sequence has at most one limit point in R , theset A cannot be the underlying set of an increasing rational sequence. As every generatingset of N contains A , we conclude that N is not an increasing Puiseux monoid, whichcompletes the proof. 33s a direct consequence of Theorem 3.4.9, one obtains the following corollary. Corollary 3.4.10.
Being atomic, increasing, and strongly increasing are hereditaryproperties on the class of strongly increasing Puiseux monoids.
The next result is important as it provides a large class of Puiseux monoids that areFFMs.
Theorem 3.4.11.
Every increasing Puiseux monoid is an FFM.Proof.
See [46, Theorem 5.6].On the other hand, the converse of Theorem 3.4.11 does not hold; the followingexample sheds some light upon this observation.
Example 3.4.12.
Let ( p n ) n ∈ N be a strictly increasing sequence of primes, and considerthe Puiseux monoid of Q defined as follows: M = (cid:104) A (cid:105) , where A = (cid:26) p n + 1 p n , p n +1 + 1 p n +1 : n ∈ N (cid:27) . (3-8)Since A is an unbounded subset of R having 1 as a limit point, it cannot be increasing. Inaddition, as d ( a ) (cid:54) = d ( a (cid:48) ) for all a, a (cid:48) ∈ A such that a (cid:54) = a (cid:48) , every element of A is an atomof M . As each generating set of M must contain A (which is not increasing), M is not anincreasing Puiseux monoid.To verify that M is an FFM, fix x ∈ M and then take D x to be the set of primenumbers dividing d ( x ). Now choose N ∈ N such that N > max { x, d ( x ) } . For each a ∈ A with d ( a ) > N , the number of copies α of the atom a appearing in any z ∈ Z ( x ) mustbe a multiple of d ( a ) because d ( a ) / ∈ D x . Then α = 0; otherwise, we would have that x ≥ αa ≥ d ( a ) a > d ( a ) > x . Thus, if an atom a divides x in M , then d ( a ) ≤ N . As aresult, only finitely many elements of A ( M ) divide x in M and so | Z ( x ) | < ∞ . Hence M isan FFM that is not increasing. Remark 3.4.13.
For an ordered field F , a positive monoid of F is an additive submonoidof the nonnegative cone of F . As for Puiseux monoids, a positive monoid is increasing ifit can be generated by an increasing sequence. Increasing positive monoids are FFMs [46,34heorem 5.6], but the proof of this general version of Theorem 3.4.11 is much moreinvolved. Now that we have explored the structure of increasing Puiseux monoids, we will focuson the study of their decreasing counterpart.A Puiseux monoid M is said to be bounded if M can be generated by a boundedsubset of rational numbers. Besides, we say that M is strongly bounded if M can begenerated by a set of rationals R such that n ( R ) is bounded. If a Puiseux monoid isdecreasing, then it is obviously bounded. On the other hand, there are bounded Puiseuxmonoids that are not even monotone; see Example 3.4.2. However, every strongly boundedPuiseux monoid is decreasing, as we will show in Proposition 3.4.18.By contrast to the results we obtained in the previous section, the next propositionwill show that being decreasing is almost never hereditary. In fact, we prove that beingdecreasing is hereditary only on those Puiseux monoids that are isomorphic to numericalmonoids. Lemma 3.4.14. If M is a nontrivial decreasing Puiseux monoid, then exactly one of thefollowing conditions holds:1. M is isomorphic to a numerical monoid;2. M contains infinitely many limit points in R .Proof. Suppose that M is not isomorphic to a numerical monoid. Since M is not trivial,it fails to be finitely generated. Therefore it can be generated by a strictly decreasingsequence ( a n ) n ∈ N . The sequence ( a n ) n ∈ N must converge to a non-negative real number (cid:96) .Since ( ka n ) n ∈ N ⊆ M converges to k(cid:96) for every k ∈ N , if (cid:96) (cid:54) = 0, then every element of theinfinite set { k(cid:96) : k ∈ N } is a limit point of M . On the other hand, if (cid:96) = 0, then everyterm of the sequence ( a n ) n ∈ N is a limit point of M ; this is because for every fixed k ∈ N the sequence ( a k + a n ) n ∈ N ⊆ M converges to a k . Hence M has infinitely many limit pointsin R . 35ow let us verify that at most one of the above two conditions can hold. For this,assume that M is isomorphic to a numerical monoid. So M is finitely generated, namely, M = (cid:104) r , . . . , r n (cid:105) , where n ∈ N and r i ∈ Q > for i = 1 , . . . , n . For every r ∈ R the interval[0 , r ] contains only finitely many elements of M . Since M ∩ [0 , r ] is finite for all r ∈ R , itfollows that M cannot have any limit points in the real line. Proposition 3.4.15.
Let M be a nontrivial decreasing Puiseux monoid. Then exactly oneof the following conditions holds:1. M is isomorphic to a numerical monoid;2. M contains a submonoid that is not decreasing.Proof. Suppose that M is not isomorphic to a numerical monoid. Let us construct asubmonoid of M that fails to be decreasing. Lemma 3.4.14 implies that M has a nonzerolimit point (cid:96) . Since M is closed under addition, 2 (cid:96) and 3 (cid:96) are both limit points of M .An argument as the one given in the proof of Theorem 3.4.9 will guarantee the existenceof sequences ( a n ) n ∈ N and ( b n ) n ∈ N in M having infinite underlying sets such that ( a n ) n ∈ N converges to 2 (cid:96) , ( b n ) n ∈ N converges to 3 (cid:96) , and the submonoid N = (cid:104) a n , b n : n ∈ N (cid:105) of M is atomic with A ( M ) = { a n , b n : n ∈ N } . Since every decreasing sequence of Q containsat most one limit point, A ( M ) cannot be the underlying set of a decreasing sequence ofrationals. As every generating set of N must contain A ( M ), we can conclude that N isnot decreasing. Hence at least one of the given conditions must hold.To see that both conditions cannot hold simultaneously, it suffices to observe thatif M is isomorphic to a numerical monoid, then every nontrivial submonoid of M is alsoisomorphic to a numerical monoid and, therefore, decreasing.Similarly, as we did in the case of increasing Puiseux monoids, we will split the familyof decreasing Puiseux monoids into two fundamental subfamilies, depending on whether0 is or is not a limit point. We say that a non-negative sequence of rationals is strongly ecreasing if it is decreasing and it converges to zero. A non-negative decreasing sequenceof rationals converging to a positive real is called weakly decreasing . Definition 3.4.16.
A Puiseux monoid is strongly decreasing if it can be generated by astrongly decreasing sequence of rational numbers. On the other hand, a Puiseux monoidis said to be weakly decreasing if it can be generated by a weakly decreasing sequence ofrationals.Observe that if a Puiseux monoid M is weakly decreasing, then it has a generatingsequence decreasing to a positive real number and, therefore, 0 is not in the closure of M .As a consequence, every weakly decreasing Puiseux monoid must be atomic. The nextproposition describes those Puiseux monoids that are both strongly and weakly decreasing. Proposition 3.4.17.
A decreasing Puiseux monoid is either strongly or weakly decreasing.A Puiseux monoid is both strongly and weakly decreasing if and only if it is isomorphic toa numerical monoid.Proof.
As in the case of increasing Puiseux monoids, the first statement follows immediately.Now suppose that M is a Puiseux monoid that is both strongly and weakly decreasing.Since M is weakly decreasing, 0 is not a limit point of M • . Let ( a n ) n ∈ N be a sequencedecreasing to zero such that M = (cid:104) a n : n ∈ N (cid:105) . Because 0 is not a limit point of M • ,there exists n ∈ N such that a n = 0 for all n ≥ n . Hence M is isomorphic to a numericalmonoid. As in the increasing case, it is easily seen that every numerical monoid is bothstrongly and weakly decreasing.We mentioned at the beginning of this section that every strongly bounded Puiseuxmonoid is decreasing. Indeed, a stronger statement holds. Proposition 3.4.18.
Every strongly bounded Puiseux monoid is strongly decreasing.Proof.
Let M be a strongly bounded Puiseux monoid. Since the trivial monoid is bothstrongly bounded and strongly decreasing, for this proof we will assume that M (cid:54) = { } .Let S ⊂ Q > be a generating set of M such that n ( S ) is bounded. Since n ( S ) is finite, we37an take m to be the least common multiple of the elements of n ( S ). The map x (cid:55)→ m x is an order-preserving isomorphism from M to M (cid:48) = m M . Consequently, M (cid:48) is stronglydecreasing if and only if M is strongly decreasing. In addition, S (cid:48) = m S generates M (cid:48) .Since n ( S (cid:48) ) = { } , it follows that S (cid:48) is the underlying set of a strongly decreasing sequenceof rationals. Hence M (cid:48) is a strongly decreasing Puiseux monoid, which implies that M isstrongly decreasing as well.Recall that a Puiseux monoid M is finite if v p ( d ( M • )) = { } for all but finitely manyprimes p . Strongly decreasing Puiseux monoids are not always strongly bounded, even ifwe require them to be finite. For example, if r ∈ Q such that 0 < r < n ( r ) and d ( r ) are different from 1, then we have seen in Proposition 3.1.4 that the Puiseux monoid M r = (cid:104) r n : n ∈ N (cid:105) is atomic and A ( M r ) = { r n : n ∈ N } . As a result, M r is finite andstrongly decreasing. However, M r fails to be strongly bounded. On the other hand, notevery bounded Puiseux monoid is decreasing, as illustrated in Example 3.4.2.Because numerical monoids are finitely generated, they are both increasing anddecreasing Puiseux monoids. We end this section showing that numerical monoids are theonly such Puiseux monoids. Proposition 3.4.19.
A nontrivial Puiseux monoid M is isomorphic to a numericalmonoid if and only if M is both increasing and decreasing.Proof. If M is isomorphic to a numerical monoid, then it is finitely generated and,consequently, increasing and decreasing.Conversely, suppose that M is a nontrivial Puiseux monoid that is increasing anddecreasing. Proposition 3.4.3 implies that M is atomic and, moreover, A ( M ) is theunderlying set of an increasing sequence (because A ( M ) (cid:54) = ∅ ). Suppose, by way ofcontradiction, that A ( M ) is not finite. In this case, A ( M ) does not contain a largestelement. Since M is decreasing, there exists D = { d n : n ∈ N } ⊂ Q > such that d > d > · · · and M = (cid:104) D (cid:105) . Let m = min { n ∈ N : d n ∈ A ( M ) } , which must existbecause A ( M ) ⊆ D . Since A ( M ) is contained in D , the minimality of m implies that d m
38s the largest element of A ( M ), which is a contradiction. Hence A ( M ) is finite. Since M isatomic and A ( M ) is finite, M is isomorphic to a numerical monoid. The only Puiseux monoid that is a UFM (or even an HFM) is, up to isomorphism,( N , +). The following proposition formalizes this observation. Proposition 3.5.1.
For a nontrivial atomic Puiseux monoid M , the following statementsare equivalent.1. M is a UFM.2. M is a HFM.3. M ∼ = ( N , +) .4. M contains a prime element.Proof. Clearly, (3) ⇒ (1) ⇒ (2). To argue (2) ⇒ (3), assume that M is an HFM. Since M is an atomic nontrivial Puiseux monoid, A ( M ) is not empty. Let a and a be twoatoms of M . Then z := n ( a ) d ( a ) a and z := n ( a ) d ( a ) a are two factorizationsof the element n ( a ) n ( a ) ∈ M . As M is an HFM, it follows that | z | = | z | and so n ( a ) d ( a ) = n ( a ) d ( a ). Then a = a , which implies that |A ( M ) | = 1. As a consequence, M ∼ = ( N , +). Because (3) ⇒ (4) holds trivially, we only need to argue (4) ⇒ (3). Fix aprime element p ∈ M and take a ∈ A ( M ). Since p | M n ( p ) d ( a ) a , one finds that p | M a .This, in turn, implies that a = p . Hence A ( M ) = { p } , and so M ∼ = ( N , +). Example 3.5.2.
For general monoids, the property of being an HFM is strictly weakerthan that of being a UFM. For instance, the submonoid (cid:104) (1 , n ) | n ∈ N (cid:105) of N is an HFMthat is not a UFM (see [48, Propositions 5.1 and 5.4] for more details).A dual notion of being an HFM was introduced in [27] by Coykendall and Smith. Definition 3.5.3.
An atomic monoid M is an OHFM (or an other-half-factorial monoid )if for all x ∈ M \ U ( M ) and z, z (cid:48) ∈ Z ( x ) with | z | = | z (cid:48) | , we have z = z (cid:48) .39learly, every UFM is an OHFM. Although the multiplicative monoid of an integraldomain is a UFM if and only if it is an OHFM [27, Corollary 2.11], OHFMs are not alwaysUFMs or HFMs, even in the class of Puiseux monoids. Proposition 3.5.4.
For a nontrivial atomic Puiseux monoid M , the following conditionsare equivalent.1. M is an OHFM.2. |A ( M ) | ≤ .3. M is isomorphic to a numerical monoid with embedding dimension in { , } .Proof. To prove (1) ⇒ (2), let M be an OHFM. If M is factorial, then M ∼ = ( N , +), andwe are done. Then suppose that M is not factorial. In this case, |A ( M ) | ≥
2. Assume fora contradiction that |A ( M ) | ≥
3. Take a , a , a ∈ A ( M ) satisfying that a < a < a .Let d = d ( a ) d ( a ) d ( a ), and set a (cid:48) i = da i for each i ∈ { , . . . , } . Since a (cid:48) , a (cid:48) , and a (cid:48) areintegers satisfying that a (cid:48) < a (cid:48) < a (cid:48) , there exist m, n ∈ N such that m ( a (cid:48) − a (cid:48) ) = n ( a (cid:48) − a (cid:48) ) . (3-9)Clearly, z := ma + na and z := ( m + n ) a are two distinct factorizations in Z ( M ) satisfying that | z | = m + n = | z | . In addition, after dividing both sides of theequality (3-9) by d , one obtains ma + na = ( m + n ) a , which means that z and z arefactorizations of the same element. However, this contradicts that M is an OHFM. Hence |A ( M ) | ≤
2, as desired.To show that (2) ⇒ (3), suppose that |A ( M ) | ≤
2. By Proposition 2.2.6, M isisomorphic to a numerical monoid N . As |A ( M ) | ≤
2, the embedding dimension of N belongs to { , } , as desired.To show (3) ⇒ (1), suppose that either M ∼ = ( N , +) or M ∼ = (cid:104) a, b (cid:105) for a, b ∈ N ≥ withgcd( a, b ) = 1. If M ∼ = ( N , +), then M is factorial and, in particular, an OHFM. On theother hand, if M ∼ = (cid:104) a, b (cid:105) , then it is an OHFM by [27, Example 2.13].40 emark 3.5.5. There are Puiseux monoids that are FFMs but neither HFMs norOHFMs. As a direct consequence of Theorem 3.4.11 and Propositions 3.5.1 and 3.5.4, onefinds that (cid:104) p − p : p ∈ P (cid:105) is one of such monoids.We started this chapter exhibiting the following chain of implications, which holds forall monoids: UFM ⇒ HFM reduced ⇒ FFM ⇒ BFM ⇒ ACCP ⇒ atomic monoid This chain was first introduced by Anderson et al. in the context of integral domains [2].As we mentioned at the beginning of the chapter, none of the implications above isreversible in the context of integral domains. In the context of Puiseux monoids, we haveestablished the following chain of implications:(
UFM ⇔ HFM ) ⇒ OHFM ⇒ FGM ⇒ FFM ⇒ BFM ⇒ ACCP ⇒ AM , where FGM and AM stand for finitely generated monoid and atomic monoid, respectively.We have also provided examples to illustrate that, except UFM ⇔ HFM, none of theimplications in the previous chain is reversible in the context of Puiseux monoids.41HAPTER 4BOUNDED AND FINITE PUISEUX MONOIDS
We begin this chapter exploring two properties of Puiseux monoids: being boundedand being strongly bounded. Although these two properties are neither atomic noralgebraic, they often help to understand the structure of Puiseux monoids. For instance,they can be used to give another characterization of finitely generated Puiseux monoids(see Theorem 4.2.1 below).Recall that a Puiseux monoid M is said to be bounded if M can be generated by abounded subset of rational numbers and strongly bounded if M can be generated by aset of rationals R such that n ( R ) is bounded. There are Puiseux monoids that are notbounded, as we shall see below. Let us proceed to study the boundedness of prime reciprocal Puiseux monoids.Obviously, every finitely generated Puiseux monoid is strongly bounded. However, thereare strongly bounded Puiseux monoids that fail to be finitely generated. Clearly, thefamily of strongly bounded Puiseux monoids is strictly contained in that of boundedPuiseux monoids. The following example illustrates these observations.
Example 4.1.1.
Consider the Puiseux monoid M = (cid:104) A (cid:105) , where A = (cid:26) p − p : p ∈ P (cid:27) . An elementary divisibility argument will reveal that A ( M ) = A . Because A is anunbounded set, it follows that M is not a bounded Puiseux monoid. On the other hand,let us consider the strongly bounded Puiseux monoid M = (cid:104) A (cid:105) , where A = (cid:26) p : p ∈ P (cid:27) .
42s in the previous case, it is not hard to verify that A ( M ) = A . Therefore M is astrongly bounded Puiseux monoid that is not finitely generated. Finally, consider thebounded Puiseux monoid M = (cid:104) A (cid:105) , where A = (cid:26) p − p : p ∈ P (cid:27) . Once again, A ( M ) = A follows from an elementary divisibility argument. As a result, M cannot be strongly bounded.Let M be a Puiseux monoid, and let N be a submonoid of M . If M is finitelygenerated, then N is also finitely generated. Thus, being finitely generated is hereditary onthe class of finitely generated Puiseux monoids. As we should expect, not every propertyof a Puiseux monoid is inherited by all its submonoids. For example, being antimatter isnot hereditary on the class of antimatter Puiseux monoids; for instance, Q ≥ is antimatter,but it contains the atomic additive submonoid N , which satisfies A ( N ) = { } . Moreover,as Corollary 4.1.3 indicates, boundedness and strong boundedness are not hereditary, evenon the class of prime reciprocal Puiseux monoids.Let S be a set of naturals. If the series (cid:80) s ∈ S /s diverges, S is said to be substantial .If S is not substantial, it is said to be insubstantial (see [24, Chapter 10]). For example, itis well known that the set of prime numbers is substantial as it was first noticed by Eulerthat the series of reciprocal primes is divergent. Proposition 4.1.2.
Let P be a set of primes, and let M be the prime reciprocal Puiseuxmonoid (cid:104) /p : p ∈ P (cid:105) . If every submonoid of M is bounded, then P is insubstantial.Proof. Suppose, by way of contradiction, that P is substantial. Then P must containinfinitely many primes. Let ( p n ) n ∈ N be a strictly increasing enumeration of the elements in P . Take N to be the submonoid of M generated by A = { a n : n ∈ N } , where a n = n (cid:88) i =1 p i . P is substantial, A is unbounded. We will show that N fails to be bounded. For thispurpose, we verify that A ( N ) = A , which implies that every generating set of N contains A and, therefore, must be unbounded. Suppose that a n = a n + · · · + a n (cid:96) (4-1)for some (cid:96), n, n , . . . , n (cid:96) ∈ N such that n ≤ · · · ≤ n (cid:96) . Since ( a n ) n ∈ N is an increasingsequence, n ≥ n (cid:96) . After multiplying the equation (4-1) by m = p . . . p n and moving everysummand but m/p n to the right-hand side, we obtain p . . . p n − = (cid:96) (cid:88) j =1 n j (cid:88) i =1 mp i − n − (cid:88) i =1 mp i . (4-2)Now we observe that if n were strictly greater than n (cid:96) , then m/p i would be an integerdivisible by p n for each i = 1 , . . . , n j and j = 1 , . . . , (cid:96) , which would imply that theright-hand side of (4-2) is divisible by p n . This cannot be possible because p . . . p n − isnot divisible by p n . Thus, n = n (cid:96) and so (cid:96) = 1. Since ( a n ) n ∈ N is an increasing sequencesatisfying that a n / ∈ (cid:104) a , . . . , a n − (cid:105) , Proposition 3.4.3 ensures that A ( N ) = A . As a result, M contains a submonoid that fails to be bounded; but this is a contradiction. Hence theset P is insubstantial.For m, n ∈ N such that n > m, n ) = 1, Dirichlet’s theorem states that theset P of all primes p satisfying that p ≡ m (mod n ) is infinite. For a relatively elementaryproof of Dirichlet’s theorem, see [63]. Furthermore, it is also known that the set P issubstantial; indeed, as indicated in [6, page 156], there exists a constant A for which (cid:88) p ∈ P,p ≤ x p = 1 ϕ ( n ) log log x + A + O (cid:18) x (cid:19) , (4-3)where ϕ is the Euler totient function. In particular, the set comprising all primes of theform 4 k + 1 (or 4 k + 3) is substantial. The next corollary follows immediately fromProposition 4.1.2 and equation (4-3). 44 orollary 4.1.3. Let m, n ∈ N such that n > and gcd( m, n ) = 1 , and let P be theset of all primes p satisfying p ≡ m (mod n ) . Then the prime reciprocal Puiseux monoid M = (cid:104) /p : p ∈ P (cid:105) contains an unbounded submonoid. Let us turn to study the boundedness of the multiplicatively cyclic Puiseux monoids.
Proposition 4.1.4.
For r ∈ Q > , let M r be the multiplicatively r -cyclic Puiseux monoid.Then the following statements hold.1. If n ( r ) = 1 or d ( r ) = 1 , then M r is strongly bounded.2. If n ( r ) , d ( r ) > and r < , then M r is bounded but not strongly bounded.3. If n ( r ) , d ( r ) > and r > , then M r is not bounded.Proof.
1. Set A r := { r n : n ∈ N } . If n ( r ) = 1, then n ( A r ) = { } and, therefore, M r isstrongly bounded. On the other hand, if d ( r ) = 1, then M r is finitely generated, and sostrongly bounded.2. Suppose that n ( r ) , d ( r ) > r <
1. It follows from Proposition 3.1.4 that M r isatomic with A ( M r ) = A r . Since r < A r is bounded, which means that M r is bounded. However M r cannot be strongly bounded because every generating set of M r must contain the set A n , whose numerator set n ( A r ) = { n ( r ) n : n ∈ N } is unbounded.3. Finally, suppose that n ( r ) , d ( r ) > r >
1. As in the previous part, it followsfrom Proposition 3.1.4 that M r is atomic with A ( M r ) = A r . As A r is unbounded and anygenerating set of M r must contain A r , we obtain that M r is not bounded.As illustrated by Corollary 4.1.3, being bounded (or strongly bounded) is nothereditary on the class of prime reciprocal Puiseux monoids. Additionally, boundedness(resp., strong boundedness) is not hereditary on the class of bounded (resp., stronglybounded) multiplicatively cyclic Puiseux monoids.45 xample 4.1.5. Let M be the multiplicatively (1/2)-cyclic Puiseux monoid, that is, M = (cid:104) / n : n ∈ N (cid:105) . It is strongly bounded, and yet its submonoid N = (cid:28) n (cid:88) i =1 i : n ∈ N (cid:29) = (cid:28) n − n : n ∈ N (cid:29) (4-4)is not strongly bounded; to see this, it is enough to verify that A ( N ) = S , where S isthe generating set defining N in (4-4). Note that the sum of any two elements of thegenerating set S is at least one, while every element of S is less than one. Therefore eachelement of S must be an atom of N , and so A ( N ) = S .We argue now that boundedness (resp., strong boundedness) is almost neverhereditary on the class of bounded (resp., strongly bounded) multiplicatively cyclicPuiseux monoids. Proposition 4.1.6.
For r ∈ Q > , let M r be the multiplicatively r -cyclic Puiseux monoid.Then every submonoid of M is bounded (or strongly bounded) if and only if M r is isomor-phic to a numerical monoid.Proof. Let a and b denote n ( r ) and d ( r ), respectively. To prove the direct implication,suppose, by way of contradiction, that M r is not isomorphic to a numerical monoid. Inthis case, b >
1. Consider the submonoid N = (cid:104) s , s , . . . (cid:105) of M , where s n = ( nb n + 1) a n b n for every natural n . Proving the forward implication amounts to verifying that N isnot bounded and, as a consequence, not strongly bounded. First, let us check that A ( N ) = { s n : n ∈ N } . Note that s n +1 = (( n + 1) b n +1 + 1) a n +1 b n +1 > ( nb n +1 + b ) a n b n +1 = s n for each n ∈ N , and so ( s n ) n ∈ N is an increasing sequence. Moreover, it is easy to see that s n > n for every n . Thus, ( s n ) n ∈ N is unbounded. Suppose that there exist k, α k ∈ N and α i ∈ N for every i = 1 , . . . , k − s n = α s + · · · + α k s k . Since ( s n ) n ∈ N is46ncreasing and α k >
0, we have k ≤ n . Let p be a prime divisor of b , and let m = v p ( b ).The fact that p (cid:45) ( nb n + 1) a n for every natural n implies v p ( s n ) = − mn . Therefore − mn = v p ( s n ) ≥ min ≤ i ≤ k { v p ( α i s i ) } ≥ min ≤ i ≤ k { v p ( s i ) } = − mk, which implies that k ≥ n . Thus, k = n and then α = · · · = α n − = 0 and α n = 1. So s n / ∈ (cid:104) s , . . . , s n − (cid:105) for every n ∈ N and, by Proposition 3.4.3, A ( N ) = { s n : n ∈ N } . Since A ( N ) is unbounded, N cannot be a bounded Puiseux monoid.On the other hand, if M r is isomorphic to a numerical monoid, then it is finitelygenerated and, hence, bounded and strongly bounded. This gives us the converseimplication. A Puiseux monoid M over P is a Puiseux monoid such that v p ( m ) ≥ m ∈ M and p / ∈ P . If P is finite, then we say that M is a finite Puiseux monoid over P . The Puiseux monoid M is said to be finite if there exists a finite set of primes P suchthat M is finite over P . One can use strongly boundedness and finiteness to characterizefinitely generated Puiseux monoids. The following result was established in [43]. Theorem 4.2.1. [43, Theorem 5.8] For a Puiseux monoid M , the following statementsare equivalent.1. M is atomic, strongly bounded, and finite.2. M is finitely generated. No two of the three conditions in statement (1) of Theorem 4.2.1 imply the third one.In addition, the three conditions are required for the equivalence to hold. The followingexample illustrates these observations.
Example 4.2.2.
For q ∈ P , consider the Puiseux monoid M = (cid:28) q n : n ∈ N (cid:29) .
47t follows immediately that M is a nontrivial Puiseux monoid that is finite, stronglybounded, and antimatter. In particular, M is not atomic. Since M is not atomic, itcannot be finitely generated.Now consider the Puiseux monoid M = (cid:28)(cid:18) q + 1 q (cid:19) n : n ∈ N (cid:29) . It is clear that M is finite. It follows from Proposition 3.1.4 that M is an atomic Puiseuxmonoid with set of atoms A ( M ) = (cid:8)(cid:0) q +1 q (cid:1) n : n ∈ N (cid:9) . As a result, M is not stronglybounded. Therefore M is an atomic and finite Puiseux monoid that is not stronglybounded. Since M is not strongly bounded, it is not finitely generated.Finally, we have already seen that the prime reciprocal Puiseux monoid M = (cid:28) p : p ∈ P (cid:29) is atomic with sets of atoms A ( M ) = { /p : p ∈ P } . Then M is a strongly boundedatomic Puiseux monoid, but it is not finite. In particular, it is not finitely generated,which concludes our set of examples. p -adic Puiseux Monoids Let M be a Puiseux monoid. We have seen in Chapter 3 that when 0 is not a limitpoint of M • , then M is a BFM and, in particular, an atomic monoid. When 0 is a limitpoint of M • , the situation is significantly more subtle, and there is not a general criterionto decide whether such Puiseux monoids are atomic (or BFMs). Some of the simplestrepresentatives of this class are the Puiseux monoids M p = (cid:104) /p n : n ∈ N (cid:105) for p ∈ P , whichhappen to be antimatter. However, M p contain plenty of submonoids with a very diverseatomic structure. In this section we delve into the atomicity of submonoids of M p . Definition 4.2.3.
Let p be a prime. We say that a Puiseux monoid M is p -adic if d ( x ) isa power of p for all x ∈ M • . 48e use the term p -adic monoid as a short for p -adic Puiseux monoid. Throughoutthis section, every time that we define a p -adic monoid by specifying a sequence ofgenerators ( r n ) n ∈ N , we shall implicitly assume that ( d ( r n )) n ∈ N increases to infinity; thisassumption comes without loss of generality because in order to generate a Puiseuxmonoid we only need to repeat each denominator finitely many times. On the other hand,lim d ( r n ) = ∞ does not affect the generality of the results we prove in this section for if( d ( r n )) n ∈ N is a bounded sequence, then the p -adic monoid generated by ( r n ) n ∈ N is finitelygenerated and, therefore, isomorphic to a numerical monoid.Strongly bounded p -adic monoids happen to have only finitely many atoms(cf. Theorem 4.2.1), as revealed by the next proposition. Proposition 4.2.4.
A strongly bounded p -adic monoid has only finitely many atoms.Proof. For p ∈ P , let M be a strongly bounded p -adic monoid. Let ( r n ) n ∈ N be a generatingsequence for M with underlying set R satisfying that n ( R ) = { n , . . . , n k } for some k, n , . . . , n k ∈ N . For each i ∈ { , . . . , k } , take R i = { r n : n ( r n ) = n i } and M i = (cid:104) R i (cid:105) . Thefact that R ⊆ M ∪ · · · ∪ M k , along with A ( M ) ∩ M i ⊆ A ( M i ), implies that A ( M ) ⊆ k (cid:91) i =1 A ( M i ) . Thus, showing that A ( M ) is finite amounts to verifying that |A ( M i ) | < ∞ for each i = 1 , . . . , k . Fix i ∈ { , . . . , k } . If M i is finitely generated, then |A ( M i ) | < ∞ . Letus assume, therefore, that M i is not finitely generated. This means that there existsa strictly increasing sequence ( α n ) n ∈ N such that M i = (cid:104) n i /p α n : n ∈ N (cid:105) . Because n i /p α n = p α n +1 − α n ( n i /p α n +1 ), the monoid M i satisfies that |A ( M i ) | = 0. Hence we concludethat A ( M ) is finite.We are now in a position to give a necessary condition for the atomicity of p -adicmonoids. 49 heorem 4.2.5. Let p ∈ P , and let M be an atomic p -adic monoid satisfying that A ( M ) = { r n : n ∈ N } . If lim r n = 0 , then lim n ( r n ) = ∞ .Proof. Set a n = n ( r n ) and p α n = d ( r n ) for every natural n . Suppose, by way ofcontradiction, that lim a n (cid:54) = ∞ . Then there exists m ∈ N such that a n = m forinfinitely many n ∈ N . For each positive divisor d of m we define the Puiseux monoid M d = (cid:104) S d (cid:105) , where S d = (cid:26) a k n p α kn : a k n = m or gcd( m, a k n ) = d (cid:27) . Observe that A ( M ) is included in the union of the M d . On the other hand, the fact that A ( M ) ∩ M d ⊆ A ( M d ) for every d dividing m implies that A ( M ) ⊆ (cid:91) d | m A ( M d ) . (4-5)Because A ( M ) contains infinitely many atoms, the inclusion (4-5) implies the existenceof a divisor d of m such that |A ( M d ) | = ∞ . Set N d = d M d . Since d divides n ( q ) forall q ∈ M d , it follows that N d is also a p -adic monoid. In addition, the fact that N d isisomorphic to M d implies that |A ( N d ) | = |A ( M d ) | = ∞ . After setting b n = a k n /d and β n = α k n for every natural n such that either a k n = m or gcd( m, a k n ) = d , we have N d = (cid:28) b n p β n : n ∈ N (cid:29) . As a n = m for infinitely many n ∈ N , the sequence ( β n ) n ∈ N is an infinite subsequence of( α n ) n ∈ N and, therefore, it increases to infinity. In addition, as lim a n /p α n = 0, it followsthat lim b n /p β n = 0.Now we argue that A ( N d ) is finite, which will yield the desired contradiction. Take m (cid:48) = m/d . Since there are infinitely many n ∈ N such that b n = m (cid:48) , it is guaranteed that m (cid:48) /p n ∈ N d for every n ∈ N . In addition, gcd( m (cid:48) , b n ) = 1 for each b n (cid:54) = m (cid:48) . If b n (cid:54) = m (cid:48) for only finitely many n , then N d is strongly bounded and Proposition 4.2.4 ensures that A ( N d ) is finite. Suppose otherwise that gcd( b n , m (cid:48) ) = 1 (i.e., b n (cid:54) = m (cid:48) ) for infinitely many n ∈ N . For a fixed i with b i (cid:54) = m (cid:48) take j ∈ N satisfying that gcd( b j , m (cid:48) ) = 1 and large50nough so that b i p β j − β i > b j m (cid:48) ; the existence of such an index j is guaranteed by the factthat lim b n /p β n = 0. As b i p β j − β i > b j m (cid:48) > f ( (cid:104) b j , m (cid:48) (cid:105) ), there exist positive integers x and y such that b i p β j − β i = xb j + ym (cid:48) , that is b i p β i = x b j p β j + y m (cid:48) p β j . As b j /p β j , m (cid:48) /p β j ∈ N • d , it follows that b i /p β i / ∈ A ( N d ). Because i was arbitrarily taken, N d is antimatter. In particular, A ( N d ) is finite, which leads to a contradiction.The conditions lim r n = 0 and lim n ( r n ) = ∞ are not enough to guarantee that thenon-finitely generated p -adic monoid M is atomic. The next example sheds some lightupon this observation. Example 4.2.6.
For an odd prime p , consider the p -adic monoid M = (cid:28) p n − p n +1 , p n + 1 p n +1 : n ∈ N (cid:29) . (4-6)Observe that the sequence of numerators ( p n − , p n + 1) n ∈ N increases to infinity whilethe sequence of generators of M converges to zero. Also, notice that for every n ∈ N ,2 p n = p n − p n +1 + p n + 1 p n +1 ∈ M. Now we can see that M is not atomic; indeed, M is antimatter, which immediately followsfrom the fact that p n ± p n +1 = p n ±
12 2 p n +1 . The next proposition yields a necessary and a sufficient condition for the atomicity of p -adic monoids having generating sets whose numerators are powers of the same prime. Proposition 4.2.7.
Let p and q be two different primes, and let M = (cid:104) r n : n ∈ N (cid:105) be a p -adic monoid such that n ( r n ) is a power of q for every n ∈ N . Then1. if M is atomic, then lim n ( r n ) = ∞ ;2. if lim n ( r n ) = ∞ and ( r n ) n ∈ N is decreasing, then M is atomic. roof. Define the sequences ( α n ) n ∈ N and ( β n ) n ∈ N such that p α n = d ( r n ) and q β n = n ( r n ).To check condition (1), suppose, by way of contradiction, that lim n ( r n ) (cid:54) = ∞ . Thereforethere is a natural j such that n ( r n ) = q j for infinitely many n ∈ N . This implies that q j /p n ∈ M for every n ∈ N . Thus, for every x ∈ M • such that n ( x ) = q m ≥ q j , one canwrite x = q m d ( x ) = pq m − j q j p d ( x ) / ∈ A ( M ) . As a result, every a ∈ A ( M ) satisfies that n ( a ) < q j . This immediately implies that A ( M )is finite. As M is atomic with |A ( M ) | < ∞ , it must be finitely generated, which is acontradiction.Let us verify condition (2). Consider the subsequence ( k n ) n ∈ N of naturals satisfyingthat n ( r k n ) < n ( r i ) for every i > k n . It follows immediately that the sequence ( n ( r k n )) n ∈ N is increasing. We claim that M = (cid:104) r k n : n ∈ N (cid:105) . Suppose that j / ∈ ( k n ) n ∈ N . Becauselim n ( r n ) = ∞ there are only finitely many indices i ∈ N such that n ( r i ) ≤ n ( r j ),and it is easy to see that the maximum of such indices, say m , belongs to ( k n ) n ∈ N . As r i = p α m − α i q β i − β m r m , it follows that r i ∈ (cid:104) r k n : n ∈ N (cid:105) . Hence M = (cid:104) r k n : n ∈ N (cid:105) .Therefore it suffices to show that r k n ∈ A ( M ) for every n ∈ N . If q β kn p α kn = t (cid:88) i =1 c i q β ki p α ki , (4-7)for some t, c , . . . , c t ∈ N , then t ≥ n , c = · · · = c n − = 0, and c n ∈ { , } . If c n = 0,then by applying the q -adic valuation map to both sides of (4-7) we immediately obtain acontradiction. Thus, c n = 1, which implies that r k n is an atom. Hence M is atomic.52HAPTER 5FACTORIZATION INVARIANTS Let M be an atomic monoid. Recall from Chapter 2 that for each x ∈ M , the set oflengths of x is defined by L ( x ) := {| z | : z ∈ Z ( x ) } . The system of sets of lengths of M isdefined by L ( M ) := { L ( x ) : x ∈ M } . In [35] the interested reader can find a friendly introduction to sets of lengths and the rolethey play in factorization theory. In general, sets of lengths and systems of sets of lengthsare arithmetic invariants of atomic monoids that have received significant attention inrecent years (see, for instance, [1, 29, 40]).
If a monoid M is a BFM, then it is not hard to verify that the set of lengths of eachelement of M belongs to the collection S = (cid:8) { } , { } , S : S ⊆ Z ≥ and | S | < ∞ (cid:9) , i.e., L ( M ) ⊆ S . We say that M has full system of sets of lengths provided that L ( M ) = S .The first class of BFMs with full system of sets of lengths was found by Kainrath [58]; heproved that Krull monoids having infinite class groups, with primes in each class, havefull systems of sets of lengths. On the other hand, Frisch [30] proved that the subdomainInt( Z ) of Q [ x ] consisting of all integer-valued polynomials also has full system of sets oflengths; this result has been generalized for Dedekind domains [31].In the context of numerical monoids, Geroldinger and Schmid have proved thefollowing realization theorem for sets of lengths.53 heorem 5.1.1. [41, Theorem 3.3] For every nonempty finite subset S of Z ≥ , thereexists a numerical monoid N and x ∈ N such that L ( x ) = S . Theorem 5.1.1 was a crucial tool to construct the first Puiseux monoid with fullsystem of sets of lengths.
Theorem 5.1.2. [47, Theorem 3.6] There is an atomic Puiseux monoid with full systemof sets of lengths.
In this subsection we show that the set of lengths of each element in an atomicmultiplicatively cyclic Puiseux monoid M r is an arithmetic sequence. First, we describethe minimum-length and maximum-length factorizations for elements of M r . We startwith the case where 0 < r < Lemma 5.1.3.
Take r ∈ (0 , ∩ Q such that M r is atomic, and for x ∈ M • r consider thefactorization z = (cid:80) Ni =0 α i r i ∈ Z ( x ) , where N ∈ N and α , . . . , α N ∈ N . The followingstatements hold.1. min L ( x ) = | z | if and only if α i < d ( r ) for i ∈ { , . . . , N } .2. There exists exactly one factorization in Z ( x ) of minimum length.3. sup L ( x ) = ∞ if and only if α i ≥ n ( r ) for some i ∈ { , . . . , N } .4. | Z ( x ) | = 1 if and only if | L ( x ) | = 1 , in which case, α i < n ( r ) for i ∈ { , . . . , N } .Proof. To verify the direct implication of (1), we only need to observe that if α i ≥ d ( r )for some i ∈ { , . . . , N } , then the identity α i r i = ( α i − d ( r )) r i + n ( r ) r i − would yielda factorization z (cid:48) in Z ( x ) with | z (cid:48) | < | z | . To prove the reverse implication, suppose that w := (cid:80) Ki =0 β i r i ∈ Z ( x ) has minimum length. By the implication already proved, β i < d ( r )for i ∈ { , . . . , N } . Insert zero coefficients if necessary and assume that K = N . Suppose,by way of contradiction, that there exists m ∈ { , . . . , N } such that β m (cid:54) = α m and assume This is a simplified version of the original theorem, where the number of factorizationscan be specified for each length. 54hat such index m is as large as possible. Since z, w ∈ Z ( x ) we can write( α m − β m ) r m = m − (cid:88) i =0 ( β i − α i ) r i . After multiplying the above equality by d ( r ) m , it is easy to see that d ( r ) | α m − β m , whichcontradicts the fact that 0 < | α m − β m | ≤ d ( r ). Hence β i = α i for i ∈ { , . . . , N } and,therefore, w = z . As a result, | z | = | w | = min L ( x ). In particular, there exists only onefactorization in Z ( x ) having minimum length, and (2) follows.For the direct implication of (3), take a factorization w = (cid:80) Ni =0 β i r i ∈ Z ( x ) whoselength is not the minimum of L ( x ); such a factorization exists because sup L ( x ) = ∞ . Bypart (1), there exists i ∈ { , . . . , N } such that β i ≥ d ( r ). Now we can use the identity β i r i = ( β i − d ( r )) r i + n ( r ) r i − to obtain w ∈ Z ( x ) with | w | < | w | . Notice that there isan atom (namely r i − ) appearing at least n ( r ) times in w . In a similar way we can obtainfactorizations w = w , w , . . . , w n in Z ( x ), where w n =: (cid:80) Ni =0 β (cid:48) i r i ∈ Z ( x ) satisfies β (cid:48) i < d ( r )for i ∈ { , . . . , N } . By (1) we have that w n is a factorization of minimum length and,therefore, z = w n by (2). Hence α i ≥ n ( r ) for some i ∈ { , . . . , N } , as desired. For thereverse implication, it suffices to note that given a factorization w = (cid:80) Ni =0 β i r i ∈ Z ( x )with β i ≥ n ( r ) we can use the identity β i r i = ( β i − n ( r )) r i + d ( r ) r i +1 to obtain anotherfactorization w (cid:48) = (cid:80) N +1 i =0 β (cid:48) i r i ∈ Z ( x ) (perhaps β (cid:48) N +1 = 0) with | w (cid:48) | > | w | and satisfying β i +1 > n ( r ).Finally, we argue the reverse implication of (4) as the direct implication is trivial. Todo this, assume that L ( x ) is a singleton. Then each factorization of x has minimum length.By (2) there exists exactly one factorization of minimum length in Z ( x ). Thus, Z ( x ) is alsoa singleton. The last statement of (4) is straightforward.We continue with the case of r > Lemma 5.1.4.
Take r ∈ Q > \ N such that M r is atomic, and for x ∈ M • r consider thefactorization z = (cid:80) Ni =0 α i r i ∈ Z ( x ) , where N ∈ N and α , . . . , α N ∈ N . The followingstatements hold. . min L ( x ) = | z | if and only if α i < n ( r ) for i ∈ { , . . . , N } .2. There exists exactly one factorization in Z ( x ) of minimum length.3. max L ( x ) = | z | if and only if α i < d ( r ) for i ∈ { , . . . , N } .4. There exists exactly one factorization in Z ( x ) of maximum length.5. | Z ( x ) | = 1 if and only if | L ( x ) | = 1 , in which case α < n ( r ) and α i < d ( r ) for i ∈ { , . . . , N } .Proof. To argue the direct implication of (1) it suffices to note that if α i ≥ n ( r ) forsome i ∈ { , . . . , N } , then we can use the identity α i r i = ( α i − n ( r )) r i + d ( r ) r i +1 toobtain a factorization z (cid:48) in Z ( x ) satisfying | z (cid:48) | < | z | . For the reverse implication, supposethat w = (cid:80) Ki =0 β i r i is a factorization in Z ( x ) of minimum length. There is no loss inassuming that K = N . Note that β i < n ( r ) for each i ∈ { , . . . , N } follows from thedirect implication. Now suppose for a contradiction that w (cid:54) = z , and let m be the smallestnonnegative integer satisfying that α m (cid:54) = β m . Then( α m − β m ) r m = N (cid:88) i = m +1 ( β i − α i ) r i . (5-1)After clearing the denominators in (5-1), it is easy to see that n ( r ) | α m − β m , whichimplies that α m = β m , a contradiction. Hence w = z and so | z | = | w | = min L ( x ). We havealso proved that there exists a unique factorization of x of minimum length, which is (2).For the direct implication of (3), it suffices to observe that if α i ≥ d ( r ) for some i ∈ { , . . . , N } , then we can use the identity α i r i = (cid:0) α i − d ( r ) (cid:1) r i + n ( r ) r i − to obtaina factorization z (cid:48) in Z ( x ) satisfying | z (cid:48) | > | z | . For the reverse implication of (3), take w = (cid:80) Ki =0 β i r i to be a factorization in Z ( x ) of maximum length ( M r is a BFM because 0is not a limit point of M • r ). Once again, there is no loss in assuming that K = N . Themaximality of | w | now implies that β i < d ( r ) for i ∈ { , . . . , N } . Suppose, by way ofcontradiction, that z (cid:54) = u . Then take m be the smallest index such that α m (cid:54) = β m . Clearly,56 ≥ α m − β m ) r m = m − (cid:88) i =0 ( β i − α i ) r i . After clearing denominators, it is easy to see that d ( r ) | α m − β m , which contradicts that0 < | α M − β M | < d ( r ). Hence α i = β i for each i ∈ { , . . . , N } , which implies that z = w .Thus, max L ( x ) = | z | . In particular, there exists only one factorization of x of maximumlength, which is condition (4).The direct implication of (5) is trivial. For the reverse implication of (5), supposethat L ( x ) is a singleton. Then any factorization in Z ( x ) is a factorization of minimumlength. Since we proved in the first paragraph that Z ( x ) contains only one factorizationof minimum length, we have that Z ( x ) is also a singleton. The last statement of (5) is animmediate consequence of (1) and (3).We are in a position now to describe the sets of lengths of any atomic multiplicativelycyclic Puiseux monoid. Theorem 5.1.5.
Take r ∈ Q > such that M r is atomic.1. If r < , then for each x ∈ M r with | Z ( x ) | > , L ( x ) = (cid:8) min L ( x ) + k (cid:0) d ( r ) − n ( r ) (cid:1) : k ∈ N (cid:9) .
2. If r ∈ N , then | Z ( x ) | = | L ( x ) | = 1 for all x ∈ M r .3. If r ∈ Q > \ N , then for each x ∈ M r with | Z ( x ) | > , L ( x ) = (cid:26) min L ( x ) + k (cid:0) n ( r ) − d ( r ) (cid:1) : 0 ≤ k ≤ max L ( x ) − min L ( x ) n ( r ) − d ( r ) (cid:27) . Thus, L ( x ) is an arithmetic progression with difference | n ( r ) − d ( r ) | for all x ∈ M r .Proof. To argue (1), take x ∈ M r such that | Z ( x ) | >
1. Let z := (cid:80) Ni =0 α i r i be afactorization in Z ( x ) with | z | > min L ( x ). Lemma 5.1.3 guarantees that α i ≥ d ( r ) forsome i ∈ { , . . . , N } . Then one can use the identity α i r i = ( α i − d ( r )) r i + n ( r ) r i − to find a factorization z ∈ Z ( x ) with | z | = | z | − ( d ( r ) − n ( r )). Carrying out this57rocess as many times as necessary, we can obtain a sequence z , . . . , z n ∈ Z ( x ), where z n =: (cid:80) Ki =0 α (cid:48) i r i satisfies that α (cid:48) i < d ( r ) for i ∈ { , . . . , K } and | z j | = | z | − j ( d ( r ) − n ( r ))for j ∈ { , . . . , n } . By Lemma 5.1.3(1), the factorization z n has minimum length and,therefore, | z | ∈ { min L ( x ) + k (cid:0) d ( r ) − n ( r ) (cid:1) : k ∈ N } . Then L ( x ) ⊆ (cid:8) min L ( x ) + k (cid:0) d ( r ) − n ( r ) (cid:1) : k ∈ N (cid:9) . For the reverse inclusion, we check inductively that min L ( x ) + k ( d ( r ) − n ( r )) ∈ L ( x )for every k ∈ N . Since | Z ( x ) | >
1, Lemma 5.1.3(2) guarantees that | L ( x ) | >
1. Thenthere exists a factorization of length strictly greater than min L ( x ), and we have alreadyseen that such a factorization can be connected to a minimum-length factorization of Z ( x ) by a chain of factorizations in Z ( x ) with consecutive lengths differing by d ( r ) − n ( r ).Therefore min L ( x ) + ( d ( r ) − n ( r )) ∈ L ( x ). Suppose now that z = (cid:80) Ni =0 β i r i is afactorization in Z ( x ) with length min L ( x ) + k ( d ( r ) − n ( r )) for some k ∈ N . Then byLemma 5.1.3(1), there exists i ∈ { , . . . , N } such that β i ≥ d ( r ) > n ( r ). Now using theidentity β i r i = ( β i − n ( r )) r i + d ( r ) r i +1 , one can produce a factorization z (cid:48) ∈ Z ( x ) such that | z (cid:48) | = min L ( x ) + ( k + 1)( d ( r ) − n ( r )). Hence the reverse inclusion follows by induction.Clearly, statement (2) is a direct consequence of the fact that r ∈ N implies that M r = ( N , +).To prove (3), take x ∈ S • r . Since M r is a BFM, there exists z ∈ Z ( x ) such that | z | = max L ( x ). Take N ∈ N and α , . . . , α N ∈ N such that z = (cid:80) Ni =0 α i r i . If α i ≥ n ( r )for some i ∈ { , . . . , N } , then we can use the identity α i r i = ( α i − n ( r )) r i + d ( r ) r i +1 to find a factorization z ∈ Z ( x ) such that | z | = | z | − ( n ( r ) − d ( r )). Carrying out thisprocess as many times as needed, we will end up with a sequence z , . . . , z n ∈ Z ( x ), where z n =: (cid:80) Ki =0 β i r i satisfies that β i < n ( r ) for i ∈ { , . . . , K } and | z j | = | z | − j ( n ( r ) − d ( r )) for j ∈ { , . . . , n } . Lemma 5.1.4(1) ensures that | z n | = min L ( x ). Then (cid:26) min L ( x ) + j ( n ( r ) − d ( r )) : 0 ≤ j ≤ max L ( x ) − min L ( x ) n ( r ) − d ( r ) (cid:27) ⊆ L ( x ) . (5-2)58n the other hand, we can connect any factorization w ∈ Z ( x ) to the minimum-lengthfactorization w (cid:48) ∈ Z ( x ) by a chain w = w , . . . , w t = w (cid:48) of factorizations in Z ( x ) so that | w i | − | w i +1 | = n ( r ) − d ( r ). As a result, both sets involved in the inclusion (5-2) are indeedequal.We conclude this section collecting some immediate consequences of Theorem 5.1.5. Corollary 5.1.6.
Take r ∈ Q > such that M r is atomic.1. M r is a BFM if and only if r ≥ .2. If r ∈ N , then M r = N and, as a result, ∆( x ) = ∅ .3. If r / ∈ N , then ∆( x ) = {| n ( r ) − d ( r ) |} for all x ∈ M r such that | Z ( x ) | > . Therefore ∆( M r ) = {| n ( r ) − d ( r ) |} . Similar to the system of sets of lengths, the elasticity is another arithmetical invariantused to measure up to what extent factorizations in monoids (or domains) fail to beunique. The elasticity was introduced by R. Valenza [64] as a tool to measure thephenomenon of non-unique factorizations in the context of algebraic number theory.The elasticity of numerical monoids has been successfully studied in [23]. In addition,the elasticity of atomic monoids naturally generalizing numerical monoids has receivedsubstantial attention in the literature in recent years (see, for instance, [50, 54, 55, 65]).
Definition 5.2.1.
The elasticity ρ ( M ) of an atomic monoid M is given by ρ ( M ) = sup { ρ ( x ) : x ∈ M } , where ρ ( x ) = sup L ( x )min L ( x ) . The following formula for the elasticity of an atomic Puiseux monoid in terms of theinfimum and supremum of its set of atoms was established in [54].
Theorem 5.2.2. [54, Theorem 3.2] Let M be an atomic Puiseux monoid. If is a limitpoint of M • , then ρ ( M ) = ∞ . Otherwise, ρ ( M ) = sup A ( M )inf A ( M ) . A ( M ) is finite, then M is a numerical monoid; in this case, Theorem 5.2.2coincides with the elasticity formula given in [23, Theorem 2.1].The elasticity of M is accepted if there exists x ∈ M with ρ ( x ) = ρ ( M ). Theorem 5.2.3. [54, Theorem 3.4] For any atomic Puiseux monoid M such that ρ ( M ) < ∞ , the elasticity of M is accepted if and only if A ( M ) has both a maximumand a minimum. For an atomic monoid M the set R ( M ) = { ρ ( x ) : x ∈ M } is called the set of elasticities of M , and M is called fully elastic if R ( M ) = Q ∩ [1 , ρ ( M )]when ∞ / ∈ R ( M ) and R ( M ) \ {∞} = Q ∩ [1 , ∞ ) when ∞ ∈ R ( M ). The sets of elasticitiesof prime reciprocal Puiseux monoids was described in [54, Section 4]. For a nontrivial reduced monoid M and k ∈ N , we let U k ( M ) denote the union ofsets of lengths containing k , that is, U k ( M ) is the set of (cid:96) ∈ N for which there exist atoms a , . . . , a k , b , . . . , b (cid:96) such that a . . . a k = b . . . b (cid:96) . The set U k ( M ) is known as the union ofsets of lengths of M containing k . In addition, we set λ k ( M ) := min U k ( M ) and ρ k ( M ) := sup U k ( M ) , and we call ρ k ( M ) the k-th local elasticity of M . Unions of sets of lengths have receiveda great deal of attention in recent literature; see, for example, [9, 10, 28]. By [37,Section 1.4], the elasticity of an atomic monoid can be expressed in terms of its localelasticities as follows ρ ( M ) = sup (cid:26) ρ k ( M ) k : k ∈ N (cid:27) = lim k →∞ ρ k ( M ) k . k ∈ N , we define L − ( k ) := { x ∈ M : k ∈ L ( x ) } . It is easy to verify that U k ( M ) = {| z | : z ∈ Z ( x ) for some x ∈ L − ( k ) } . For a numerical monoid N with minimal generating set A , it was proved in [23,Section 2] that the elasticity of N is given by max A/ min A . On the other hand, it is nothard to verify that U n ( N ) is bounded and, therefore, every local elasticity of N is finite.In the next two sections, we will generalize this fact in two different ways to Puiseuxmonoids.Now we will propose a sufficient condition under which most of the local elasticities ofan atomic Puiseux monoid have infinite cardinality. On the other hand, we will describea subclass of Puiseux monoids (containing isomorphic copies of each numerical monoid)whose local k -elasticities are finite.If P is a Puiseux monoid, then we say that a ∈ A ( P ) is stable provided that the set { a ∈ A ( P ) : n ( a ) = n ( a ) } is infinite. Proposition 5.2.4.
Let P be an atomic Puiseux monoid. If P contains a stable atom,then ρ k ( P ) is infinite for all sufficiently large k .Proof. Suppose that for some m ∈ N the set A := { a ∈ A ( P ) : n ( a ) = m } containsinfinitely many elements. Let ( a n ) n ∈ N be an enumeration of the elements of A . Becausethe elements of A have the same numerator, namely m , we can assume that the sequence( a n ) n ∈ N is decreasing. Setting d = d ( a ), we can easily see that da = m = d ( a j ) a j for each j ∈ N . Therefore d ( a j ) ∈ U d ( P ) for each j ∈ N . As d ( A ) is an infinite set so is U d ( P ).The fact that |U d ( P ) | = ∞ immediately implies that |U k ( P ) | = ∞ for all k ≥ d . Hence ρ k ( P ) = sup U k ( P ) = ∞ for every k ≥ d .Recall that a Puiseux monoid P is strongly bounded if it can be generated by aset of rationals A whose numerator set n ( A ) is bounded. As a direct consequence ofProposition 5.2.4 we obtain the following result.61 orollary 5.2.5. If P is a non-finitely generated strongly bounded atomic Puiseuxmonoid, then ρ k ( P ) is infinite for all k sufficiently large. In contrast to the previous proposition, the next result gives a condition under whichPuiseux monoids have finite k -elasticity for each k ∈ N . Proposition 5.2.6.
Let P be a Puiseux monoid that does not contain as a limit point.If P is bounded, then ρ k ( P ) < ∞ for every k ∈ N .Proof. Because 0 is not a limit point of P • , the Puiseux monoid P is atomic As P is abounded Puiseux monoid, A ( P ) is a bounded set of rational numbers. Take q, Q ∈ Q suchthat 0 < q < a < Q for all a ∈ A ( P ). Now fix k ∈ N , and suppose that (cid:96) ∈ U k ( P ). Thenthere exists x ∈ L − ( k ) such that (cid:96) ∈ L ( x ). Because x has a factorization of length k , itfollows that x < kQ . Taking a , . . . , a (cid:96) ∈ A ( P ) such that x = a + · · · + a (cid:96) , we find that q(cid:96) < a + · · · + a (cid:96) = x < kQ. Therefore (cid:96) < kQ/q . Because neither q nor Q depends on the choice of x , one obtains that U k ( P ) is bounded from above by kQ/q . Hence ρ k ( P ) = sup U k ( P ) is finite, and the prooffollows.With the following two examples, we shall verify that the conditions of containing astable atom and not having 0 as a limit point are not superfluous in Proposition 5.2.4 andProposition 5.2.6, respectively. Example 5.2.7.
Let ( p n ) n ∈ N be a strictly increasing enumeration of the prime numbers,and consider the following Puiseux monoid: P = (cid:104) A (cid:105) , where A = (cid:26) p n − p n : n ∈ N (cid:27) . As the denominators of elements in A are pairwise distinct primes, it immediately followsthat A ( P ) = A . Therefore P is atomic. Clearly, P does not contain stable atoms. Because A is bounded so is P (as a Puiseux monoid). On the other hand, 0 is not a limit point of62 . Thus, it follows by Proposition 5.2.6 that ρ k ( P ) is finite for every k ∈ N . Notice alsothat1. if q ∈ P has at least two factorizations with no atoms in common, then q ∈ N ;2. by Proposition 5.2.6, we have both a lower and an upper bound for any q ∈ L − ( k ).Using the previous two observations, we have created an R-script that generates thesets U k for k ∈ { , . . . , } . Each U k appears as the k -th column in Table 5-1.Table 5-1. U k for k ∈ { , . . . , } . U U U U U U U U U U U U U U U Example 5.2.8.
Let ( p n ) n ∈ N be an enumeration of the prime numbers, and consider thePuiseux monoid P = (cid:10) /p n : n ∈ N (cid:11) . It is not difficult to argue that P is atomic with A ( P ) = { /p n : n ∈ N } . As A ( P ) is a bounded subset of positive rationals, the Puiseuxmonoid P is bounded. Notice, however, that 0 is a limit point of P . By Proposition 5.2.4,it follows that the local elasticities ρ k ( P ) are infinite for all k sufficiently large.The condition of boundedness on Proposition 5.2.6 is also required, as shown by thefollowing proposition. Proposition 5.2.9.
There exist infinitely many non-isomorphic Puiseux monoids without as a limit point that have no finite local elasticities. roof. Let P = { S n : n ∈ N } be a family of disjoint infinite sets of odd prime numbers.For each set S n , we will construct an atomic Puiseux monoid M n . Then we will show that M i ∼ = M j implies i = j .Fix j ∈ N and take p ∈ S j . To construct the Puiseux monoid M j , let us inductivelycreate a sequence ( A n ) n ∈ N of finite subsets of positive rationals with A (cid:40) A (cid:40) · · · suchthat, for each k ∈ N , the following three conditions hold:1. d ( A k ) consists of odd prime numbers;2. d (max A k ) = max d ( A k );3. A k minimally generates the Puiseux monoid P k = (cid:104) A k (cid:105) .Take A = { /p } , with p an odd prime number, and assume we have already constructedthe sets A , . . . , A n for some n ∈ N satisfying our three conditions. To construct A n +1 , wetake a = max A n and let b = n ( a ) (cid:98) q/ (cid:99) q and b = n ( a ) (cid:0) q − (cid:98) q/ (cid:99) (cid:1) q , where q is an odd prime in S j satisfying q > max d ( A n ) and q (cid:45) n ( a ). Using the fact that q ≥ d ( a ) ≥
3, one obtains that b > b = (cid:98) q/ (cid:99) q n ( a ) > n ( a ) ≥ a. Now set A n +1 = A n ∪ { b , b } . Notice that b + b = n ( a ). Clearly, A n (cid:40) A n +1 , andcondition (1) is an immediate consequence of our inductive construction. In addition, d (max A n +1 ) = d ( b ) = q = max d ( A n +1 ) , which is condition (2). Therefore it suffices to verify that A n +1 minimally generates P n +1 = (cid:104) A n +1 (cid:105) . Because both b and b are greater than every element in A n , we onlyneed to check that b / ∈ P n and b / ∈ (cid:104) A n ∪ { b }(cid:105) . Let d be the product of all the elementsin d ( A n ). Assuming that b = a + · · · + a r for some a , . . . , a r ∈ A n , and multiplyingboth sides of the same equality by qd , we would obtain that q | n ( b ), which contradicts64hat q (cid:45) n ( a ). Hence b / ∈ P n . Similarly, one finds that b / ∈ P n . Suppose, again bycontradiction, that b ∈ (cid:104) A n ∪ { b }(cid:105) . Then there exist a (cid:48) , . . . , a (cid:48) s ∈ A n and m ∈ N such that b = mb + a (cid:48) + · · · + a (cid:48) s . Notice that 2 b = n ( a )( q − /q > b , which implies that m ≤ b / ∈ P n , it follows that m = 1. Then we can write n ( a ) q = b − b = s (cid:88) i =1 a (cid:48) i . (5-3)Once again, we can multiply the extreme parts of the equality (5-3) by q d ( { a (cid:48) , . . . , a (cid:48) s } ), toobtain that q | n ( a ), a contradiction. As a result, condition (3) follows.Now set M j := ∪ n ∈ N P n . As P (cid:40) P (cid:40) . . . , the set M j is, indeed, a Puiseux monoid.We can easily see that M j is generated by the set A := ∪ n ∈ N A n . Let us verify now that A ( M j ) = A . It is clear that A ( M j ) ⊆ A . To check the reverse inclusion, suppose that a ∈ A is the sum of atoms a , . . . , a r ∈ A ( M j ). Take t ∈ N such that a, a , . . . , a r ∈ A t .Because A t minimally generates P t it follows that r = 1 and a = a and, therefore, that a ∈ A ( M j ). Hence A ( M j ) = A , which implies that M j is an atomic monoid.To disregard 0 as a limit point of M j , it is enough to observe that min A ( M j ) = 1 /p .We need to show then that ρ k ( M j ) = ∞ for k ≥
2. Set a n = max A n . When constructingthe sequence ( A n ) n ∈ N , we observed that n ( a n ) = b n + b n , where { b n , b n } = A n +1 \ A n .Because n ( a n ) ∈ M j and b n + b n = n ( a n ) = d ( a n ) a n , one has that the factorizations z = b n + b n and z (cid:48) = d ( a n ) a n are both in Z ( n ( a n )).Since | z | = 2 and | z (cid:48) | = d ( a n ) it follows that d ( a n ) ∈ U ( M j ). By condition (2) above, d ( a n ) = d (max A n ) = max d ( A n ). This implies that the set { d ( a n ) : n ∈ N } containsinfinitely many elements. As { d ( a n ) : n ∈ N } ⊆ U ( M j ), we obtain that ρ ( M j ) = ∞ .Hence ρ k ( M j ) = ∞ for all k ≥ F := { M n : n ∈ N } of atomic Puiseuxmonoids with infinite k -elasticities. Let us show now that the monoids in F are pairwisenon-isomorphic. To do this we use the fact that the only homomorphisms between Puiseux65onoids are given by rational multiplication [53, Lemma 3.3]. Take i, j ∈ N such that M i ∼ = M j . Then there exists r ∈ Q such that M i = rM j . Let m ∈ M j such that d ( m ) = p and p (cid:45) n ( r ) for some prime p in S j . Since the element rm ∈ M i and p | d ( rm ), wemust have that the prime p belongs to S i . Because the sets in P are pairwise disjoint, weconclude that i = j . This completes the proof.Proposition 5.2.4 (respectively, Proposition 5.2.6) establishes sufficient conditionsunder which a Puiseux monoid has most of its local elasticities infinite (respectively,finite). In addition, we have verified that such conditions are not necessary. For the sakeof completeness, we now exhibit a Puiseux monoid that does not satisfy the conditions ofeither of the propositions above and has no finite k -elasticity for any k ≥ Example 5.2.10.
Consider the Puiseux monoid P = (cid:28)(cid:18) (cid:19) n : n ∈ N (cid:29) . It was proved in [51, Theorem 6.2] that P is atomic and A ( P ) = { (2 / n : n ∈ N } . Inaddition, it is clear that P is bounded, has 0 as a limit point, and does not contain anystable atoms. So neither Proposition 5.2.4 nor Proposition 5.2.6 applies to P . Now weargue that ρ k ( P ) = ∞ for each k ∈ N such that k ≥ k ≥ x = k ∈ P . Notice that, by definition, x ∈ L − ( k ). We canconveniently rewrite x as x = (cid:0) ( k −
2) + 2 (cid:1)
23 = ( k −
2) 23 + 3 · (cid:18) (cid:19) , which reveals that z = ( k − + 3( ) is a factorization of x with | z | = k + 1. Taking k (cid:48) = 3 to play the role of k and repeating this process as many times as needed, one canobtain factorizations of x of lengths as large as one desires. The fact that k was chosenarbitrarily implies now that ρ k ( P ) = ∞ for each k ≥ .2.2 Prime Reciprocal Puiseux Monoids We proceed to study the local elasticity of prime reciprocal Puiseux monoids.Recall from Section 3.2 that a Puiseux monoid is said to be prime reciprocal if it canbe generated by a subset of positive rational numbers whose denominators are pairwisedistinct primes. We have seen before that every prime reciprocal Puiseux monoid isatomic.In Proposition 5.2.6, we established a sufficient condition on Puiseux monoids toensure that all their local k -elasticities are finite. Here we restrict our study to the case ofprime reciprocal Puiseux monoids, providing two more sufficient conditions to guaranteethe finiteness of all the local k -elasticities. Theorem 5.2.11.
For a prime reciprocal Puiseux monoid P , the following two conditionshold.1. If is not a limit point of P , then ρ k ( P ) < ∞ for every k ∈ N .2. If P is bounded and has no stable atoms, then ρ k ( P ) < ∞ for every k ∈ N .Proof. Because every finitely generated Puiseux monoid is isomorphic to a numericalmonoid, and numerical monoids have finite k -elasticities, we can assume, without loss ofgenerality, that P is not finitely generated.To prove condition (1), suppose, by way of contradiction, that ρ k ( P ) = ∞ for some k ∈ N . Because 0 is not a limit point of P there exists q ∈ Q such that 0 < q < a for each a ∈ A ( P ). Let (cid:96) = min { n ∈ N : |U n ( P ) | = ∞} . Clearly, (cid:96) ≥
2. Let m = max U (cid:96) − ( P ). Now take N ∈ N sufficiently large such that, foreach a ∈ A ( P ), a > N implies that d ( a ) > (cid:96) . As U (cid:96) ( P ) contains infinitely many elements,there exists k ∈ U (cid:96) ( P ) such that k > max (cid:26) (cid:96)q N, m + 1 (cid:27) .
67n particular, k − U (cid:96) − ( P ). As k ∈ U (cid:96) ( P ), we can choosean element x ∈ P such that { k, (cid:96) } ⊆ L ( x ). Take A = { a , . . . , a k } (cid:40) A ( P ) and B = { b , . . . , b (cid:96) } (cid:40) A ( P ) with a + · · · + a k = x = b + · · · + b (cid:96) . (5-4)Observe that the sets A and B must be disjoint, for if a ∈ A ∩ B , canceling a in (5-4)would yield that { (cid:96) − , k − } ⊆ L ( x − a ), which contradicts that k − U (cid:96) − ( P ). Because k > ( (cid:96)/q ) N , it follows that x > kq > (cid:96)N. Therefore b := max { b , . . . , b (cid:96) } > N , which implies that p = d ( b ) > (cid:96) . Since a i (cid:54) = b for each i = 1 , . . . , k , it follows that p / ∈ d ( { a , . . . , a k } ). We can assume, without loss of generality,that there exists j ∈ { , . . . , (cid:96) } such that b i (cid:54) = b for every i ≤ j and b j +1 = · · · = b (cid:96) = b .This allows us to rewrite (5-4) as( (cid:96) − j ) b = k (cid:88) i =1 a i − j (cid:88) i =1 b i . (5-5)After multiplying 5-5 by p times the product d of all the denominators of the atoms { a , . . . , a k , b , . . . , b j } , we find that p divides d ( (cid:96) − j ) b . As gcd( p, d ) = 1 and (cid:96) − j < p , itfollows that p divides n ( b ), which is a contradiction. Hence we conclude that ρ k ( P ) < ∞ for every k ∈ N .Now we argue the second condition. Let ( a n ) n ∈ N be an enumeration of the elementsof A ( P ) such that ( d ( a n )) n ∈ N is an increasing sequence. Set p n = d ( a n ). Since P has nostable atoms, lim n ( a n ) = ∞ . Let B be an upper bound for A ( P ).Suppose, by way of contradiction, that ρ n ( P ) = ∞ for some n ∈ N . Let k be thesmallest natural number such that |U k ( P ) | = ∞ . Now take (cid:96) ∈ U k ( P ) large enough suchthat (cid:96) − > max U k − ( P ) and for each a ∈ A ( P ) satisfying a ≤ Bk/(cid:96) we have that n ( a ) > Bk . Take x ∈ L − ( k ) such that a + · · · + a k = x = b + · · · + b (cid:96) for some68 , . . . , a k , b , . . . , b (cid:96) ∈ A ( P ). Now set b = min { b , . . . , b (cid:96) } . Then b ≤ b + · · · + b (cid:96) (cid:96) = a + · · · + a k (cid:96) ≤ Bk(cid:96) .
Therefore n ( b ) > Bk . We claim that d ( b ) / ∈ d ( { a , . . . , a k } ). Suppose by contradiction thatthis is not the case. Then b = a i for some i ∈ { , . . . , k } . This implies that { k − , (cid:96) − } ⊆ L ( x − b ), contradicting that (cid:96) − > max U k − ( P ). Hence d ( b ) / ∈ d ( { a , . . . , a k } ). Nowassume, without loss of generality, that there exists j ∈ { , . . . , (cid:96) } such that b i (cid:54) = b for each i ≤ j and b j +1 = · · · = b (cid:96) = b . Write( (cid:96) − j ) b = k (cid:88) i =1 a i − j (cid:88) i =1 b i . (5-6)From (5-6) we obtain that p (cid:96) divides (cid:96) − j . As a consequence, Bk ≥ k (cid:88) i =1 a i ≥ (cid:96) − jp (cid:96) n ( b ) ≥ n ( b ) > Bk, which is a contradiction. Hence ρ k ( P ) < ∞ for every k ∈ N .The sufficient conditions in part (1) of Theorem 5.2.11 and the condition ofboundedness in part (2) of Theorem 5.2.11 are not necessary, as the following exampleillustrates. Example 5.2.12.
1. Consider the prime reciprocal Puiseux monoid P = (cid:28) np n : n ∈ N (cid:29) , where ( p n ) n ∈ N is the increasing sequence of all prime numbers. Since A ( P ) = { n/p n : n ∈ N } , it follows that P does not contain any stable atom. It is well known thatthe sequence ( n/p n ) n ∈ N converges to 0, which implies that P is bounded. Hencepart (2) of Theorem 5.2.11 ensures that ρ k ( P ) < ∞ for all k ∈ N . Thus, the reverseimplication of part (1) in Theorem 5.2.11 does not hold.69. Consider now the Puiseux monoid P = (cid:28) p n − p n : n ∈ N (cid:29) , where ( p n ) n ∈ N is any enumeration of the prime numbers. Since 0 is not a limit pointof P , we can apply part (1) of Theorem 5.2.11 to conclude that ρ k ( P ) < ∞ for all k ∈ N . Notice, however, that P is not bounded. Therefore, the boundedness inpart (2) of Theorem 5.2.11 is not a necessary condition. On this subsection, we focus on the elasticity of multiplicatively cyclic Puiseuxmonoids.
Proposition 5.2.13.
Take r ∈ Q > such that M r is atomic. Then the following state-ments are equivalent.1. r ∈ N .2. ρ ( M r ) = 1 .3. ρ ( M r ) < ∞ .Hence, if M r is atomic, then either ρ ( M r ) = 1 or ρ ( M r ) = ∞ .Proof. To prove that (1) implies (2), suppose that r ∈ N . In this case, M r ∼ = N . Since N is a factorial monoid, ρ ( M r ) = ρ ( N ) = 1. Clearly, (2) implies (3). Now assume (3) andthat r / ∈ N . If r <
1, then 0 is a limit point of M • r as lim n →∞ r n = 0. Therefore it followsby Theorem 5.2.2 that ρ ( M r ) = ∞ . If r >
1, then lim n →∞ r n = ∞ and, as a result,sup A ( M r ) = ∞ . Then Theorem 5.2.2 ensures that ρ ( M r ) = ∞ . Thus, (3) implies (1). Thefinal statement now easily follows.Recall that the elasticity of an atomic monoid M is said to be accepted if there exists x ∈ M such that ρ ( M ) = ρ ( x ). Proposition 5.2.14.
Take r ∈ Q > such that M r is atomic. Then the elasticity of M r isaccepted if and only if r ∈ N or r < . roof. For the direct implication, suppose that r ∈ Q > \ N . Proposition 5.2.13 ensuresthat ρ ( M r ) = ∞ . However, as 0 is not a limit point of M • r , it follows from Theorem 3.3.1that M r is a BFM, and, therefore, ρ ( x ) < ∞ for all x ∈ M r . As a result, M r cannot haveaccepted elasticityFor the reverse implication, assume first that r ∈ N and, therefore, that M r = N .In this case, M r is a factorial monoid and, as a result, ρ ( M r ) = ρ (1) = 1. Now supposethat r <
1. Then it follows by Proposition 5.2.13 that ρ ( M r ) = ∞ . In addition, for x = n ( r ) ∈ M r Lemma 5.1.3(1) and Theorem 5.1.5(1) guarantee that L ( x ) = (cid:8) n ( r ) + k (cid:0) d ( r ) − n ( r ) (cid:1) : k ∈ N (cid:9) . Because L ( x ) is an infinite set, we have ρ ( M r ) = ∞ = ρ ( x ). Hence M r has acceptedelasticity, which completes the proof.Let us proceed to describe the sets of elasticities of atomic multiplicatively cyclicPuiseux monoids. Proposition 5.2.15.
Take r ∈ Q > such that M r is atomic.1. If r < , then R ( M r ) = { , ∞} and, therefore, M r is not fully elastic.2. If r ∈ N , then R ( M r ) = { } and, therefore, M r is fully elastic.3. If r ∈ Q > \ N and n ( r ) = d ( r ) + 1 , then M r is fully elastic, in which case R ( M r ) = Q ≥ .Proof. First, suppose that r <
1. Take x ∈ M r such that | Z ( x ) | >
1. It follows byTheorem 5.1.5(1) that L ( x ) is an infinite set, which implies that ρ ( x ) = ∞ . As a result, ρ ( M r ) = { , ∞} and then M r is not fully elastic.To argue (2), it suffices to observe that r ∈ N implies that M r = ( N , +) is a factorialmonoid and, therefore, ρ ( M r ) = { } .Finally, let us argue that M r is fully elastic when n ( r ) = d ( r ) + 1. To do so, fix q ∈ Q > . Take m ∈ N such that m d ( q ) > d ( r ), and set k = m (cid:0) n ( q ) − d ( q ) (cid:1) . Let t = m d ( q ) − d ( r ), and consider the factorizations z = d ( r ) r k + (cid:80) ti =1 r k + i ∈ Z ( M r ) and71 (cid:48) = d ( r ) · (cid:80) k − i =0 r i + (cid:80) ti =1 r k + i ∈ Z ( M r ). Since n ( r ) = d ( r ) + 1, it can be easily checkedthat r − = d ( r ). As d ( r ) + k − (cid:88) i =0 r i + t (cid:88) i =1 r k + i = d ( r ) + r k − r − t (cid:88) i =1 r k + i = d ( r ) r k + t (cid:88) i =1 r k + i , there exists x ∈ M r such that z, z (cid:48) ∈ Z ( x ). By Lemma 5.1.4 it follows that z is afactorization of x of minimum length and z (cid:48) is a factorization of x of maximum length.Thus, ρ ( x ) = | z (cid:48) || z | = d ( r ) + k + t d ( r ) + t = m n ( q ) m d ( q ) = q. As q was arbitrarily taken in Q > , it follows that R ( M r ) = Q ≥ . Hence M r is fully elasticwhen n ( r ) = d ( r ) + 1.We were unable to determine in Proposition 5.2.15 whether M r is fully elastic when r ∈ Q > \ N with n ( r ) (cid:54) = d ( r ) + 1. However, we proved in Proposition 5.2.16 that the set ofelasticities of M r is dense in R ≥ . Proposition 5.2.16. If r ∈ Q > \ N , then the set R ( M r ) is dense in R ≥ .Proof. Since sup A ( M r ) = ∞ , it follows by Theorem 5.2.2 that ρ ( M r ) = ∞ . This, alongwith the fact that M r is a BFM (because of Theorem 3.3.1, ensures the existence of asequence ( x n ) n ∈ N of elements of M r such that lim n →∞ ρ ( x n ) = ∞ . Then it follows by [54,Lemma 5.6] that the set S := (cid:26) n ( ρ ( x n )) + k d ( ρ ( x n )) + k : n, k ∈ N (cid:27) is dense in R ≥ . Fix n, k ∈ N . Take m ∈ N such that r m is the largest atom dividing x n in M r . Now take K := k gcd(min L ( x n ) , max L ( x n )). Consider the element y n,k := x n + (cid:80) Ki =1 r m + i ∈ M r . It follows by Lemma 5.1.4 that x n has a unique minimum-lengthfactorization and a unique maximum-length factorization; let them be z and z ,respectively. Now consider the factorizations w := z + (cid:80) Ki =1 r m + i ∈ Z ( y n,k ) and w := z + (cid:80) Ki =1 r m + i ∈ Z ( y n,k ). Once again, we can appeal to Lemma 5.1.4 to ensure that w and w are the minimum-length and maximum-length factorizations of y n,k . Therefore72in L ( y n,k ) = min L ( x n ) + K and max L ( y n,k ) = max L ( x n ) + K . Then we have ρ ( y n,k ) = max L ( y n,k )min L ( y n,k ) = max L ( x n ) + K min L ( x n ) + K = n ( ρ ( x n )) + k d ( ρ ( x n )) + k . Since n and k were arbitrarily taken, it follows that S is contained in R ( M r ). As S isdense in R ≥ so is R ( M r ), which concludes our proof. Corollary 5.2.17.
The set of elasticities of M r is dense in R ≥ if and only if r ∈ Q > \ N . Remark 5.2.18.
Proposition 5.2.16 contrasts with the fact that the elasticity of anumerical monoid is always nowhere dense in R [23, Corollary 2.3].Let us conclude this section studying the unions of sets of lengths and the localelasticities of atomic multiplicatively cyclic Puiseux monoids. Proposition 5.2.19.
Take r ∈ Q > such that M r is atomic. Then U k ( M r ) is an arith-metic progression containing k with distance | n ( r ) − d ( r ) | for every k ∈ N . Morespecifically, the following statements hold.1. If r < , then • U k ( M r ) = { k } if k < n ( r ) , • U k ( M r ) = { k + j ( d ( r ) − n ( r )) : j ∈ N } if n ( r ) ≤ k < d ( r ) , and • U k ( M r ) = { k + j ( d ( r ) − n ( r )) : j ∈ Z ≥ (cid:96) } for some (cid:96) ∈ Z < if k ≥ d ( r ) .2. If r ∈ Q > \ N , then • U k ( M r ) = { k } if k < d ( r ) , • U k ( M r ) = { k + j ( n ( r ) − d ( r )) : j ∈ N } if d ( r ) ≤ k < n ( r ) , and • U k ( M r ) = { k + j ( n ( r ) − d ( r )) : j ∈ Z ≥ (cid:96) } for some (cid:96) ∈ Z < if k ≥ n ( r ) .3. If r ∈ N , then U k ( M r ) = { k } for every k ∈ N .Proof. That U k ( M r ) is an arithmetic progression containing k with distance | n ( r ) − d ( r ) | isan immediate consequence of Theorem 5.1.5.To show (1), assume that r <
1. Suppose first that k < n ( r ). Take L ∈ L ( M r )with k ∈ L , and take x ∈ M r such that L = L ( x ). Choose z = (cid:80) Ni =0 α i r i ∈ Z ( x ) with73 Ni =0 α i = k . Since α i ≤ k < n ( r ) for i ∈ { , . . . , N } , Lemma 5.1.3 ensures that | Z ( x ) | = 1,which yields L = L ( x ) = { k } . Thus, U k ( M r ) = { k } . Now suppose that n ( r ) ≤ k < d ( r ).Notice that the element k ∈ M r has a factorization of length k , namely, k · ∈ Z ( k ). Nowwe can use Lemma 5.1.3(3) to conclude that sup L ( k ) = ∞ . Hence ρ k ( M r ) = ∞ . On theother hand, let x be an element of M r having a factorization of length k . Since k < d ( r ), itfollows by Lemma 5.1.3(1) that any length- k factorization in Z ( x ) is a factorization of x ofminimum length. Hence λ k ( M r ) = k and, therefore, U k ( M r ) = { k + j ( d ( r ) − n ( r )) : j ∈ N } . Now assume that k ≥ d ( r ). As k ≥ n ( r ), we have once again that ρ k ( M r ) = ∞ . Also,because k ≥ d ( r ) one finds that ( k − d ( r )) r + n ( r ) · Z ( kr ) of length k − ( d ( r ) − n ( r )). Then there exists (cid:96) ∈ Z < such that U k ( M r ) = { k + j ( d ( r ) − n ( r )) : j ∈ Z ≥ (cid:96) } . Suppose now that r ∈ Q > \ N . Assume first that k < d ( r ). Take L ∈ L ( M r )containing k and x ∈ M r such that L = L ( x ). If z = (cid:80) Ni =0 α i r i ∈ Z ( x ) satisfies | z | = k ,then α i ≤ k < d ( r ) for i ∈ { , . . . , N } , and Lemma 5.1.4 implies that L = L ( x ) = { k } . As aresult, U k ( M r ) = { k } . Suppose now that d ( r ) ≤ k < n ( r ). In this case, for each n > k , wecan consider the element x n = kr n ∈ M r and set L n := L ( x n ). It is not hard to check that z n := n ( r ) · (cid:18) n − (cid:88) i =1 (cid:0) n ( r ) − d ( r ) (cid:1) r i (cid:19) + (cid:0) k − d ( r ) (cid:1) r n is a factorization of x n . Therefore | z n | = k + n ( n ( r ) − d ( r )) ∈ L n . Since k ∈ L n for every n ∈ N , it follows that ρ k ( M r ) = ∞ . On the other hand, it follows by Lemma 5.1.4(1) thatany factorization of length k of an element x ∈ M r must be a factorization of minimumlength in Z ( x ). Hence λ k ( M r ) = k , which implies that U k ( M r ) = { k + j ( n ( r ) − d ( r )) : j ∈ N } . k ≥ n ( r ). As k ≥ d ( r ) we still obtain ρ k ( M r ) = ∞ . In addition, because k ≥ n ( r ), we have that ( k − n ( r )) · d ( r ) r is a factorization in Z ( k ) having length k − ( n ( r ) − d ( r )). Thus, there exists (cid:96) ∈ Z < such that U k ( M r ) = { k + j ( n ( r ) − d ( r )) : j ∈ Z ≥ (cid:96) } . Finally, condition (3) follows directly from the fact that M r = ( N , +) when r ∈ N and, therefore, for every k ∈ N there exists exactly one element in M r having a length- k factorization, namely k . Corollary 5.2.20.
Take r ∈ Q > such that M r is atomic. Then ρ ( M r ) < ∞ if and only if ρ k ( M r ) < ∞ for every k ∈ N .Proof. It follows from [37, Proposition 1.4.2(1)] that ρ k ( M r ) ≤ kρ ( M r ), which yields thedirect implication. For the reverse implication, we first notice that, by Proposition 5.2.19,if r / ∈ N and k > max { n ( r ) , d ( r ) } , then ρ k ( M r ) = ∞ . Hence the fact that ρ k ( M r ) < ∞ forevery k ∈ N implies that r ∈ N . In this case ρ ( M r ) = ρ ( N ) = 1, and so ρ ( M r ) < ∞ .As [37, Proposition 1.4.2(1)] holds for every atomic monoid, the direct implication ofCorollary 5.2.20 also holds for any atomic monoid. However, the reverse implication of thesame corollary is not true even in the context of Puiseux monoids. Example 5.2.21.
Let ( p n ) n ∈ N be a strictly increasing sequence of primes, and considerthe Puiseux monoid M := (cid:28) p n + 1 p n : n ∈ N (cid:29) . It is not hard to verify that the monoid M is atomic with set of atoms given by thedisplayed generating set. Then it follows from [54, Theorem 3.2] that ρ ( M r ) = ∞ .However, [55, Theorem 4.1(1)] guarantees that ρ k ( M ) < ∞ for every k ∈ N . As the elasticity, the tameness is an arithmetic tool to measure how far is an atomicmonoid from being a UFM. Although the tameness of many classes of atomic monoids has75een studied in the past (see [16], [13], [38]), no systematic investigation of the tamenesshas been carried out for Puiseux monoids. For the special class of strongly primaryPuiseux monoids, recent results have been achieved in [36, Section 3]. In this section, westudy the tameness of the multiplicatively cyclic Puiseux monoids.
Let M be a reduced atomic monoid. The omega function ω : M → N ∪ {∞} isdefined as follows: for each x ∈ M • we take ω ( x ) to be the smallest n ∈ N satisfyingthat whenever x | M (cid:80) ti =1 a i for some a , . . . , a t ∈ A ( M ), there exists T ⊆ { , . . . , t } with | T | ≤ n such that x | M (cid:80) i ∈ T a i . If no such n exists, then ω ( x ) = ∞ . In addition, we define ω (0) = 0. Then we define ω ( M ) := sup { ω ( a ) : a ∈ A ( M ) } . Notice that ω ( x ) = 1 if and only if x is prime in M . The omega function was introducedby Geroldinger and Hassler in [38] to measure how far in an atomic monoid an element isfrom being prime.Before proving the main results of this section, let us collect two technical lemmas. Lemma 5.3.1. If r ∈ Q > , then | M r d ( r ) r k for every k ∈ N .Proof. If r ∈ N , then M r = ( N , +) and the statement of the lemma follows straightforwardly.Then we assume that r ∈ Q > \ N . For k = 0, the statement of the lemma holds trivially.For k ∈ N , consider the factorization z k := d ( r ) r k ∈ Z ( M r ). The factorization z := n ( r ) + k − (cid:88) i =1 ( n ( r ) − d ( r )) r i Z ( φ ( z k )) (recall that φ : Z ( M r ) → M r is the factorization homomorphismof M r ). This is because n ( r ) + k − (cid:88) i =1 ( n ( r ) − d ( r )) r i = n ( r ) + k − (cid:88) i =1 n ( r ) r i − k − (cid:88) i =1 d ( r ) r i = n ( r ) + k − (cid:88) i =1 n ( r ) r i − k − (cid:88) i =1 n ( r ) r i − = d ( r ) r k . Hence 1 | M r d ( r ) r k Lemma 5.3.2.
Take r ∈ Q ∩ (0 , such that M r is atomic, and let (cid:80) Ni =0 α i r i be thefactorization in Z ( x ) of minimum length. Then α ≥ if and only if | M r x .Proof. The direct implication is straightforward. For the reverse implication, suppose that1 | M r x . Then there exists a factorization z (cid:48) := (cid:80) Ki =0 β i r i ∈ Z ( x ) such that β ≥
1. If β i ≥ d ( r ) for some i ∈ { , . . . , K } , then we can use the identity d ( r ) r i = n ( r ) r i − to findanother factorization z (cid:48)(cid:48) ∈ Z ( x ) such that | z (cid:48)(cid:48) | < | z (cid:48) | . Notice that the atom 1 appears in z (cid:48)(cid:48) .Then we can replace z (cid:48) by z (cid:48)(cid:48) . After carrying out such a replacement as many times aspossible, we can guarantee that β i < d ( r ) for i ∈ { , . . . , K } . Then Lemma 5.1.3(1) ensuresthat z (cid:48) is a minimum-length factorization of x . Now Lemma 5.1.3(2) implies that z (cid:48) = z .Finally, α = β ≥ z (cid:48) . Proposition 5.3.3.
Take r ∈ Q > such that M r is atomic.1. If r < , then ω (1) = ∞ .2. If r ∈ N , then ω (1) = 1 .3. If r ∈ Q > \ N , then ω (1) = d ( r ) .Proof. To verify (1), suppose that r <
1. Then set x = n ( r ) ∈ M r and note that 1 | M r x .Fix an arbitrary N ∈ N . Take now n ∈ N such that d ( r ) + n ( d ( r ) − n ( r )) ≥ N . It is nothard to check that z := d ( r ) r n +1 + n (cid:88) i =1 ( d ( r ) − n ( r )) r i
77s a factorization in Z ( x ). Suppose that z (cid:48) = (cid:80) Ki =1 α i r i is a sub-factorization of z suchthat 1 | M r x (cid:48) := φ ( z (cid:48) ). Now we can move from z (cid:48) to a factorization z (cid:48)(cid:48) of x (cid:48) of minimumlength by using the identity d ( r ) r i +1 = n ( r ) r i finitely many times. As 1 | M r x (cid:48) , itfollows by Lemma 5.3.2 that the atom 1 appears in z (cid:48)(cid:48) . Therefore, when we obtained z (cid:48)(cid:48) from z (cid:48) (which does not contain 1 as a formal atom), we must have applied the identity d ( r ) r = n ( r ) · z (cid:48)(cid:48) contains at least n ( r ) copies of the atom 1.This implies that x (cid:48) = φ ( z (cid:48)(cid:48) ) ≥ n ( r ) = x . Thus, x (cid:48) = x , which implies that z (cid:48) is thewhole factorization z . As a result, ω (1) ≥ | z | ≥ N . Since N was arbitrarily taken, we canconclude that ω (1) = ∞ , as desired.Notice that (2) is a direct consequence of the fact that 1 is a prime element in M r = ( N , +).Finally, we prove (3). Take z = (cid:80) Ni =0 α i r i ∈ Z ( x ) for some x ∈ M r such that1 | M r x . We claim that there exists a sub-factorization z (cid:48) of z such that | z (cid:48) | ≤ d ( r ) and1 | M r φ ( z (cid:48) ), where φ is the factorization homomorphism of M r . If α >
0, then 1 is oneof the atoms showing in z and our claim follows trivially. Therefore assume that α = 0.Since 1 | M r x and 1 does not show in z , we have that | Z ( x ) | >
1. Then conditions (1)and (3) in Lemma 5.1.4 cannot be simultaneously true, which implies that α i ≥ d ( r ) forsome i ∈ { , . . . , N } . Lemma 5.3.1 ensures now that 1 | M r φ ( z (cid:48) ) for the sub-factorization z (cid:48) := d ( r ) r i of z . This proves our claim and implies that ω (1) ≤ d ( r ). On the other hand,take w to be a strict sub-factorization of d ( r ) r . Note that the atom 1 does not appear in w . In addition, it follows by Lemma 5.1.4 that | Z ( φ ( w )) | = 1. Hence 1 (cid:45) M r φ ( w ). As aresult, we have that ω (1) ≥ d ( r ), and (3) follows. For an atom a ∈ A ( M ), the local tame degree t ( a ) ∈ N is the smallest n ∈ N ∪ {∞} such that in any given factorization of x ∈ a + M at most n atoms have to be replaced byat most n new atoms to obtain a new factorization of x that contains a . More specifically,78t means that t ( a ) is the smallest n ∈ N ∪ {∞} with the following property: if Z ( x ) ∩ ( a + Z ( M )) (cid:54) = ∅ and z ∈ Z ( x ), then there exists a z (cid:48) ∈ Z ( x ) ∩ ( a + Z ( M )) such that d ( z, z (cid:48) ) ≤ n . Definition 5.3.4.
An atomic monoid M is said to be locally tame provided that t ( a ) < ∞ for all a ∈ A ( M ).Every factorial monoid is locally tame (see [37, Theorem 1.6.6 and Theorem 1.6.7]).In particular, ( N , +) is locally tame. The tame degree of numerical monoids was firstconsidered in [16]. The factorization invariant τ : M → N ∪ {∞} , which was introducedin [38], is defined as follows: for k ∈ N and b ∈ M , we take Z min ( k, b ) := (cid:26) j (cid:88) i =1 a i ∈ Z ( M ) : j ≤ k, b | M j (cid:88) i =1 a i , and b (cid:45) M (cid:88) i ∈ I a i for any I (cid:40) { , . . . , j } (cid:27) and then we set τ ( b ) = sup k sup z (cid:8) min L (cid:0) φ ( z ) − b (cid:1) : z ∈ Z min ( k, b ) (cid:9) . The monoid M is called (globally) tame provided that the tame degree t ( M ) = sup { t ( a ) : a ∈ A ( M ) } < ∞ . The following result will be used in the proof of Theorem 5.3.6.
Theorem 5.3.5. [38, Theorem 3.6] Let M be a reduced atomic monoid. Then M is locallytame if and only if ω ( a ) < ∞ and τ ( a ) < ∞ for all a ∈ A ( M ) . We conclude this section by characterizing the multiplicatively cyclic Puiseux monoidsthat are locally tame.
Theorem 5.3.6.
Take r ∈ Q > such that M r is atomic. Then the following conditions areequivalent:1. r ∈ N ;2. ω ( M r ) < ∞ ;3. M r is globally tame;4. M r is locally tame. roof. That (1) implies (2) follows from Proposition 5.3.3(2). Now suppose that (2) holds.Then [39, Proposition 3.5] ensures that t ( M r ) ≤ ω ( M r ) < ∞ , which implies (3). Inaddition, (3) implies (4) trivially.To prove that (4) implies (1) suppose, by way of contradiction, that r ∈ Q > \ N . Letus assume first that r <
1. In this case, ω (1) = ∞ by Proposition 5.3.3(3). Then it followsby Theorem 5.3.5 that M r is not locally tame, which is a contradiction. For the rest of theproof, we assume that r ∈ Q > \ N .We proceed to show that τ (1) = ∞ . For k ∈ N such that k ≥ d ( r ), consider thefactorization z k = d ( r ) r k ∈ Z ( M r ). Since any strict sub-factorization z (cid:48) k of z k is of the form βr k for some β < d ( r ), it follows by Lemma 5.1.4 that | Z ( z (cid:48) k ) | = 1. On the other hand,1 | M r d ( r ) r k by Lemma 5.3.1. Therefore z k ∈ Z min ( k, z (cid:48) k := ( n ( r ) − · k − (cid:88) i =1 ( n ( r ) − d ( r )) r i . Proceeding as in the proof of Lemma 5.3.1, one can verify that φ ( z (cid:48) k ) = d ( r ) r k −
1. Inaddition, the coefficients of the atoms 1 , . . . , r k − in z (cid:48) k are all strictly less than n ( r ). Thenit follows from Lemma 5.1.4(1) that z (cid:48) k is a factorization of d ( r ) r k − | z (cid:48) k | = k ( n ( r ) − d ( r )) + d ( r ) −
1, one has that τ (1) = sup k sup z (cid:8) min L (cid:0) φ ( z ) − (cid:1) : z ∈ Z min ( k, (cid:9) ≥ sup k min L (cid:0) φ ( z k ) − (cid:1) = sup k | z (cid:48) k | = lim k →∞ k ( n ( r ) − d ( r )) + d ( r ) − ∞ . Hence τ (1) = ∞ . Then it follows by Theorem 5.3.5 that M r is not locally tame, whichcontradicts condition (3). Thus, (3) implies (1), as desired.80HAPTER 6FACTORIAL ELEMENTS OF PUISEUX MONOIDS The elements having exactly one factorization are crucial in the study of factorizationtheory of commutative cancellative monoids and integral domains. Aiming to avoidrepeated long descriptions, we call such elements molecules . Molecules were first studiedin the context of algebraic number theory by W. Narkiewicz and other authors in the1960’s. For instance, in [59] and [61] Narkiewicz studied some distributional aspects of themolecules of quadratic number fields. In addition, he gave an asymptotic formula for thenumber of (non-associated) integer molecules of any algebraic number field [60]. In thischapter, we study the molecules of submonoids of ( Q ≥ , +), including numerical monoids,and the molecules of their corresponding monoid algebras.If a numerical monoid N satisfies that N (cid:54) = N , then it contains only finitelymany molecules. Notice, however, that every positive integer is a molecule of ( N , +).Figure 6-1 shows the distribution of the sets of molecules of four numerical monoids. Webegin Section 6.2 pointing out how the molecules of numerical monoids are related tothe Betti elements. Then we show that each element in the set N ≥ ∪ {∞} (and onlysuch elements) can be the number of molecules of a numerical monoid. We conclude ourstudy of molecules of numerical monoids exploring the possible cardinalities of the sets ofreducible molecules (i.e., molecules that are not atoms).The class of Puiseux monoids, on the other hand, contains members having infinitelymany atoms and, consequently, infinitely many molecules. In Section 6.3, we study thesets of molecules of Puiseux monoids, finding infinitely many non-isomorphic Puiseuxmonoids all whose molecules are atoms (in contrast to the fact that the set of molecules ofa numerical monoid always differs from its set of atoms).We conclude with Section 6.4, where we construct infinitely many non-isomorphicPuiseux monoids having infinitely many molecules that are not atoms (in contrast to81igure 6-1. The dots on the horizontal line labeled by N i represent the nonzero elementsof the numerical monoid N i ; here we are setting N = (cid:104) , (cid:105) , N = (cid:104) , , (cid:105) , N = (cid:104) , , , , (cid:105) , and N = (cid:104) , (cid:105) . Atoms are represented in blue, moleculesthat are not atoms in red, and non-molecules in black.the fact that the set of molecules of a nontrivial numerical monoid is always finite).Special attention is given in this section to prime reciprocal Puiseux monoids and acharacterization of their molecules. In this section we study the sets of molecules of numerical monoids, putting particularemphasis on their possible cardinalities.
As one of the main purposes of this chapter is to study elements with exactly onefactorization in Puiseux monoids (in particular, numerical monoids), we introduce thefollowing definition.
Definition 6.2.1.
Let M be a monoid. We say that an element x ∈ M \ U ( M ) is a molecule provided that | Z ( x ) | = 1. The set of all molecules of M is denoted by M ( M ).It is clear that the set of atoms of any monoid is contained in the set of molecules.However, such an inclusion might be proper (consider, for instance, the additive monoid N ). In addition, for any atomic monoid M the set M ( M ) is divisor-closed in the sensethat if x ∈ M ( M ) and x (cid:48) | M x for some x (cid:48) ∈ M \ U ( M ), then x (cid:48) ∈ M ( M ). If the conditionof atomicity is dropped, then this observation is not necessarily true (see Example 6.3.1).82 xample 6.2.2. For k ≥
1, consider the numerical monoid N = (cid:104) , (cid:105) , whose moleculesare depicted in Figure 6-1. It is not hard to see that x ∈ N • is a molecule if and only ifevery factorization of x contains at most one copy of 21. Therefore M ( N ) = (cid:8) m + 21 n : 0 ≤ m < , n ∈ { , } , and ( m, n ) (cid:54) = (0 , (cid:9) . In addition, if 2 m + 21 n = 2 m (cid:48) + 21 n (cid:48) for some m, m (cid:48) ∈ { , . . . , } and n, n (cid:48) ∈ { , } , thenone can readily check that m = m (cid:48) and n = n (cid:48) . Hence |M ( N ) | = 41. Let N = (cid:104) a , . . . , a n (cid:105) be a minimally generated numerical monoid. We alwaysrepresent an element of Z ( N ) with an n -tuple z = ( c , . . . , c n ) ∈ N n , where the entry c i specifies the number of copies of a i that appear in z . Clearly, | z | = c + · · · + c n . Givenfactorizations z = ( c , . . . , c n ) and z (cid:48) = ( c (cid:48) , . . . , c (cid:48) n ), we definegcd( z, z (cid:48) ) = (min { c , c (cid:48) } , . . . , min { c n , c (cid:48) n } ) . The factorization graph of x ∈ N , denoted by ∇ x ( N ) (or just ∇ x when no risk of confusionexists), is the graph with vertices Z ( x ) and edges between those z, z (cid:48) ∈ Z ( x ) satisfyingthat gcd( z, z (cid:48) ) (cid:54) = 0. The element x is called a Betti element of N provided that ∇ x isdisconnected. The set of Betti elements of N is denoted by Betti( N ). Example 6.2.3.
Take N to be the numerical monoid (cid:104) , , , , (cid:105) . A computationin SAGE using the numericalsgps GAP package immediately reveals that N has nineBetti elements. In particular, 90 ∈ Betti( N ). In Figure 6-2 one can see the disconnectedfactorization graph of the Betti element 90 on the left and the connected factorizationgraph of the non-Betti element 84 on the right.Observe that 0 / ∈ Betti( N ) since | Z (0) | = 1. It is well-known that every numericalmonoid has finitely many Betti elements. Betti elements play a fundamental role inthe study of uniquely-presented numerical monoids [33] and the study of delta sets ofBFMs [17]. In a numerical monoid, Betti elements and molecules are closely related.83 B Figure 6-2. Factorization graphs of one Betti element and one non-Betti element in thenumerical monoid N = (cid:104) , , , , (cid:105) : A) the element 90 ∈ Betti( N )and B) the element 84 / ∈ Betti( N ). Remark 6.2.4.
Let N be a numerical monoid. An element m ∈ N is a molecule if andonly if β (cid:45) N m for any β ∈ Betti( N ). Proof.
For the direct implication, suppose that m is a molecule of N and take α ∈ N suchthat α | N m . As the set of molecules is closed under division, | Z ( α ) | = 1. This implies that ∇ α is connected and, therefore, α cannot be a Betti element. The reverse implication isjust a rephrasing of [33, Lemma 1]. Obviously, for every n ∈ N there exists a numerical monoid having exactly n atoms.The next proposition answers the same realization question replacing the concept ofan atom by that one of a molecule. Recall that N • denotes the class of all nontrivialnumerical monoids. Proposition 6.2.5. {|M ( N ) | : N ∈ N • } = N ≥ .Proof. Let N be a nontrivial numerical monoid. Then N must contain at least two atoms.Let a and b denote the two smallest atoms of N , and assume that a < b . Note that 2 a and a + b are distinct molecules that are not atoms. Hence |M ( N ) | ≥
4. As a result, {|M ( N ) | : N ∈ N • } ⊆ N ≥ ∪ {∞} . Now take x ∈ N with x > f ( N ) + ab . Since x (cid:48) := x − ab > f ( N ), we have that x (cid:48) ∈ N and, therefore, Z ( x (cid:48) ) contains at least one84actorization, namely z . So we can find two distinct factorizations of x by adding to z either a copies of b or b copies of a . Thus, f ( N ) + ab is an upper bound for M ( N ), whichmeans that |M ( N ) | ∈ N ≥ . Thus, {|M ( N ) | : N ∈ N • } ⊆ N ≥ .To argue the reverse inclusion, suppose that n ∈ N ≥ , and let us find N ∈ N with |M ( N ) | = n . For n = 4, we can take the numerical monoid (cid:104) , (cid:105) (see Figure 6-1). For n >
4, consider the numerical monoid N = (cid:104) n − , n − , . . . , n − − (cid:105) . It follows immediately that A ( N ) = { n − , n − , . . . , n − − } . In addition, it is nothard to see that 2( n − , n −
2) + 1 ∈ M ( N ) while k / ∈ M ( N ) for any k > n −
2) + 1.Consequently, M ( N ) = A ( N ) ∪ { n − , n −
2) + 1 } , which implies that |M ( N ) | = n .Therefore {|M ( N ) | : N ∈ N } ⊇ N ≥ , which completes the proof. Corollary 6.2.6.
The monoid ( N , +) is the only numerical monoid having infinitelymany molecules. In Proposition 6.2.5 we have fully described the set {|M ( N ) | : N ∈ N } . A fulldescription of the set {|M ( N ) \ A ( N ) | : N ∈ N } seems to be significantly more involved.However, the next theorem offers some evidence to believe that {|M ( N ) \ A ( N ) | : N ∈ N } = N ≥ ∪ {∞} . Theorem 6.2.7.
The following statements hold.1. {|M ( N ) \ A ( N ) | : N ∈ N • } ⊆ N ≥ .2. |M ( N ) \A ( N ) | = 2 for infinitely many numerical monoids N .3. For each k ∈ N , there is a numerical monoid N k with |M ( N ) \A ( N ) | > k .Proof. To prove (1), take N ∈ N • . Then we can assume that N has embedding dimension n with n ≥
2. Take a , . . . , a n ∈ N such that a < · · · < a n such that N = (cid:104) a , . . . , a n (cid:105) .Since a < a < a j for every j = 3 , . . . , n , the elements 2 a and a + a are two distinct85olecules of N that are not atoms. Hence M ( N ) \A ( N ) ⊆ N ≥ ∪{∞} . On the other hand,Proposition 6.2.5 guarantees that |M ( N ) | < ∞ , which implies that |M ( N ) \ A ( N ) | < ∞ .As a result, the statement (1) follows.To verify the statement (2), one only needs to consider for every n ∈ N the numericalmonoid N n := { } ∪ N ≥ n − . The minimal set of generators of N n is the ( n − { n − , n − , . . . , n − − } and, as we have already argued in the proof ofProposition 6.2.5, the set M ( N n ) \A ( N n ) consists precisely of two elements.Finally, let us prove condition (3). We first argue that for any a, b ∈ N ≥ withgcd( a, b ) = 1 the numerical monoid (cid:104) a, b (cid:105) has exactly ab − a < b , take N := (cid:104) a, b (cid:105) , and set M = { ma + nb : 0 ≤ m < b, ≤ n < a, and ( m, n ) (cid:54) = (0 , } . Now take x ∈ N to be a molecule of N . As | Z ( x ) | = 1, the unique factorization z := ( c , c ) ∈ Z ( x ) (with c , c ∈ N ) satisfies that c < b ; otherwise, we couldexchange b copies of the atom a by a copies of the atom b to obtain another factorizationof x . A similar argument ensures that c < a . As a consequence, M ( N ) ⊆ M . On theother hand, if ma + nb = m (cid:48) a + n (cid:48) b for some m, m (cid:48) , n, n (cid:48) ∈ N , then gcd( a, b ) = 1 impliesthat b | m − m (cid:48) and a | n − n (cid:48) . Because of this observation, the element ( b − a + ( a − b has only the obvious factorization, namely ( b − , a − b − a + ( a − b is amolecule satisfying that y | N ( b − a + ( a − b for every y ∈ M , the inclusion M ⊆ M ( N )holds. Hence |M ( N ) | = |M| = ab −
1. To argue the statement (3) now, it suffices to take N k := (cid:104) , k + 1 (cid:105) . In this section we study the sets of molecules of the general class of Puiseux monoids.We will argue that there are infinitely many non-finitely generated atomic Puiseuxmonoids P such that |M ( P ) \ A ( P ) | = ∞ . On the other hand, we will prove that, unlike86he case of numerical monoids, there are infinitely many non-isomorphic atomic Puiseuxmonoids all whose molecules are, indeed, atoms.In Section 6.2 we mentioned that the set of molecules of an atomic monoid isdivisor-closed. The next example indicates that this property might not hold fornon-atomic monoids. Example 6.3.1.
Consider the Puiseux monoid P = (cid:28) , , n : n ∈ N (cid:29) . First, observe that 0 is not a limit point of P • , and so P cannot be finitely generated.After a few easy verifications, one can see that A ( P ) = { / , / } . On the other hand, itis clear that 1 / / ∈ (cid:104) / , / (cid:105) , so P is not atomic. Observe now that Z (1) contains only onefactorization, namely 2 / /
5. Therefore 1 ∈ M ( P ). Since Z (1 /
2) is empty, 1 / P . However, 1 / | P
1. As a result, M ( P ) is not divisor-closed.Although the additive monoid N contains only one atom, it has infinitely manymolecules. The next result implies that N is basically the only atomic Puiseux monoidhaving finitely many atoms and infinitely many molecules. Proposition 6.3.2.
Let P be a Puiseux monoid. Then |M ( P ) | ∈ N ≥ if and only if |A ( P ) | ∈ N ≥ .Proof. Suppose first that |M ( P ) | ∈ N ≥ . As every atom is a molecule, A ( P ) is finite.Furthermore, note that if A ( P ) = { a } , then every element of the set S = { na : n ∈ N } would be a molecule, which is not possible as | S | = ∞ . As a result, |A ( P ) | ∈ N ≥ .Conversely, suppose that |A ( P ) | ∈ N ≥ . Since the elements in P \ (cid:104)A ( P ) (cid:105) have nofactorizations, M ( P ) = M ( (cid:104)A ( P ) (cid:105) ). Therefore there is no loss in assuming that P isatomic. As 1 < |A ( P ) | < ∞ , the monoid P is isomorphic to a nontrivial numericalmonoid. The proposition now follows from the fact that nontrivial numerical monoids havefinitely many molecules. Corollary 6.3.3. If P is a Puiseux monoid, then |M ( P ) | (cid:54) = 1 . P and P (cid:48) are isomorphic, then there exists q ∈ Q > such that P (cid:48) = qP ; this is aconsequence of Proposition 2.2.2.11. Theorem 6.3.4 (cf. Theorem 6.2.7(1)) . There are infinitely many non-isomorphic atomicPuiseux monoids P satisfying that M ( P ) = A ( P ) .Proof. Let S = { S n : n ∈ N } be a collection of infinite and pairwise-disjoint sets ofprimes. Now take S = S n for some arbitrary n ∈ N , and label the primes in S strictlyincreasingly by p , p , . . . . Recall that D S ( r ) denotes the set of primes in S dividing d ( r )and that D S ( R ) = ∪ r ∈ R D S ( r ) for R ⊆ Q > . We proceed to construct a Puiseux monoid P S satisfying that D S ( P S ) = S .Take P := (cid:104) /p (cid:105) and P := (cid:104) P , / ( p p ) (cid:105) . In general, suppose that P k is a finitelygenerated Puiseux monoid such that D S ( P k ) ⊂ S , and let r , . . . , r n k be all the elements in P k which can be written as a sum of two atoms. Clearly, n k ≥
1. Because | S | = ∞ , onecan take p (cid:48) , . . . , p (cid:48) n k to be primes in S \ D S ( P k ) satisfying that p (cid:48) i (cid:45) n ( r i ). Now consider thefollowing finitely generated Puiseux monoid P k +1 := (cid:28) P k ∪ (cid:26) r p (cid:48) , . . . , r n k p (cid:48) n k (cid:27)(cid:29) . For every i ∈ { , . . . , n k } , there is only one element in P k ∪ { r /p (cid:48) , . . . , r n k /p (cid:48) n k } whosedenominator is divisible by p (cid:48) i , namely r i /p (cid:48) i . Therefore r i /p (cid:48) i ∈ A ( P k +1 ) for i = 1 , . . . , n k .To check that A ( P k ) ⊂ A ( P k +1 ), fix a ∈ A ( P k ) and take z := m (cid:88) i =1 α i a i + n k (cid:88) i =1 β i r i p (cid:48) i ∈ Z P k +1 ( a ) , (6-1)where a , . . . , a k are pairwise distinct atoms in A ( P k +1 ) ∩ P k and α i , β j are nonnegativecoefficients for i = 1 , . . . , m and j = 1 , . . . , n k . In particular, a , . . . , a k ∈ A ( P k ). For each i = 1 , . . . , n k , the fact that the p (cid:48) i -adic valuation of a is nonnegative implies that p (cid:48) i | β i .88ence a = m (cid:88) i =1 α i a i + n k (cid:88) i =1 β (cid:48) i r i , where β (cid:48) i = β i /p (cid:48) i ∈ N for i = 1 , . . . , n k . Since r i ∈ A ( P k ) + A ( P k ) and ( β i /p (cid:48) i ) r i | P k a for every i = 1 , . . . , n k , one obtains that β = · · · = β n k = 0. As a result, a = (cid:80) mi =1 α i a i .Because a ∈ A ( P k ), the factorization (cid:80) mi =1 α i a i in Z P k ( a ) must have length 1, i.e, (cid:80) mi =1 α i = 1. Thus, (cid:80) mi =1 α i + (cid:80) n k i =1 β i = 1, which means that z has length 1 and so a ∈ A ( P k +1 ). As a result, the inclusion A ( P k ) ⊆ A ( P k +1 ) holds. Observe that because n k ≥
1, the previous containment must be strict. Now set P S = (cid:91) k ∈ N P k . Let us verify that P S is an atomic monoid satisfying that A ( P S ) = ∪ k ∈ N A ( P k ). Since P k is atomic for every k ∈ N , the inclusion chain A ( P ) ⊂ A ( P ) ⊂ . . . implies that P ⊂ P ⊂ . . . . In addition, if a = a + · · · + a m for m ∈ N and a , a , . . . , a m ∈ P S ,then a = a + · · · + a m will also hold in P k for some k ∈ N large enough. Thisimmediately implies that ∪ k ∈ N A ( P k ) ⊆ A ( P S ). Since the reverse inclusion follows trivially, A ( P S ) = ∪ k ∈ N A ( P k ). To check that P S is atomic, take x ∈ P • S . Then there exists k ∈ N such that x ∈ P k and, because P k is atomic, x ∈ (cid:104)A ( P k ) (cid:105) ⊆ (cid:104)A ( P S ) (cid:105) . Hence P S is atomic.To check that M ( P S ) = A ( P S ), suppose that m is a molecule of P S , and thentake K ∈ N such that m ∈ P k for every k ≥ K . Since A ( P k ) ⊂ A ( P k +1 ) ⊂ . . . , wehave that Z P k ( m ) ⊆ Z P k +1 ( m ) ⊆ . . . . Moreover, ∪ k ≥ K A ( P k ) = A ( P S ) implies that ∪ k ≥ K Z P k ( m ) = Z P S ( m ). Now suppose for a contradiction that m = (cid:80) ij =1 a j for i ∈ N ≥ ,where a , . . . , a i ∈ A ( P S ). Take j ∈ N ≥ K such that a , . . . , a i ∈ A ( P j ). Then theway in which P j +1 was constructed ensures that | Z P j +1 ( a + a ) | ≥ | Z P j +1 ( m ) | ≥
2. As Z P j +1 ( m ) ⊆ Z P S ( m ), it follows that | Z P S ( m ) | ≥
2, which contradictsthat m is a molecule. Hence M ( P S ) = A ( P S ).Finally, we argue that the monoids constructed are not isomorphic. Let S and S (cid:48) be two distinct members of the collection S and suppose, by way of contradiction,89hat ψ : P S → P S (cid:48) is a monoid isomorphism. Because the only homomorphisms ofPuiseux monoids are given by rational multiplications, there exists q ∈ Q > such that P S (cid:48) = qP S . In this case, all but finitely many primes in D P ( P S ) belong to D P ( P S (cid:48) ). Since D P ( P S ) ∩ D P ( P S (cid:48) ) = ∅ when S (cid:54) = S (cid:48) , we get a contradiction. In this section, we focus our attention on the class consisting of all prime reciprocalPuiseux monoids.
Proposition 6.4.1 (cf. Theorem 6.2.7(1)) . There exist infinitely many non-finitelygenerated atomic Puiseux monoids P such that |M ( P ) \A ( P ) | = ∞ .Proof. As in the proof of Theorem 6.3.4, let S = { S n : n ∈ N } be a collection of infiniteand pairwise-disjoint subsets of P \ { } . For every n ∈ N , let P n be a prime reciprocalPuiseux monoid over S n . Fix a ∈ A ( P n ), and take a factorization z := k (cid:88) i =1 α i a i ∈ Z (2 a ) , for some k ∈ N , pairwise distinct atoms a , . . . , a k , and α , . . . , α k ∈ N . Since d ( a ) (cid:54) = 2,after applying the d ( a )-adic valuation on both sides of the equality 2 a = (cid:80) ti =1 α i a i , oneobtains that z = 2 a . So 2 a ∈ M ( P n ) \ A ( P n ) and, as a result, |M ( P n ) \ A ( P n ) | = ∞ .Now suppose, by way of contradiction, that P i ∼ = P j for some i, j ∈ N with i (cid:54) = j . Since theonly isomorphisms of Puiseux monoids are given by rational multiplication, there exists q ∈ Q > such that P j = qP i . However, this implies that only finitely many primes in d ( P i )are not contained in d ( P j ), which contradicts that S i ∩ S j = ∅ . Hence no two monoids in { P n : n ∈ N } are isomorphic, and the proposition follows.Before characterizing the molecules of prime reciprocal monoids, let us introduce theconcept of maximal multiplicity. Let P be a Puiseux monoid. For x ∈ P and a ∈ A ( P ) wedefine the maximal multiplicity of a in x to be m ( a, x ) := max { n ∈ N : na | P x } . roposition 6.4.2. Let P be a prime reciprocal monoid, and let x ∈ P . If m ( a, x ) < d ( a ) for all a ∈ A ( P ) , then x ∈ M ( P ) .Proof. Suppose, by way of contradiction, that x / ∈ M ( P ). Then there exist k ∈ N ,elements α i , β i ∈ N (for i = 1 , . . . , k ), and pairwise distinct atoms a , . . . , a k such that z := k (cid:88) i =1 α i a i and z (cid:48) := k (cid:88) i =1 β i a i are two distinct factorizations in Z ( x ). As z (cid:54) = z (cid:48) , there is an index i ∈ { , . . . , k } such that α i (cid:54) = β i . Now we can apply the d ( a i )-adic valuation to both sides of the equality k (cid:88) i =1 α i a i = k (cid:88) i =1 β i a i to verify that d ( a i ) | β i − α i . As α i (cid:54) = β i , we obtain that m ( a i , x ) ≥ max { α i , β i } ≥ d ( a i ) . However, this contradicts the fact that m ( a, x ) < d ( a ) for all a ∈ A ( P ). As a consequence, x ∈ M ( P ).For S ⊆ P , we call the monoid E S := (cid:104) /p : p ∈ S (cid:105) the elementary prime reciprocalmonoid over S ; if S = P we say that E S is the elementary prime reciprocal monoid. It wasproved in [52, Section 5] that every submonoid of the elementary prime reciprocal monoidis atomic. This gives a large class of non-finitely generated atomic Puiseux monoids, whichcontains each prime reciprocal monoid. Proposition 6.4.3.
Let S be an infinite set of primes, and let E S be the elementary primereciprocal monoid over S . For x ∈ E S , the following conditions are equivalent:1. x ∈ M ( E S ) ;2. does not divide x in E S ;3. m ( a, x ) < d ( a ) for all a ∈ A ( E S ) ; . If a , . . . , a n ∈ A ( E S ) are distinct atoms and α , . . . , α n ∈ N satisfy that (cid:80) nj =1 α i a i ∈ Z ( x ) , then α j < d ( a j ) for each j = 1 , . . . , n .Proof. First, let us recall that since E S is atomic, M ( E S ) is divisor-closed. On the otherhand, note that for any two distinct atoms a, a (cid:48) ∈ A ( E S ), both factorizations d ( a ) a and d ( a (cid:48) ) a (cid:48) are in Z (1). Therefore 1 / ∈ M ( E S ). Because the set of molecules of E S isdivisor-closed, 1 (cid:45) E S m for any m ∈ M ( E S ); in particular, 1 (cid:45) E S x . Thus, (1) implies (2). If m ( a, x ) ≥ d ( a ) for a ∈ A ( E S ), then x = m ( a, x ) a + y = 1 + ( m ( a, x ) − d ( a )) a + y for some y ∈ E S . As a result, 1 | E S x , from which we can conclude that (2) implies (3). Itis obvious that (3) and (4) are equivalent conditions. Finally, the fact that (3) implies (1)follows from Proposition 6.4.2. Corollary 6.4.4.
Let S be an infinite set of primes, and let E S be the elementary primereciprocal monoid over S . Then | Z ( x ) | = ∞ for all x / ∈ M ( E S ) . In order to describe the set of molecules of an arbitrary prime reciprocal monoid, weneed to cast its atoms into two categories.
Definition 6.4.5.
Let P be a prime reciprocal monoid. We say that a ∈ A ( P ) is stable ifthe set { a (cid:48) ∈ A ( P ) : n ( a (cid:48) ) = n ( a ) } is infinite, otherwise we say that a is unstable . If everyatom of P is stable (resp., unstable), then we call P stable (resp., unstable ).For a prime reciprocal monoid P , we let S ( P ) denote the submonoid of P generatedby the set of stable atoms. Similarly, we let U ( P ) denote the submonoid of P generated bythe set of unstable atoms. Clearly, P is stable (resp., unstable) if and only if P = S ( P )(resp., P = U ( P )). In addition, P = S ( P ) + U ( P ), and S ( P ) ∩ U ( P ) is trivial only wheneither S ( P ) or U ( P ) is trivial. Clearly, if P is stable, then it cannot be finitely generated.Finally, we say that u ∈ U ( P ) is absolutely unstable provided that u is not divisible byany stable atom in P , and we let U a ( P ) denote the set of all absolutely unstable elementsof P . 92 xample 6.4.6. Let ( p n ) n ∈ N be the strictly increasing sequence with underlying set P \ { } , and consider the prime reciprocal monoid P defined as P := (cid:28) − n p n − , p n − p n : n ∈ N (cid:29) . Set a n = − n p n − and b n = p n − p n . One can readily verify that P is an atomic monoid with A ( P ) = { a n , b n : n ∈ N } . As both sets { n ∈ N : n ( a n ) = 2 } and { n ∈ N : n ( a n ) = 4 } have infinite cardinality, a n is a stable atom for every n ∈ N . In addition, since ( n ( b n )) n ∈ N is a strictly increasing sequence bounded below by n ( b ) = 4 and n ( a n ) ∈ { , } , we havethat b n is an unstable atom for every n ∈ N ≥ . Also, notice that 4 / a ∈ S ( P ), but4 / / ∈ U ( P ) because d (4 /
3) = 3 / ∈ d ( U ( P )). Furthermore, for every n ∈ N the element u n := ( p n − b n ∈ U ( P ) is not in S ( P ) because p n = d ( u n ) / ∈ d ( S ( P )). However, S ( P ) ∩ U ( P ) (cid:54) = ∅ since the element 4 = 6 a = 5 b belongs to both S ( P ) and U ( P ).Finally, we claim that 2 b n is absolutely unstable for every n ∈ N . If this were not the case,then 2 b k / ∈ M ( P ) for some k ∈ N . By Proposition 6.4.2 there exists a ∈ A ( P ) such that m ( a, b k ) ≥ d ( a ). In this case, one would obtain that 2 b k ≥ m ( a, b k ) a ≥ d ( a ) a = n ( a ) ≥ b n < n ∈ N . Thus, 2 b n ∈ U a ( P ) for every n ∈ N . Proposition 6.4.7.
Let P be a prime reciprocal monoid that is stable, and let x ∈ P .Then x ∈ M ( P ) if and only if n ( a ) does not divide x in P for any a ∈ A ( P ) .Proof. For the direct implication, assume that x ∈ M ( P ) and suppose, by way ofcontradiction, that n ( a ) | P x for some a ∈ A ( P ). Since a is a stable atom, there exist p , p ∈ P with p (cid:54) = p such that gcd( p p , n ( a )) = 1 and n ( a ) /p , n ( a ) /p ∈ A ( P ). As n ( a ) | P x , we can take a , . . . , a k ∈ A ( P ) such that x = n ( a ) + a + · · · + a k . Therefore p n ( a ) p + a + · · · + a k and p n ( a ) p + a + · · · + a k Z ( x ), contradicting that x is a molecule. Conversely,suppose that x is not a molecule. Consider two distinct factorizations z := (cid:80) ki =1 α i a i and z (cid:48) := (cid:80) ki =1 β i a i in Z ( x ), where k ∈ N , α i , β i ∈ N , and a , . . . , a k ∈ A ( P ) are pairwisedistinct atoms. Pick an index j ∈ { , . . . , k } such that α j (cid:54) = β j and assume, without lossof generality, that α j < β j . After applying the d ( a j )-adic valuations on both sides of theequality k (cid:88) i =1 α i a i = k (cid:88) i =1 β i a i one finds that the prime d ( a j ) divides β j − α j . Therefore β j > d ( a j ) and so x = n ( a j ) + ( β j − d ( a j )) a j + (cid:88) i (cid:54) = j α i a i . Hence n ( a j ) | P x , which concludes the proof.Observe that the reverse implication of Proposition 6.4.7 does not require S ( P ) = P .However, the stability of P is required for the direct implication to hold as the followingexample illustrates. Example 6.4.8.
Let ( p n ) n ∈ N be the strictly increasing sequence with underlying set P \ { } , and consider the unstable prime reciprocal monoid P := (cid:28) , p n − p n : n ∈ N (cid:29) . Because the smallest two atoms of P are 1 / /
3, it immediately follows that m :=2(1 /
2) + 8 / / ∈ (cid:104) / (cid:105) must be a molecule of P . In addition, notice that 1 = n (1 /
2) divides m in P .We conclude this section characterizing the molecules of prime reciprocal monoids. Theorem 6.4.9.
Let P be a prime reciprocal monoid. Then x ∈ P is a molecule if andonly if x = s + u for some s ∈ S ( P ) ∩ M ( P ) and u ∈ U a ( P ) ∩ M ( P ) .Proof. First, suppose that x is a molecule. As P = S ( P ) + U ( P ), there exist s ∈ S ( P ) and u ∈ U ( P ) such that x = s + u . The fact that x ∈ M ( P ) guarantees that s, u ∈ M ( P ).94n the other hand, since | Z ( u ) | = 1 and u can be factored using only unstable atoms, u cannot be divisible by any stable atom in P . Thus, u ∈ U a ( P ), and the direct implicationfollows.For the reverse implication, assume that x = s + u , where s ∈ S ( P ) ∩ M ( P )and u ∈ U a ( P ) ∩ M ( P ). We first check that x can be uniquely expressed as a sumof two elements s and u contained in the sets S ( P ) ∩ M ( P ) and U a ( P ) ∩ M ( P ),respectively. To do this, suppose that x = s + u = s (cid:48) + u (cid:48) , where s (cid:48) ∈ S ( P ) ∩ M ( P )and u (cid:48) ∈ U a ( P ) ∩ M ( P ). Take pairwise distinct stable atoms a , . . . , a k of P for some k ∈ N such that z = (cid:80) ki =1 α i a i ∈ Z P ( s ) and z (cid:48) = (cid:80) ki =1 α (cid:48) i a i ∈ Z P ( s (cid:48) ), where α j , α (cid:48) j ∈ N for j = 1 , . . . , k . Because u and u (cid:48) are absolutely unstable elements, they are not divisiblein P by any of the atoms a i ’s. Thus, d ( a j ) (cid:45) d ( u ) and d ( a j ) (cid:45) d ( u (cid:48) ) for any j ∈ { , . . . , k } .Now for each j = 1 , . . . , k we can apply the d ( a j )-adic valuation in both sides of theequality u + k (cid:88) i =1 α i a i = u (cid:48) + k (cid:88) i =1 α (cid:48) i a i to conclude that the prime d ( a j ) must divide α j − α (cid:48) j . Therefore either z = z (cid:48) or thereexists j ∈ { , . . . , k } such that | α j − α (cid:48) j | > d ( a j ). Suppose that | α j − α (cid:48) j | > d ( a j ) forsome j , and say α j > α (cid:48) j . As α j > d ( a j ), one can replace α j a j by ( α j − d ( a j )) a j + n ( a j )in s = φ ( z ) = α a + · · · + α k a k to find that n ( a j ) divides s in S ( P ), which contradictsProposition 6.4.7. Then we have z = z (cid:48) . Therefore s (cid:48) = s and u (cid:48) = u .Finally, we argue that x ∈ M ( P ). Write x = (cid:80) (cid:96)i =1 γ i a i + (cid:80) (cid:96)i =1 β i b i for (cid:96) ∈ N ≥ k ,pairwise distinct stable atoms a , . . . , a (cid:96) (where a , . . . , a k are the atoms showing upin z ), pairwise distinct unstable atoms b , . . . , b (cid:96) , and coefficients γ i , β i ∈ N for every i = 1 , . . . , (cid:96) . Set z (cid:48)(cid:48)(cid:48) := (cid:80) (cid:96)i =1 γ i a i and w (cid:48)(cid:48)(cid:48) = (cid:80) (cid:96)i =1 β i b i . Note that, a priori , φ ( z (cid:48)(cid:48)(cid:48) )and φ ( w (cid:48)(cid:48)(cid:48) ) are not necessarily molecules. As in the previous paragraph, we can apply d ( a j )-adic valuation to both sides of the equality u + k (cid:88) i =1 α i a i = (cid:96) (cid:88) i =1 γ i a i + (cid:96) (cid:88) i =1 β i b i
95o find that z (cid:48)(cid:48)(cid:48) = z . Hence φ ( z (cid:48)(cid:48)(cid:48) ) = s and φ ( w (cid:48)(cid:48)(cid:48) ) = u are both molecules. Therefore z (cid:48)(cid:48)(cid:48) must be the unique factorization of s , while w (cid:48)(cid:48)(cid:48) must be the unique factorization of u . Asa result, x ∈ M ( P ). 96HAPTER 7PUISEUX ALGEBRAS Let M be a monoid and let R be a commutative ring with identity. Then R [ X ; M ]denotes the ring of all functions f : M → R having finite support , which means that Supp ( f ) := { s ∈ M : f ( s ) (cid:54) = 0 } is finite. We represent an element f ∈ R [ X ; M ] by f ( X ) = n (cid:88) i =1 f ( s i ) X s i , where s , . . . , s n are the elements in Supp ( f ). The ring R [ X ; M ] is called the monoid ring of M over R , and the monoid M is called the exponent monoid of R [ X ; M ]. For a field F , we will say that F [ X ; M ] is a monoid algebra . As we are primarily interested in themolecules of monoid algebras of Puiseux monoids, we introduce the following definition. Definition 7.1.1. If F is a field and P is a Puiseux monoid, then we say that F [ X ; P ] isa Puiseux algebra . If N is a numerical monoid, then F [ X ; N ] is called a numerical monoidalgebra .Let F [ X ; P ] be a Puiseux algebra. We write any element f ∈ F [ X ; P ] \ { } in canonical representation , that is, f ( X ) = α X q + · · · + α k X q k with α i (cid:54) = 0 for every i = 1 , . . . , k and q > · · · > q k . It is clear that any element of F [ X ; P ] \ { } has a uniquecanonical representation. In this case, deg( f ) := q is called the degree of f , and we seethat the degree identity deg( f g ) = deg( f ) + deg( g ) holds for all f, g ∈ F [ X ; P ] \ { } .As for polynomials, we say that f is a monomial if k = 1. It is not hard to verify that F [ X ; P ] is an integral domain with set of units F × , although this follows from [42,Theorem 8.1] and [42, Theorem 11.1]. Finally, note that, unless P ∼ = ( N , +), no monomialof F [ X ; P ] can be a prime element; this is a consequence of the trivial fact that non-cyclicPuiseux monoids do not contain prime elements. Puiseux algebras have been consideredin [3, 25, 44, 49]. 97or an integral domain R , we let R red denote the reduced monoid of the multiplicativemonoid of R . We proceed to study the factorial elements of a given Puiseux algebra.
Definition 7.2.1.
Let R be an integral domain. We call a nonzero non-unit r ∈ R a molecule if rR × is a molecule of R red .Let R be an integral domain. By simplicity, we let A ( R ), M ( R ), Z ( R ), and φ R denote A ( R red ), M ( R red ), Z ( R red ), and φ R red , respectively. In addition, for a nonzero non-unit r ∈ R , we let Z R ( r ) and L R ( r ) denote Z R red ( rR × ) and L R red ( rR × ), respectively. Proposition 7.2.2.
Let F be a field, and let P be a Puiseux monoid. For a nonzero α ∈ F , a monomial X q ∈ M ( F [ X ; P ]) if and only if q ∈ M ( P ) .Proof. Consider the canonical monoid monomorphism µ : P → F [ X ; P ] \ { } given by µ ( q ) = X q . It follows from [26, Lemma 3.1] that an element a ∈ P is an atom if and onlyif the monomial X a is irreducible in F [ X ; P ] (or, equivalently, an atom in the reducedmultiplicative monoid of F [ X ; P ]). Therefore µ lifts canonically to the monomorphism¯ µ : Z ( P ) → Z ( F [ X ; P ]) determined by the assignments a (cid:55)→ X a for each a ∈ A ( P ),preserving not only atoms but also factorizations of the same element. Put formally, thismeans that the diagram Z ( P ) ¯ µ −−−→ Z ( F [ X ; P ]) φ P (cid:121) φ F [ X ; P ] (cid:121) P µ −−−→ F [ X ; P ] red commutes, and the (fiber) restriction maps ¯ µ q : Z P ( q ) → Z F [ X ; P ] ( X q ) of ¯ µ are bijectionsfor every q ∈ P . Hence | Z P ( q ) | = 1 if and only if | Z F [ X ; P ] ( X q ) | = 1 for all q ∈ P • , whichconcludes our proof. Corollary 7.2.3.
For each field F , there exists an atomic Puiseux monoid P whosePuiseux algebra satisfies that |M ( F [ X ; P ]) \ A ( F [ X ; P ]) | = ∞ .Proof. It is an immediate consequence of Proposition 6.4.1 and Proposition 7.2.2.98n element x ∈ gp ( M ) is called a root element if it is contained in the root closureof M , i.e., x ∈ (cid:102) M . Before providing a characterization for the irreducible elements of F [ X ; P ], let us argue the following two easy lemmas. Lemma 7.2.4.
Let P be a Puiseux monoid. Then d ( P • ) is closed under taking leastcommon multiples.Proof. Take d , d ∈ d ( P • ) and q , q ∈ P • with d ( q ) = d and d ( q ) = d . Now set d = gcd( d , d ) and n = gcd( n ( q ) , n ( q )). It is clear that n is the greatest common divisorof ( d /d ) n ( q ) and ( d /d ) n ( q ). So there exist m ∈ N and c , c ∈ N such that n (cid:0) m lcm( d , d ) (cid:1) = c d d n ( q ) + c d d n ( q ) . (7-1)Using the fact that d lcm( d , d ) = d d , one obtains that n (cid:0) m lcm( d , d ) (cid:1) lcm( d , d ) = c q + c q ∈ P after dividing both sides of the equality (7-1) by lcm( d , d ). In addition, note that n (1 + m lcm( d , d )) and lcm( d , d ) are relatively prime. Hence lcm( d , d ) ∈ d ( P • ), fromwhich the lemma follows. Lemma 7.2.5.
Let P be a root-closed Puiseux monoid containing . Then /d ∈ P for all d ∈ d ( P • ) .Proof. Let d ∈ d ( P • ), and take r ∈ P • such that d ( r ) = d . As gcd( n ( r ) , d ( r )) = 1, thereexist a, b ∈ N such that a n ( r ) − b d ( r ) = 1. Therefore1 d = a n ( r ) − b d ( r ) d = ar − b ∈ gp ( P ) . This, along with the fact that d (1 /d ) = 1 ∈ P , ensures that 1 /d is a root element of P .Since P is root-closed, it must contain 1 /d , which concludes our argument.We are in a position now to characterize the irreducibles of F [ X ; P ].99 roposition 7.2.6. Let F be a field, and let P be a root-closed Puiseux monoid contain-ing . Then f ∈ F [ X ; P ] \ F is irreducible in F [ X ; P ] if and only if f ( X m ) is irreduciblein F [ X ] for every m ∈ d ( P • ) that is a common multiple of the elements of d ( Supp ( f )) .Proof. Suppose first that f ∈ F [ X ; P ] \ F is an irreducible element of F [ X ; P ], andlet m ∈ d ( P • ) be a common multiple of the elements of d (cid:0) Supp ( f ) (cid:1) . Then f ( X m )is an element of F [ X ]. Take g, h ∈ F [ X ] such that f ( X m ) = g ( X ) h ( X ). As P is aroot-closed and m ∈ d ( P • ), Lemma 7.2.5 ensures that g ( X /m ) , h ( X /m ) ∈ F [ X ; P ]. Thus, f ( X ) = g ( X /m ) h ( X /m ) in F [ X ; P ]. Since f is irreducible in F [ X ; P ] either g ( X /m ) ∈ F or h ( X /m ) ∈ F , which implies that either g ∈ F or h ∈ F . Hence f ( X m ) is irreducible in F [ X ].Conversely, suppose that f ∈ F [ X ; P ] satisfies that f ( X m ) is an irreduciblepolynomial in F [ X ] for every m ∈ d ( P • ) that is a common multiple of the elementsof the set d ( Supp ( f )). To argue that f is irreducible in F [ X ; P ] suppose that f = g h for some g, h ∈ F [ X ; P ]. Let m be the least common multiple of the elements of d ( Supp ( g )) ∪ d ( Supp ( h )). Lemma 7.2.4 guarantees that m ∈ d ( P • ). Moreover, f = g h implies that m is a common multiple of the elements of d ( Supp ( f )). As a result, theequality f ( X m ) = g ( X m ) h ( X m ) holds in F [ X ]. Since f ( X m ) is irreducible in F [ X ],either g ( X m ) ∈ F or h ( X m ) ∈ F and, therefore, either g ∈ F or h ∈ F . This implies that f is irreducible in F [ X ; P ], as desired.We proceed to show the main result of this section. Theorem 7.2.7.
Let F be a field, and let P be a root-closed Puiseux monoid. Then M ( F [ X ; P ]) = (cid:104)A ( F [ X ; P ]) (cid:105) . Proof.
As each molecule of F [ X ; P ] is a product of irreducible elements in F [ X ; P ], theinclusion M ( F [ X ; P ]) ⊆ (cid:104)A ( F [ X ; P ]) (cid:105) holds trivially. For the reverse inclusion, supposethat f ∈ F [ X ; P ] \ F can be written as a product of irreducible elements in F [ X ; P ]. Asa result, there exist k, (cid:96) ∈ N and irreducible elements g , . . . , g k and h , . . . , h (cid:96) in F [ X ; P ]100atisfying that g ( X ) · · · g k ( X ) = f ( X ) = h ( X ) · · · h (cid:96) ( X ) . (7-2)Let m be the least common multiple of all the elements of the set (cid:18) k (cid:91) i =1 d (cid:0) Supp ( g i ) (cid:1)(cid:19) (cid:91) (cid:18) (cid:96) (cid:91) j =1 d (cid:0) Supp ( h j ) (cid:1)(cid:19) . Note that f ( X m ), g i ( X m ) and h j ( X m ) are polynomials in F [ X ] for i = 1 , . . . , k and j = 1 , . . . , (cid:96) . Lemma 7.2.4 ensures that m ∈ d ( P • ). On the other hand, m is a commonmultiple of all the elements of d ( Supp ( g i )) (or all the elements of d ( Supp ( h i ))). ThereforeProposition 7.2.6 guarantees that the polynomials g i ( X m ) and h j ( X m ) are irreducible in F [ X ] for i = 1 , . . . , k and j = 1 , . . . , (cid:96) . After substituting X by X m in (7-2) and usingthe fact that F [ X ] is a UFD, one finds that (cid:96) = k and g i ( X m ) = h σ ( i ) ( X m ) for somepermutation σ ∈ S k and every i = 1 , . . . , k . This, in turns, implies that g i = h σ ( i ) for i = 1 , . . . , k . Hence | Z F [ X ; P ] ( f ) | = 1, which means that f is a molecule of F [ X ; P ].As we have seen before, Corollary 7.2.3 guarantees the existence of a Puiseux algebra F [ X ; P ] satisfying that |M ( F [ X ; P ]) \ A ( F [ X ; P ]) | = ∞ . Now we use Theorem 7.2.7 toconstruct an infinite class of Puiseux algebras satisfying a slightly more refined condition. Proposition 7.2.8.
For any field F , there exist infinitely many Puiseux monoids P suchthat the algebra F [ X ; P ] contains infinitely many molecules that are neither atoms normonomials.Proof. Let ( p j ) j ∈ N be the strictly increasing sequence with underlying set P . Then foreach j ∈ N consider the Puiseux monoid P j = (cid:104) /p nj : n ∈ N (cid:105) . Fix j ∈ N , and take P := P j . The fact that gp ( P ) = P ∪ − P immediately implies that P is a root-closedPuiseux monoid containing 1. Consider the Puiseux algebra Q [ X ; P ] and the element X + p ∈ Q [ X ; P ], where p ∈ P . To argue that X + p is an irreducible element in Q [ X ; P ],write X + p = g ( X ) h ( X ) for some g, h ∈ Q [ X ; P ]. Now taking m to be the maximumpower of p j in the set d ( Supp ( g ) ∪ Supp ( h )), one obtains that X m + p = g ( X m ) h ( X m ) in101 [ X ]. Since Q [ X ] is a UFD, it follows by Eisenstein’s criterion that X m + p is irreducibleas a polynomial over Q . Hence either g ( X ) ∈ Q or h ( X ) ∈ Q , which implies that X + p is irreducible in Q [ X ; P ]. Now it follows by Theorem 7.2.7 that ( X + p ) n is a molecule in Q [ X ; P ] for every n ∈ N . Clearly, the elements ( X + p ) n are neither atoms nor monomials.Finally, we prove that the algebras we have defined in the previous paragraph arepairwise non-isomorphic. To do so suppose, by way of contradiction, that Q [ X ; P j ] and Q [ X ; P k ] are isomorphic algebras for distinct j, k ∈ N . Let ψ : Q [ X ; P j ] → Q [ X ; P k ] bean algebra isomorphism. Since ψ fixes Q , it follows that ψ ( X q ) / ∈ Q for any q ∈ P • j .This implies that deg( ψ ( X )) ∈ P • k . As d ( P • j ) is unbounded there exists n ∈ N such that p nj > n (deg( ψ ( X ))). Observe thatdeg (cid:0) ψ ( X ) (cid:1) = deg (cid:0) ψ (cid:0) X pnj (cid:1) p nj (cid:1) = p nj deg (cid:0) ψ (cid:0) X pnj (cid:1)(cid:1) . (7-3)Because gcd( p j , d ) = 1 for every d ∈ d ( P • k ), from (7-3) one obtains that p nj divides n (deg ψ ( X )), which contradicts that p nj > n (deg( ψ ( X ))). Hence the Puiseux algebras in { P j : j ∈ N } are pairwise non-isomorphic, which completes our proof.102EFERENCES[1] J. Amos, S. T. Chapman, N. Hine, and J. Paixao. Sets of lengths do not characterizenumerical monoids. Integers , 7:A50, 2007.[2] D. D. Anderson, D. F. Anderson, and M. Zafrullah. Factorizations in integraldomains.
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