Length-factoriality in commutative monoids and integral domains
Scott T. Chapman, Jim Coykendall, Felix Gotti, William W. Smith
aa r X i v : . [ m a t h . A C ] J a n LENGTH-FACTORIALITY IN COMMUTATIVE MONOIDS ANDINTEGRAL DOMAINS
SCOTT T. CHAPMAN, JIM COYKENDALL, FELIX GOTTI, AND WILLIAM W. SMITH
Abstract.
An atomic monoid M is called a length-factorial monoid (or an other-half-factorial monoid)if for each non-invertible element x ∈ M no two distinct factorizations of x have the same length.The notion of length-factoriality was introduced by Coykendall and Smith in 2011 as a dual of thewell-studied notion of half-factoriality. They proved that in the setting of integral domains, length-factoriality can be taken as an alternative definition of a unique factorization domain. However, being alength-factorial monoid is in general weaker than being a factorial monoid (i.e., a unique factorizationmonoid). Here we further investigate length-factoriality. First, we offer two characterizations of alength-factorial monoid M , and we use such characterizations to describe the set of Betti elementsand obtain a formula for the catenary degree of M . Then we study the connection between length-factoriality and purely long (resp., purely short) irreducibles, which are irreducible elements that appearin the longer (resp., shorter) part of any unbalanced factorization relation. Finally, we prove that anintegral domain cannot contain purely short and a purely long irreducibles simultaneously, and weconstruct a Dedekind domain containing purely long (resp., purely short) irreducibles but not purelyshort (resp., purely long) irreducibles. Introduction
An atomic monoid M is called half-factorial if for all non-invertible x ∈ M , any two factorizationsof x have the same length. In contrast to this, we say that M is length-factorial if for all non-invertible x ∈ M , any two distinct factorizations of x have different lengths. An integral domain is called half-factorial if its multiplicative monoid is half-factorial. Half-factorial monoids and domains have beensystematically investigated during the last six decades in connection with algebraic number theory,combinatorics, and commutative algebra: from work that appeared more than two decades ago, suchas [9,17,20,51], to more recent literature, including [26,32,35,42,46–48]. The term “half-factorial” wascoined by Zaks in [51]. On the other hand, length-factorial monoids were first investigated in 2011 bythe second and fourth authors [22]. As their main result, they proved that unique factorization domainscan be characterized as integral domains whose multiplicative monoids are length-factorial. Recently,length-factorial monoids have been classified in the class of torsion-free rank-1 monoids [40], in the classof submonoids of finite-rank free monoids [38], and in the class of monoids of the form N [ α ], where α is a positive algebraic numbers [16].Here we offer a deeper investigation of length-factoriality in atomic monoids and integral domains aswell as some connections between length-factoriality and the existence of certain extremal irreducibleelements, which when introduced were called purely long and purely short irreducibles [22]. We saythat a monoid satisfies the PLS property if it contains both purely short and purely long irreducibles.Every length-factorial monoid satisfies the PLS property, and here we determine classes of small-rankmonoids where every monoid satisfying the PLS property is length-factorial. We will also establish thatthe multiplicative monoid of an atomic domain never satisfies the PLS property. As a result, we will Date : January 15, 2021.2010
Mathematics Subject Classification.
Primary: 13F15, 13A05; Secondary: 20M13, 13F05.
Key words and phrases. length-factoriality, factorization, unique factorization, other-half-factoriality, finite-rankmonoid, Dedekind domain. S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH rediscover that the multiplicative monoid of an integral domain is length-factorial if and only if theintegral domain is a unique factorization domain, which was the main result in [22].In Section 3, which is the first section of content, we offer two characterizations of length-factorialmonoids. The first of such characterizations is given in terms of the integral independence of the set ofirreducibles and the set of irreducibles somehow shifted. The second characterization states that a non-factorial monoid is length-factorial if and only if the kernel congruence of its factorization homomorphismis nontrivial and can be generated by a single factorization relation. This second characterization willallow us to recover [22, Proposition 2.9]. In addition, we use the second characterization to determinethe set of Betti elements and study the catenary degree of a length-factorial monoid.In Section 4, we delve into the study of purely long and purely short irreducibles. For an element x of a monoid M , a pair of factorizations ( z , z ) of x is called irredundant if they have no irreducibles incommon and is called unbalanced if | z | 6 = | z | . An irreducible a of M is called purely long (resp., purelyshort) provided that for any pair of irredundant and unbalanced factorizations of the same element, thelonger (resp., shorter) factorization contains a . We prove that the set of purely long (and purely short)irreducibles of an atomic monoid is finite, and we use this result to decompose any atomic monoid as adirect sum of a half-factorial monoid and a length-factorial monoid.Section 5 is devoted to the study of length-factoriality in connection with the PLS property on theclass consisting of finite-rank atomic monoids. Observe that this class comprises all finitely generatedmonoids, all additive submonoids of Z n , and a large class of Krull monoids. We start by counting thenumber of non-associated irreducibles of a finite-rank length-factorial monoid. Then we show that formonoids of rank at most 2, being a length-factorial monoid is equivalent to satisfying the PLS property.We conclude the section by offering further characterizations of length-factoriality for rank-1 atomicmonoids.In Section 6, we investigate the existence of purely long and purely short irreducibles in the setting ofintegral domains, arriving to the surprising fact that an integral domain cannot simultaneously containa purely long irreducible and a purely short irreducible. As a consequence of this fact, we rediscover themain result of [22], that the multiplicative monoid of an integral domain is length-factorial if and onlyif the integral domain is a unique factorization domain (a shorter proof of this result was later givenin [1, Theorem 2.3]). We also exhibit examples of Dedekind domains containing purely long (resp.,purely short) irreducibles, but not purely short (resp., purely long) irreducibles.2. Fundamentals
General Notation.
Throughout this paper, we let N denote the set of positive integers, and weset N := N ∪ { } . For a, b ∈ Z with a ≤ b , we let J a, b K be the discrete interval from a to b , that is, J a, b K = { n ∈ Z : a ≤ n ≤ b } . In addition, for S ⊆ R and r ∈ R , we set S ≤ r := { s ∈ S : s ≤ r } and,with similar meaning, we use the symbols S ≥ r , S
We tacitly assume that each monoid (i.e., a semigroup with an identityelement) we treat here is cancellative and commutative. As all monoids we shall be dealing with arecommutative, we will use additive notation unless otherwise specified. For the rest of this section, let M be a monoid. We let M • denote the set M \{ } , and we let U ( M ) denote the group consisting of allthe units (i.e., invertible elements) of M . We say that M is reduced if U ( M ) = { } .For the monoid M there exist an abelian group gp( M ) and a monoid homomorphism ι : M → gp( M )such that any monoid homomorphism M → G , where G is an abelian group, uniquely factors through ι . ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS M ), which is unique up to isomorphism, is called the Grothendieck group of M . Themonoid M is torsion-free if nx = ny for some n ∈ N and x, y ∈ M implies that x = y . A monoid istorsion-free if and only if its Grothendieck group is torsion-free (see [8, Section 2.A]). If M is torsion-free,then the rank of M , denoted by rank( M ), is the rank of the Z -module gp( M ), that is, the dimension ofthe Q -vector space Q ⊗ Z gp( M ).An equivalence relation ρ on M is called a congruence provided that it is compatible with the operationof M , that is, for all x, y, z ∈ M the inclusion ( y, z ) ∈ ρ implies that ( x + y, x + z ) ∈ ρ . The elementsof a congruence are called relations . Let ρ be a congruence. Clearly, the set M/ρ of congruence classes(i.e., the equivalence classes) naturally turns into a commutative semigroup with identity (it may notbe cancellative). The subset { ( x, x ) : x ∈ M } of M × M is the smallest congruence of M , and is calledthe trivial (or diagonal ) congruence. Every relation in the trivial congruence is called diagonal , while(0 ,
0) is called the trivial relation. We say that σ ⊆ M × M generates the congruence ρ provided that ρ is the smallest (under inclusion) congruence on M containing σ . A congruence on M is cyclic if it canbe generated by one element.For x, y ∈ M , we say that y divides x in M and write y | M x provided that x = y + y ′ for some y ′ ∈ M . If x | M y and y | M x , then x and y are said to be associated elements (or associates ) and, in thiscase, we write x ∼ y . Being associates determines a congruence on M , and M red := M/ ∼ is called the reduced monoid of M . When M is reduced, we identify M red with M . For S ⊆ M , we let h S i denote thesmallest (under inclusion) submonoid of M containing S , and we say that S generates M if M = h S i .An element a ∈ M \ U ( M ) is an irreducible (or an atom ) if for each pair of elements u, v ∈ M suchthat a = u + v either u ∈ U ( M ) or v ∈ U ( M ). We let A ( M ) denote the set of irreducibles of M . Themonoid M is called atomic if every element in M \ U ( M ) can be written as a sum of atoms. Clearly, M is atomic if and only if M red is atomic. Each finitely generated monoid is atomic [30, Proposition 2.7.8].2.3. Factorizations.
The free commutative monoid on the set A ( M red ) is denoted by Z ( M ), and theelements of Z ( M ) are called factorizations . If z ∈ Z ( M ) consists of ℓ irreducibles of M red (counting rep-etitions), then we call ℓ the length of z and write | z | := ℓ . We say that a ∈ A ( M ) appears in z providedthat a + U ( M ) is one of the ℓ irreducibles of z . The unique monoid homomorphism π M : Z ( M ) → M red satisfying π ( a ) = a for all a ∈ A ( M red ) is called the factorization homomorphism of M . When thereseems to be no risk of ambiguity, we write π instead of π M . The kernelker π := { ( z, z ′ ) ∈ Z ( M ) : π ( z ) = π ( z ′ ) } of π is a congruence on Z ( M ), which we call the factorization congruence of M . In addition, we callan element ( z, z ′ ) ∈ ker π a factorization relation . Let ( z, z ′ ) be a factorization relation of M . We saythat a ∈ A ( M ) appears in ( z, z ′ ) if a appears in either z or z ′ . We call ( z, z ′ ) balanced if | z | = | z ′ | and unbalanced otherwise. Also, we say that ( z, z ′ ) is irredundant provided that no irreducible of M appearsin both z and z ′ . For each x ∈ M we set Z ( x ) := Z M ( x ) := π − ( x + U ( M )) ⊆ Z ( M ) . Observe that Z ( u ) = { } if and only if u ∈ U ( M ). In addition, note that M is atomic if and only if π is surjective, that is Z ( x ) = ∅ for all x ∈ M . For each x ∈ M , we set L ( x ) := L M ( x ) := {| z | : z ∈ Z ( x ) } ⊂ N . The monoid M is called a factorial monoid (or a unique factorization monoid ) if | Z ( x ) | = 1 for all x ∈ M . On the other hand, M is called a half-factorial monoid if | L ( x ) | = 1 for all x ∈ M . Let R be anintegral domain (i.e., a commutative ring with identity and without nonzero zero-divisors). We let R • denote the multiplicative monoid R \ { } and, to simplify notation, we write π R and Z ( R ) instead of π R • The Grothendieck group of a monoid is often called the difference or the quotient group depending on whether themonoid is written additively or multiplicatively.
S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH and Z ( R • ), respectively. In addition, for each x ∈ R • , we set Z R ( x ) := Z R • ( x ) and L R ( x ) := L R • ( x ). Itis clear that R is atomic (resp., a unique factorization domain) if and only if the monoid R • is atomic(resp., factorial). We say that R is a half-factorial domain provided that R • is a half-factorial monoid.See [10] for a survey on half-factorial domains.The notion of a half-factorial monoid is therefore obtained from that of a factorial monoid by keepingthe existence and weakening the uniqueness of factorizations, i.e., replacing | Z ( x ) | = 1 by | L ( x ) | = 1for every x ∈ M . In [22] the second and fourth authors proposed a dual way to weaken the uniquefactorization property and obtain a natural relaxed version of a factorial monoid, which they called alength-factorial monoid. Definition 2.1.
Let M be an atomic monoid. We say that M is length-factorial if for all x ∈ M and z , z ∈ Z ( x ) the equality | z | = | z | implies that z = z .Before proceeding, we make the following observation. Remark 2.2.
The term “length-factorial” seems like a natural choice as for every element x of alength-factorial monoid M and every ℓ ∈ L ( x ) there is a unique factorization in Z ( x ) of length ℓ . Weemphasize, however, that the monoids we study here under the term “length-factorial monoids” werefirst investigated in [22] under the term “other-half-factorial monoids”; observe that the later termhighlights the contrast with the half-factorial property.Notice that a monoid is length-factorial if and only if its reduced monoid is length-factorial. It is clearthat every factorial monoid is a length-factorial monoid. We say that a length-factorial monoid is proper if it is not factorial. The study of length-factoriality will be our primary focus of attention here. It hasbeen proved in [22] that the multiplicative monoid of an integral domain is a length-factorial monoidif and only if the integral domain is a unique factorization domain, i.e., the multiplicative monoid ofan integral domain cannot be a proper length-factorial monoid. We will obtain this result, along withseveral additional fundamental results, as a consequence of our investigation.3. Characterizations of Length-factorial Monoids
The main purpose of this section is to provide characterizations of a proper length-factorial monoid interms of the integral dependence of its set of irreducibles and also in terms of its factorization congruence.We will use the established characterizations to describe the set of Betti elements and study the catenarydegree of a given length-factorial monoid. Throughout this section, we assume that M is an atomicmonoid.3.1. Characterizations of a Length-factorial Monoid.
The notion of integral independence playsa central role in our first characterization of a length-factorial monoid. Let S be a subset of M . We saythat S is integrally independent in M if S is linearly independent as a subset of the Z -module gp( M ),that is, for any distinct s , . . . , s n ∈ S and any c , . . . , c n ∈ Z the equality P ni =1 c i s i = 0 in gp( M ) impliesthat c i = 0 for every i ∈ J , n K . We proceed to establish two characterizations of proper length-factorialmonoids. Theorem 3.1.
Let M be an atomic monoid that is not a factorial monoid. Then the following state-ments are equivalent.(a) The monoid M is a length-factorial monoid.(b) There exists a ∈ A ( M red ) such that A ( M red ) \ { a } and a − A ( M red ) \ { a } are integrallyindependent sets in gp( M red ) .(c) The congruence ker π is nontrivial and cyclic. ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS Proof.
Since M is a length-factorial monoid if and only if M red is a length-factorial monoid and sincethe factorization homomorphisms of both M and M red are the same, there is no loss in assuming that M is a reduced monoid. Accordingly, we identify M red with M .(a) ⇒ (b): Assume that M is a length-factorial monoid. Observe that the set A ( M ) cannot beintegrally independent as, otherwise, M would be a factorial monoid. Then there exist a ∈ A ( M ) and m ∈ N such that(3.1) ma = k X i =1 m i a i for some a , . . . , a k ∈ A ( M ) \ { a } and m , . . . , m k ∈ Z . Let us verify that A ( M ) \ { a } is an integrallyindependent set in gp( M ). Suppose, for the sake of a contradiction, that this is not the case. Thenthere exist b ∈ A ( M ) \ { a } and n ∈ N satisfying(3.2) nb = ℓ X i =1 n i b i for some b , . . . , b ℓ ∈ A ( M ) \ { a, b } and n , . . . , n ℓ ∈ Z . Take c i = ( | m i | − m i ) and c ′ i = ( | m i | + m i )for every i ∈ J , k K , and also take d i = ( | n i | − n i ) and d ′ i = ( | n i | + n i ) for every i ∈ J , ℓ K . Then set z := ma + k X i =1 c i a i , z := k X i =1 c ′ i a i , w := nb + ℓ X j =1 d j b j , and w := ℓ X j =1 d ′ j b j . It follows from (3.1) and (3.2) that both ( z , z ) and ( w , w ) are irredundant factorization relations of M .Because ( z , z ) and ( w , w ) are irredundant and nontrivial, the length-factoriality of M guaranteesthat they are both unbalanced. Assume, without loss of generality, that | z | > | z | and | w | < | w | .Clearly, (( | w | − | w | ) z , ( | w | − | w | ) z ) and (( | z | − | z | ) w , ( | z | − | z | ) w ) are both factorizationrelations of M . By adding them, one can produce a new balanced factorization relation with exactlyone of its two factorization components involving the irreducible a . However, this contradicts that M is a length-factorial monoid. Thus, A ( M ) \ { a } is integrally independent in gp( M ).Let a ∈ A ( M ) be as in the previous paragraph. We proceed to argue that the set a − A ( M ) \ { a } isalso integrally independent in gp( M ). Take this time b , . . . , b ℓ ∈ A ( M ) \ { a } and n , . . . , n ℓ ∈ Z suchthat P ℓi =1 n i ( b i − a ) = 0. Then set d i = ( | n i | − n i ) and d ′ i = ( | n i | + n i ) for every i ∈ J , ℓ K , andconsider the factorizations z := ℓ X i =1 d i b i + (cid:18) ℓ X i =1 d ′ i (cid:19) a and z := ℓ X i =1 d ′ i b i + (cid:18) ℓ X i =1 d i (cid:19) a. The equality P ℓi =1 n i b i = (cid:0) P ℓi =1 n i (cid:1) a ensures that ( z , z ) is a balanced factorization relation. Since M is a length-factorial monoid, z = z and therefore n i = d i − d ′ i = 0 for every i ∈ J , ℓ K . As a consequence,we can conclude that a − A ( M ) \ { a } is an integrally independent set in gp( M ).(b) ⇒ (c): Suppose that there exists a ∈ A ( M ) such that both A ( M ) \ { a } and a − A ( M ) \ { a } areintegrally independent sets in gp( M ). Let S be the subgroup of gp( M ) generated by A ( M ) \ { a } . Wehave seen before that A ( M ) is an integrally dependent set. As a result, the annihilator Ann( a + S ) of a + S in the Z -module gp( M ) /S is not trivial. Since Ann( a + S ) is an additive subgroup of Z , there exists m ∈ N such that Ann( a + S ) = m Z . Then there is an irredundant factorization relation ( w , w ) ∈ ker π such that exactly m copies of a appear in w and no copies of a appear in w .Let us verify that ( w , w ) is unbalanced. Suppose, by way of contradiction, that | w | = | w | . Notethat π ( w ) − π ( w ) = 0 in gp( M ) ensures the existence of a , . . . , a k ∈ A ( M ) (with a = a ) and S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH m , . . . , m k ∈ Z (with m = m ) such that P ki =0 m i a i = 0. As | w | = | w | , the equality P ki =0 m i = 0holds. As a consequence, one finds that k X i =1 m i ( a − a i ) = a k X i =0 m i − k X i =0 m i a i = 0 . This, along with the fact that m i = 0 for some i ∈ J , k K , contradicts that a − A ( M ) \{ a } is an integrallyindependent set. Hence | w | 6 = | w | , and so ( w , w ) is unbalanced.We still need to show that ( w , w ) generates the congruence ker π . Towards this end, take a nontrivialirredundant factorization relation ( z , z ) ∈ ker π . As A ( M ) \ { a } is integrally independent, a mustappear in ( z , z ). Assume, without loss of generality, that exactly n copies of a appear in z for some n ∈ N . Then the equality π ( z ) = π ( z ) ensures that n ∈ Ann( a + S ), and so n = km for some k ∈ N .Then after canceling na in both sides of π ( w k z ) = π ( w k z ), we obtain two integral combinations ofirreducibles in A ( M ) \ { a } , whose corresponding coefficients must be equal. Thus, ( z , z ) = ( w , w ) k .(c) ⇒ (a): Suppose that ker π is a cyclic congruence generated by an unbalanced irredundant factor-ization relation ( w , w ). Let ∗ denote the monoid operation of the congruence ker π . Take ( z, z ′ ) ∈ ker π such that z = z ′ . Since ( w , w ) generates ker π , there exist n ∈ N and z , . . . , z n ∈ Z ( M ) with z = z and z n = z ′ such that for every i ∈ J , n K the equality(3.3) ( z i − , z i ) = ( w , w ) ∗ ( d i , d i )holds for some d i ∈ Z ( M ). After multiplying all the identities in (3.3) (for every i ∈ J , n K ), one findsthat ( z, z ′ ) ∗ ( z · · · z n − , z · · · z n − ) = ( w n , w n ) ∗ ( d, d ), where d = d · · · d n . Since z · · · z n − dividesboth w n d and w n d in the free monoid Z ( M ) and gcd( w n , w n ) = 1, there exists z ′′ ∈ Z ( M ) such that z · · · z n − z ′′ = d . As a result, ( z, z ′ ) = ( z ′′ w n , z ′′ w n ) and so ( z, z ′ ) is an unbalanced factorizationrelation. Hence M is a length-factorial monoid. (cid:3) Following [22], we call a factorization relation ( w , w ) in ker π M master if any irredundant andunbalanced factorization relation of M has the form ( w n , w n ) or ( w n , w n ) for some n ∈ N . A masterfactorization relation must be irredundant and unbalanced unless M is a half-factorial monoid. When M is a proper length-factorial monoid we have seen that ker π is a nontrivial cyclic congruence, and it isclear that ( w , w ) is a generator of ker π if and only if ( w , w ) is a master factorization relation, inwhich case, the only master factorization relations of M are ( w , w ) and ( w , w ). In this case, one canreadily verify that if | w | < | w | , then | w | < | z | < | w | for each factorization z ∈ Z ( π ( w )) \ { w , w } .As a consequence of Theorem 3.1, we obtain the following corollary, which was first established in theproof of the main theorem of [22]. Corollary 3.2.
Let M be an atomic monoid. Then M is a proper length-factorial monoid if andonly if it admits an unbalanced master factorization relation ( w , w ) , in which case the only masterfactorization relations of M are ( w , w ) and ( w , w ) . The numerical monoids that are proper length-factorial monoids have been characterized in [22] asthose having precisely two irreducibles. This was generalized in [40, Proposition 4.3], which states thatthe additive submonoids of Q ≥ that are length-factorial monoids are those generated by two elements.In general, every monoid that can be generated by two elements is a length-factorial monoid. Corollary 3.3.
Let M be a monoid generated by two elements. Then ker π is cyclic, and M is alength-factorial monoid.Proof. As M is finitely generated, it is atomic. We can assume, without loss of generality, that M isreduced. If M is a factorial monoid, then there is nothing to show. Therefore assume that M is not afactorial monoid. Then there exists a generating set A of M with | A | = 2. Because M is not a factorial ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS A ( M ) = A . As both sets A \ { a } and a − A \ { a } are singletons, the corollary follows fromTheorem 3.1. (cid:3) When a monoid cannot be generated by two elements, its factorization congruence may not be cyclic(even if the monoid is finitely generated). The next example illustrates this observation.
Example 3.4.
For n ∈ N ≥ , consider the additive submonoid M = { } ∪ N ≥ n of N . It can be readilyverified that M is atomic and A ( M ) = J n, n − K . Since 2( n + 1) = n + ( n + 2), it follows that M isnot a length-factorial monoid. Then Theorem 3.1 guarantees that the factorization congruence of M isnot cyclic.3.2. Connection with the Catenary Degree.
We call a finite sequence z , z , . . . , z k of factorizationsin Z ( M ) a chain of factorizations from z to z k if π ( z ) = π ( z ) = · · · = π ( z k ), where π is the factoriza-tion homomorphism of M . Consider the subset R of Z ( M ) defined as follows: a pair ( z, z ′ ) ∈ Z ( M ) belongs to R if there exists a chain of factorizations z , z , . . . , z k from z to z ′ such that gcd( z i − , z i ) = 1for every i ∈ J , k K , where gcd( z i − , z i ) denotes the greatest common divisor of z i − and z i as elements ofthe free commutative monoid Z ( M ). It follows immediately that R is an equivalence relation on Z ( M )that refines ker π . For each x ∈ M , we let R x denote the set of equivalence classes of R inside Z ( x ). Anelement b ∈ M is called a Betti element provided that | R x | ≥
2. Let Betti( M ) denote the set of Bettielements of M . As we proceed to show, every proper length-factorial monoid contains essentially oneBetti element. Proposition 3.5. If M is a proper length-factorial monoid, then | Betti( M red ) | = 1 .Proof. Since M is a proper length-factorial monoid, Corollary 3.2 ensures the existence of a masterfactorization relation ( w , w ). Assume that | w | < | w | . We claim that b = π ( w ) is a Betti element.To see this, take w ′ ∈ Z ( b ) with w ′ = w . As w is the minimum-length factorization of the masterrelation ( w , w ), it follows that | w | < | w ′ | . Therefore ( w , w ′ ) = ( ww n , ww n ) for some w ∈ Z ( M ) and n ∈ N , which implies that w = 1 and n = 1, that is, w ′ = w . As a result, Z ( b ) = { w , w } . This, alongwith the fact that ( w , w ) is irredundant, guarantees that | R b | = 2. Hence b ∈ Betti( M red ).Now take x ∈ M red such that x = b , and let us verify that x cannot be a Betti element of M red . If | Z ( x ) | = 1, then | R x | = 1, and so x / ∈ Betti( M red ). Assume, therefore, that | Z ( x ) | ≥
2. Take z, z ′ ∈ Z ( x )with z = z ′ and suppose, without loss of generality, that | z | < | z ′ | . Then ( z, z ′ ) = ( ww n , ww n ) forsome w ∈ Z ( M ) and n ∈ N . If w = 1, then z, z ′ is a chain of factorizations from z to z ′ such thatgcd( z, z ′ ) = 1. Otherwise, the fact that x = b ensures that n ≥
2, and after taking z i = w n − i w i for each i ∈ J , n K , one can readily see that z , z , . . . , z n is a chain of factorizations from z to z ′ satisfying thatgcd( z i − , z i ) = 1 for every i ∈ J , n K . Hence | R x | = 1, and so x / ∈ Betti( M red ). (cid:3) We will conclude this section studying the (monotone, equal) catenary degree of a length-factorialmonoid; we express the (monotone) catenary degree in terms of any of the master factorization relations.The distance d ( z, z ′ ) between two factorizations z and z ′ in Z ( M ) is defined as follows: d ( z, z ′ ) := max (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) z gcd( z, z ′ ) (cid:12)(cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12)(cid:12) z ′ gcd( z, z ′ ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:27) . It is routine to verify that d is indeed a distance function. For N ∈ N , a chain of factorizations z , z , . . . , z k is called an N - chain from z to z k if d ( z i − , z i ) ≤ N for every i ∈ J , k K . For x ∈ M , we let c ( x ) denote the smallest N ∈ N such that for every z, z ′ ∈ Z ( x ) there exists an N -chain of factorizationsfrom z to z ′ ; when such an N does not exist, we set c ( x ) = ∞ . The catenary degree of M , denoted by c ( M ), is defined by c ( M ) := sup { c ( x ) : x ∈ M } . S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH
The notion of catenary degree was introduced by Geroldinger in [27] in the context of Noetheriandomains, although the term was coined later in [28]. Since then, several variations of the catenarydegree have been investigated.An N -chain z , z , . . . , z k of factorizations in Z ( M ) is said to be monotone if | z | ≤ | z | ≤ · · · ≤ | z k | or | z | ≥ | z | ≥ · · · ≥ | z k | . For x ∈ M , we let c mon ( x ) (resp., c eq ( x )) denote the smallest N ∈ N such that for every z, z ′ ∈ Z ( x ) (resp., z, z ′ ∈ Z ( x ) with | z | = | z ′ | ) there exists a monotone N -chain offactorizations from z to z ′ ; if such an N does not exist, then we set c mon ( x ) = ∞ (resp., c eq ( x ) = ∞ ).In addition, we set c mon ( M ) := sup { c mon ( x ) : x ∈ M } and c eq ( M ) := sup { c eq ( x ) : x ∈ M } , and call them the monotone catenary degree and the equal catenary degree of M , respectively. It isclear from the definition that c ( x ) ≤ c mon ( x ) and c eq ( x ) ≤ c mon ( x ) for all x ∈ M and, therefore, c ( M ) ≤ c mon ( M ) and c eq ( M ) ≤ c mon ( M ). For every ℓ ∈ N and x ∈ M , set Z ℓ ( x ) := { z ∈ Z ( x ) : | z | = ℓ } and define c adj ( x ) as follows: c adj ( x ) := sup (cid:8) d ( Z k ( x ) , Z ℓ ( x )) : k, ℓ ∈ L ( x ) , k < ℓ, and J k, ℓ K ∩ L ( x ) = { k, ℓ } (cid:9) , where d ( Z , Z ) = min { d ( z , z ) : z ∈ Z and z ∈ Z } for any nonempty subsets Z and Z of Z ( M ).The adjacent catenary degree of M , denoted by c adj ( M ), is then defined as c adj ( M ) := sup { c adj ( x ) : x ∈ M } . It is clear that c mon ( x ) = max { c eq ( x ) , c adj ( x ) } for all x ∈ M , and so c mon ( M ) = max { c eq ( M ) , c adj ( M ) } .The notion of monotone catenary degree was introduced by Foroutan in [24], and it has been fairlystudied in past literature (see [33] and references therein). In [45, Section 3], Philipp provides charac-terizations of the monotone, equal, and adjacent catenary degrees of M in terms of the factorizationcongruence ker π . Proposition 3.6.
Let M be a monoid, and let ( w , w ) be a master factorization relation of M . Thenthe following statements hold.(1) The monoid M is length-factorial if and only if c eq ( M ) = 0 .(2) If M is a proper length-factorial monoid, then c adj ( M ) = c mon ( M ) = c ( M ) = max {| w | , | w |} . Proof. (1) For the direct implication, assume that M is a length-factorial monoid. Since M is length-factorial, for every x ∈ M two factorizations in Z ( x ) have the same length if and only if they are equal,which immediately implies that c eq ( x ) = 0. Hence c eq ( M ) = 0. Conversely, suppose that c eq ( M ) = 0.Take x ∈ M , and let z and z ′ be two factorizations of x such that | z | = | z ′ | . Since c eq ( x ) ≤ c eq ( M ) = 0,it follows that d ( z, z ′ ) = 0, and so z = z ′ . Thus, distinct factorizations of x must have different lengths.Hence M is a length-factorial monoid.(2) Now suppose that M is a proper length-factorial monoid. In order to find the catenary degreeof M , it suffices to look at the set Betti( M ): indeed, it follows from [44, Corollary 9] that c ( M ) = sup { µ ( b ) : b ∈ Betti( M ) } , where µ ( x ) = sup { min z ∈ ρ | z | : ρ ∈ R x } . By Proposition 3.5, the monoid M contains only one Bettielement b up to associate, and we have seen that R b consists of two classes, namely, { w } and { w } .Thus, c ( M ) = µ ( b ) = max {| w | , | w |} .Since c eq ( M ) = 0, the equality c adj ( M ) = c mon ( M ) holds. Finally, let us argue that c mon ( M ) = c ( M ).If b ∈ Betti( M ), then Z ( b ) = { w , w } , as we have seen in the proof of Proposition 3.5. Clearly, w , w is a monotone N -chain of factorizations from w to w , where N = max {| w | , | w |} . Thus, c mon ( b ) ≤ max {| w | , | w |} = c ( M ). Now suppose that x ∈ M is not a Betti element. If | Z ( x ) | = 1,then c mon ( x ) = 0 ≤ c ( M ). Suppose, otherwise, that | Z ( x ) | > z, z ′ ∈ Z ( x ) such that z = z ′ . ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS M is a length-factorial monoid, we can assume that | z | < | z ′ | , so ( z, z ′ ) = ( ww n , ww n ) for some w ∈ Z ( M ) and n ∈ N . In this case, we can take z i := ww n − i w i for each i ∈ J , n K to obtain an N -chainof factorizations from z to z ′ , where N = max {| w | , | w |} = c ( M ). This implies that c mon ( x ) ≤ c ( M ).Hence c mon ( M ) ≤ c ( M ) and, therefore, the equality must hold. (cid:3) Pure Irreducibles: The PLS Property
In this section, we study the notions of purely long and purely short irreducibles (as introduced in [22])in connection with length-factoriality. Based on these notions of irreducible elements, we introduce aclass of atomic monoids that strictly contains that of length-factorial monoids. We will see that eachmonoid in this new class naturally decomposes as a sum of a half-factorial monoid and a length-factorialmonoid. Throughout this section, we let M be an atomic monoid.4.1. Pure Irreducibles.
Let ( z , z ) be an unbalanced factorization relation of M . Then we call thefactorization of bigger (resp., smaller) length between z and z the longer (resp., shorter ) factorizationof ( z , z ). Definition 4.1.
Let M be a monoid, and take a ∈ A ( M red ). We say that a is purely long (resp., purelyshort ) if a is not prime and for all irredundant and unbalanced factorization relations ( z , z ) of M , thefact that a appears in z implies that | z | > | z | (resp., | z | < | z | ). Remark 4.2.
As by definition a purely long (or short) irreducible is not prime, it must appear in atleast one nontrivial irredundant factorization relation of M .We let L ( M ) (resp., S ( M )) denote the set comprising all purely long (resp., purely short) irre-ducibles of M red . When M is a proper length-factorial monoid, it follows from Corollary 3.2 that both L ( M ) and S ( M ) are nonempty sets. More precisely, if z , z ∈ Z ( M ) satisfy | z | < | z | and ( z , z )is an irredundant factorization relation generating the factorization congruence of a length-factorialmonoid M , then L ( M ) (resp., S ( M )) consists of all irreducibles that appear in z (resp., z ).We call any element of L ( M ) ∪ S ( M ) a pure irreducible. As a consequence of the following propo-sition we will obtain that every atomic monoid contains only finitely many pure irreducibles. Proposition 4.3.
For an atomic monoid M , let a be a purely short/long irreducible, and let ( w , w ) bean irredundant factorization relation. Then a appears in ( w , w ) if and only if ( w , w ) is unbalanced.Proof. To argue the direct implication suppose, by way of contradiction, that ( w , w ) is balanced. Wealso assume, without loss of generality, that a appears in w . Suppose first that a ∈ L ( M ), and takean irredundant factorization relation ( z , z ) such that | z | > | z | and a appears in z . Then we cantake n ∈ N large enough such that the number of copies of a that appear in w n z is strictly smallerthan the number of copies of a that appear in w n z . Therefore ( w n z , w n z ) yields, after cancellations,an irredundant and unbalanced factorization relation whose shorter factorization involves a . However,this contradicts that a is purely long. Supposing that a ∈ S ( M ), one can similarly arrive to anothercontradiction.For the reverse implication, assume that ( w , w ) is unbalanced with | w | < | w | . Suppose first that a ∈ L ( M ). Take an irredundant factorization relation ( z , z ) such that a appears in ( z , z ). Thereis no loss in assuming that a appears in z and, therefore, that | z | > | z | . Then there exists n ∈ N such that | w n z | < | w n z | . Since a appears in the shorter factorization of ( w n z , w n z ), the fact that a is a purely long irreducible guarantees that a also appears in the longer factorization of ( w n z , w n z ).Hence a appears in w . For a ∈ S ( M ) the proof is similar. (cid:3) Corollary 4.4.
For an atomic monoid M , both sets L ( M ) and S ( M ) are finite. S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH
Proof. If M is a half-factorial monoid, then both sets L ( M ) and S ( M ) are empty. Otherwise, theremust exist an unbalanced factorization relation ( z , z ). It follows now from Proposition 4.3 that everypure irreducible of M appears in ( z , z ). Hence both sets L ( M ) and S ( M ) must be finite. (cid:3) Clearly, atomic monoids having both purely long and purely short irreducibles are natural gener-alizations of length-factorial monoids, and they will play an important role in the remainder of thispaper.
Definition 4.5.
If an atomic monoid M contains both purely long and purely short irreducibles, thenwe say that M has the PLS property or that M is a PLS monoid .For future reference, we highlight the following immediate corollary of Theorem 3.1.
Corollary 4.6.
Every proper length-factorial monoid is a PLS monoid.
The converse of Corollary 4.6 does not hold even for finitely generated monoids. For any subset S of R d , we let cone ( S ) and aff ( S ) denote the cone and the affine space generated by S , respectively. Example 4.7.
For a = (0 , , a = (0 , , a = (1 , , a = (2 , , a = (3 , , M = h a i : i ∈ J , K i of ( N , +). Clearly, M is atomic and it is not hard to check that A ( M ) = { a i : i ∈ J , K } . Let H be the hyperplane described by the equation y = 2. Since a / ∈ H and a i ∈ H for every i ∈ J , K , the irreducible a is purely long. Because cone ( a , a ) and aff ( a , a , a ) onlyintersect in the origin, a and a cannot be in the same part of any irredundant factorization relationof M . Thus, if a appears in an irredundant factorization relation involving a , then it must appear inits shorter part. In addition, note that because a / ∈ aff ( a , a , a ), there is no irredundant factorizationrelation of M involving a but not a . Hence a ∈ L ( M ), and so M is a PLS monoid. However, itfollows from [38, Section 5] that M is not a length-factorial monoid.None of the conditions L ( M ) = ∅ and S ( M ) = ∅ implies the other one. The following examplesheds some light upon this observation. Example 4.8.
For the set A = { (0 , , (1 , , (2 , , (3 , } , consider the submonoid M of ( N , +) gen-erated by A . It is clear that M is atomic, and one can readily check that A ( M ) = A . Since all theirreducibles of M lie in the line determined by the equation x + y = 3, it follows from [38, Corollary 5.5]that M is a half-factorial monoid.Now consider the submonoid M of ( N , +) generated by the set A = A ∪ { (1 , } . It is easy to verifythat M is atomic with A ( M ) = A . Moreover, since the irreducibles of M are not colinear, it followsfrom [38, Corollary 5.5] that M is not a half-factorial monoid. Therefore there exists an irredundantfactorization relation ( z , z ) with | z | 6 = | z | . Since M is a half-factorial monoid, (1 ,
1) must appear in( z , z ); say that (1 ,
1) appears in z . After projecting on the line determined by the equation y = x ,one can easily see that | z | > | z | . As a result, (1 ,
1) is purely long. Note that the irreducibles in A are neither purely long nor purely short because they are precisely the irreducibles of M , which is ahalf-factorial monoid. Hence M contains a purely long irreducible but no purely short irreducibles.Lastly, considering the submonoid M of ( N , +) generated by the set A ∪ { (2 , } and proceedingas we did with M , one finds that (2 ,
2) is the only purely short irreducible in M , and also that M contains no purely long irreducibles.We know that half-factorial monoids contain neither purely long nor purely short irreducibles. How-ever, there are monoids that are not half-factorial and still contain neither purely long nor purely shortirreducibles. Example 4.9.
Let M and A be as in Example 4.8, and let M be the submonoid of ( N , +) generatedby the set A = A ∪ { (0 , , (1 , , (2 , } . It is not hard to verify that M is an atomic monoid with ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS A ( M ) = A . Since the equalities 2(1 ,
1) = (0 ,
2) + (2 ,
0) and (1 ,
2) + (2 ,
1) = (0 ,
3) + (3 ,
0) give riseto two irredundant and balanced factorizations involving each irreducible of M , the sets L ( M ) and S ( M ) must be empty. Because of this, M cannot be a length-factorial monoid, which is confirmed by[38, Theorem 5.10]. In addition, as the points in A are not colinear, it follows from [38, Corollary 5.5]that M is not a half-factorial monoid.4.2. Sum Decomposition of PLS Monoids.
We proceed to show how to decompose the reducedmonoid of a PLS monoid M as the inner sum of a half-factorial monoid M and a finitely generatedlength-factorial monoid M satisfying that M ∩ M = { } . We emphasize that such a decompositiondoes not guarantee the uniqueness of the representation of an element of M as a sum of an elementof M and an element of M . Theorem 4.10.
Let M be a PLS monoid. Then there exist submonoids H and O of M red satisfying M red = H + O , where H is a half-factorial monoid and O is a finitely generated proper length-factorialmonoid such that H ∩ O = { } .Proof. Let O be the submonoid of M red generated by the set L ( M ) ∪ S ( M ). It is clear that O is anatomic monoid with A ( O ) = L ( M ) ∪ S ( M ). Moreover, note that L ( O ) = L ( M ) and S ( O ) = S ( M ).By Corollary 4.4, the monoid O is finitely generated. To verify that O is a length-factorial monoid, let( z , z ) be a nontrivial irredundant factorization relation in ker π O . Since at least one irreducible in L ( M ) ∪ S ( M ) appears in the relation ( z , z ), the latter must be unbalanced by Proposition 4.3. Asa consequence, O is a proper length-factorial monoid.Now let H be the submonoid of M red generated by A ( M ) \ ( L ( M ) ∪ S ( M )). It follows immediatelythat H is atomic with A ( H ) = A ( M ) \ ( L ( M ) ∪ S ( M )). To see that H is a half-factorial monoid, itsuffices to observe that since ker π H ⊆ ker π M , any irredundant factorization relation of ker π H must bebalanced by Proposition 4.3.Because A ( M red ) = A ( H ) ∪ A ( O ), we find that M red = H + O . To argue that H and O have trivialintersection, suppose that x ∈ H ∩ O . As both H and O are atomic monoids, one can take z ∈ Z H ( x )and z ∈ Z O ( x ). Therefore ( z , z ) ∈ ker π M . Since L ( M ) = ∅ and S ( M ) = ∅ , if a pure irreducibleappeared in z , then a pure irreducible would appear in z . As z consists of non-pure irreducibles, z must be the factorization with no irreducibles, whence x = 0. As a result, H ∩ O = { } , which impliesthat M red = H ⊕ O . (cid:3) The converse of Theorem 4.10 does not hold in general, as the following example indicates.
Example 4.11.
Consider the additive submonoid M of ( N , +) generated by the set of lattice points { (1 , , (0 , , (1 , , (2 , , (3 , } . We have already seen in the second paragraph of Example 4.8 that L ( M ) = { (1 , } and S ( M ) = ∅ . Therefore M is not a PLS monoid. The submonoid H = h (1 , , (0 , i of M is clearly a factorial monoid and, in particular, a half-factorial monoid. On the other hand, one cansee that the submonoid O = h (1 , , (2 , , (3 , i of M is a proper length-factorial monoid by applyingTheorem 3.1 with a = (1 , M = H ⊕ O even though M is not a PLSmonoid.We conclude this section with the following proposition. Proposition 4.12.
Let M be a PLS monoid. Then there exists an unbalanced factorization relation ( w , w ) ∈ ker π such that every factorization relation of ker π has the form ( w n h , w n h ) for some n ∈ N and some balanced factorization relation ( h , h ) ∈ ker π .Proof. Take a ∈ L ( M ). Set A = A ( M ) \ { a } , and let S be the subgroup of gp( M ) generated by A .Since a appears in an irredundant and unbalanced factorization relation of M , there exists m ∈ N such that Ann( a + S ) = m Z , where Ann( a + S ) is the annihilator of a + S in the Z -module gp( M ) /S .2 S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH As ma ∈ S , there is an irredundant factorization relation ( w , w ) of M such that exactly m copiesof a appear in w . It follows from Proposition 4.3 that | w | > | w | . Suppose now that ( z , z ) isan irredundant factorization relation of M with | z | > | z | , and let k ∈ N be the number of copiesof a appearing in z . Notice that k ∈ Ann( a + S ), and therefore k = nm for some n ∈ N . Then( w n z , w n z ) ∈ ker π yields, after cancellations, a factorization relation that does not involve a . Thus,such a factorization must be balanced by Proposition 4.3 and cannot involve any pure irreducible. Sothe number of copies of each irreducible b in L ( M ) (resp., S ( M )) that appear in z (resp., z ) equals n times the number of copies of b that appear in w (resp., w ). Hence ( z , z ) = ( w n h , w n h ), where h , h ∈ Z ( M ) involve no pure irreducibles. Clearly, ( h , h ) ∈ ker π , and Proposition 4.3 guaranteesthat | h | = | h | . (cid:3) Finite-Rank Monoids
In this section, we continue studying the OHF and the PLS properties, but we restrict our attentionto the class of finite-rank monoids.5.1.
Number of Irreducibles. If M is a reduced finite-rank factorial monoid, then it follows from [30,Proposition 1.2.3(2)] that | A ( M ) | = rank ( M ). In parallel with this, the cardinality of A ( M ) in afinite-rank proper length-factorial monoid M can be determined. Proposition 5.1.
Let M be a proper length-factorial monoid whose rank is finite. Then the equality | A ( M red ) | = rank ( M ) + 1 holds.Proof. As gp( M red ) ∼ = gp( M ) / U ( M ), the monoid M red has finite rank. Hence one can replace M by M red and assume that M is reduced. Set r = rank ( M ) and then embed M into the Q -vector space V := Q ⊗ Z gp( M ) ∼ = Q r via M ֒ → gp( M ) → Q ⊗ Z gp( M ), where the injectivity of the second mapfollows from the flatness of the Z -module Q . So we can think of M as an additive submonoid of thefinite-dimensional vector space Q r . By Theorem 3.1, there exists a ∈ A ( M ) such that A ( M ) \ { a } and a − A ( M ) \ { a } are integrally independent sets in gp( M ). In particular, the sets A ( M ) \ { a } and a − A ( M ) \ { a } are linearly independent inside the vector space V . Because M is atomic, gp( M ) can begenerated by A ( M ) as a Z -module and, therefore, A ( M ) is a generating set of V . Since M is a properlength-factorial monoid, the monoid M is not a factorial monoid and, consequently, A ( M ) is a linearlydependent set of V . This along with the fact that A ( M ) \ { a } is linearly independent in V implies that A ( M ) \ { a } is a basis for V . Hence | A ( M ) | = | A ( M ) \ { a }| + 1 = r + 1. (cid:3) Corollary 5.2.
Every finite-rank length-factorial monoid is finitely generated.
The condition of having finite rank in Corollary 5.2 is not superfluous. For instance, consider theadditive monoid M = h , i ⊕ N ∞ , where N ∞ is the direct sum of countably many copies of N . Since h , i is a proper length-factorial monoid and N ∞ is a factorial monoid, M is a proper length-factorialmonoid. However, M is not finitely generated because rank ( M ) = ∞ . The converse of Proposition 5.1does not hold in general, as the following example shows. Example 5.3.
For every r ∈ N , consider the submonoid M r of ( N r , +) that is generated by the set S = { v , re j : j ∈ J , r K } , where v := { e + · · · + e r } . It is not hard to verify that A ( M r ) = S , and so | A ( M r ) | = r + 1. Notice that each point in S lies in the hyperplane of R r determined by the equation x + · · · + x r = r . Hence it follows from [38, Corollary 5.5] that M r is a proper half-factorial monoid.Therefore M r cannot be a length-factorial monoid. ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS
Monoids of Small Rank.
As we have emphasized in Corollary 4.6, every proper length-factorialmonoid is a PLS monoid. We proceed to show that being a length-factorial monoid is equivalent tobeing a PSLM in the class of torsion-free monoids with rank at most 2.
Theorem 5.4.
For a torsion-free monoid M with rank ( M ) ≤ , the following statements are equivalent.(a) The monoid M is a proper length-factorial monoid.(b) The monoid M is a PLS monoid.(c) The congruence ker π can be generated by an unbalanced factorization relation.Proof. (a) ⇔ (c): This is part of Theorem 3.1.(a) ⇒ (b): This is Corollary 4.6.(b) ⇒ (a): Assume that M is a PLS monoid, and suppose for the sake of a contradiction that M isnot a proper length-factorial monoid. Since M is finitely generated, it is atomic. As M is not a factorialmonoid, | A ( M ) | ≥
2. We split the rest of the proof into three cases.CASE 1: | A ( M ) | = 2. In this case, the factorization congruence ker π is cyclic by Corollary 3.3, andthe existence of purely long/short irreducibles implies that any generator of ker π must be unbalanced,contradicting that M is not a proper length-factorial monoid.CASE 2: | A ( M ) | = 3. Take a , a , a ∈ M such that A ( M ) = { a , a , a } . Assume, without lossof generality, that a ∈ L ( M ) and a ∈ S ( M ). Now take an irredundant and balanced factorizationrelation ( z , z ) ∈ ker π . Since a and a are pure irreducibles, none of them can appear in ( z , z ).Therefore only copies of the irreducible a appear in both z and z . This implies that z = z . As( z , z ) was taking to be irredundant, it must be trivial. Hence M is a proper length-factorial monoid,a contradiction.CASE 3: | A ( M ) | ≥
4. Take a ∈ L ( M ) and a ∈ S ( M ), and then take a , a ∈ A ( M ) \ { a , a } such that a = a . Since a is a purely long irreducible, the submonoid M ′ := h a , a , a i of M must bea half-factorial monoid. Now take a nontrivial factorization relation ( z , z ) ∈ ker π M ′ . As a is a purelyshort irreducible, it does not appear in ( z , z ). Therefore either ( z , z ) or ( z , z ) equals ( ma , ma )for some n ∈ N . Now the fact that M is torsion-free, along with the equality ma = ma , guaranteesthat a = a , which is a contradiction. (cid:3) Corollary 5.5.
If a torsion-free monoid M is generated by at most three elements, then it is a properlength-factorial monoid if and only if it is a PLS monoid.Proof. There is no loss in assuming that M is reduced. Clearly, | A ( M ) | ≤
3. Consider the Q -space V = Q ⊗ Z gp( M ), and identify M with its isomorphic copy inside V provided by the embedding M ֒ → gp( M ) → Q ⊗ Z gp( M ). As M is atomic, A ( M ) is a spanning set of V , whence dim V ≤
3. Ifdim V = 3, then A ( M ) is linearly independent over Q , in which case M is the free monoid on A ( M ).In this case, M is neither a proper length-factorial monoid nor a PLS monoid. On the other hand, ifdim V ≤
2, then rank ( M ) ≤ (cid:3) However, for a finitely generated monoid containing four or more irreducibles, the PLS property maynot imply the OHF property. This has been illustrated in Example 4.7. In the same example, we haveseen that the condition of having rank at most 2 is required in Theorem 5.4. On the other hand, thefollowing example indicates that the condition of being torsion-free is also required in the statement ofTheorem 5.4.
Example 5.6.
Fix n ∈ N such that n ≥
4, and consider the submonoid M := h a k : k ∈ J , n K i ofthe additive group Z n − × Z , where a = (0 , , a = (0 , , a k = ( k − , ,
0) for every k ∈ J , n K . Since M is finitely generated, it must be atomic. In addition, it can be readily verified that A ( M ) = { a k : k ∈ J , n K } . Now suppose that ( z , z ) is an irredundant and unbalanced factorization4 S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH relation in ker π , and assume that | z | < | z | . Since the second component of both a and a is 0and the second component of a , . . . , a n is 1, the numbers of irreducibles in { a , . . . , a n } that appearin z and in z must coincide. A similar observation based on third components shows that a appearsin z but not in z and also that a appears in z but not in z . Hence a ∈ L ( M ) and a ∈ S ( M ),which implies that M is a PLS monoid. Checking that M is not a length-factorial monoid amounts toobserving that the equality ( n − a = (0 , n − ,
0) = ( n − a yields an irredundant and balancednontrivial factorization relation of M .Now we turn to characterize the PSLMs in the class consisting of all torsion-free rank-1 monoids, whichhave been recently studied under the name Puiseux monoids . Puiseux monoids have been studied inconnection with commutative algebra [21], commutative factorization theory [12], and noncommutativefactorization theory [6]. An updated survey on the atomic structure of Puiseux monoids is given in [13].Notice that a Puiseux monoid is reduced unless it is a group (see [25, Section 24] and [36, Theorem2.9]).
Proposition 5.7.
Let M be an atomic Puiseux monoid. Then the following statements are equivalent.(a) The monoid M is a proper length-factorial monoid.(b) The monoid M is a PLS monoid.(c) Both inclusions inf A ( M ) ∈ L ( M ) and sup A ( M ) ∈ S ( M ) hold.(d) At least one of the inclusions inf A ( M ) ∈ L ( M ) or sup A ( M ) ∈ S ( M ) holds.(e) The equality | A ( M ) | = 2 holds.If any of the conditions above holds, then L ( M ) and S ( M ) are singletons: L ( M ) = { inf A ( M ) } and S ( M ) = { sup A ( M ) } .Proof. (a) ⇒ (b): This is Corollary 4.6.(b) ⇒ (c): Suppose that M is a PLS monoid, and take a ℓ ∈ L ( M ) and a s ∈ S ( M ). Now take a ∈ M such that a = a ℓ . Clearly, n := n ( a ) n ( a ℓ ) ∈ M and, moreover, z := n ( a ) d ( a ℓ ) a ℓ and z := n ( a ℓ ) d ( a ) a are two factorizations in Z ( n ). Since the factorization relation ( z , z ) is irredundant and a ℓ appearsin z , one finds that | z | > | z | . Therefore n ( a ) d ( a ℓ ) > n ( a ℓ ) d ( a ), which means that a > a ℓ . Then weconclude that inf A ( M ) = a ℓ ∈ L ( M ). The equality sup A ( M ) = a s can be argued similarly, fromwhich one obtains that sup A ( M ) ∈ S ( M ).(c) ⇒ (d): This is obvious.(d) ⇒ (e): Assume now that inf A ( M ) ∈ L ( M ), and take a ℓ ∈ L ( M ). Since M is an atomic Puiseuxmonoid that is not a factorial monoid, it follows that | A ( M ) | ≥
2. Suppose, by way of contradiction,that | A ( M ) | ≥
3, and take irreducibles a , a ∈ A ( M ) \ { a ℓ } such that a = a . Consider the element n := n ( a ) n ( a ) ∈ M . It is clear that both z := n ( a ) d ( a ) a and z := n ( a ) d ( a ) a are factorizationsin Z ( n ), and they have different lengths because a = a . However, the fact that a ℓ does not appearin either z or z contradicts that a ℓ ∈ L ( M ). As a result, | A ( M ) | = 2. One can similarly obtain | A ( M ) | = 2 assuming that sup A ( M ) ∈ S ( M ).(e) ⇒ (a): If | A ( M ) | = 2, it follows from Corollary 3.3 that M is a length-factorial monoid. Taking a and a to be the two irreducibles of M , one finds that n ( a ) d ( a ) a and n ( a ) d ( a ) a are two differentfactorizations of n ( a ) n ( a ) ∈ M , and so M is not a factorial monoid. Hence M must be a properlength-factorial monoid. (cid:3) Corollary 5.8.
Let N be a numerical monoid. Then L ( N ) ∪ S ( N ) is nonempty if and only if | A ( N ) | = 2 , in which case L ( N ) = { min A ( N ) } and S ( N ) = { max A ( N ) } . ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS
Pure Irreducibles in Integral Domains
We proceed to study the existence of purely long and purely short irreducibles in the context ofintegral domains. Throughout this section, we set L ( R ) := L ( R • ) and S ( R ) := S ( R • ) for anyatomic integral domain R . In addition, when R is a Dedekind domain, we let Cl( R ) denote the divisorclass group of R .6.1. Examples of Dedekind Domains.
For a finite-rank monoid M , we have already seen in Exam-ple 4.8 that none of the conditions L ( M ) = ∅ and S ( M ) = ∅ implies the other one. In this subsection,we construct examples of Dedekind domains to illustrate that a similar statement holds in the contextof atomic integral domains.The celebrated Claborn’s class group realization theorem [23, Theorem 7] states that for every abeliangroup G there exists a Dedekind domain D such that Cl( D ) ∼ = G . The following refinement of this result,due to Gilmer, Heinzer, and the fourth author, will be crucial in our constructions. Theorem 6.1. [37, Theorem 8]
Let G be a countably generated abelian group generated by B ∪ C with B ∩ C = ∅ such that B ∗ ∪ C generates G as a monoid for each cofinite subset B ∗ of B . Then there existsa Dedekind domain D with class group G satisfying the following conditions:(1) the set B ∪ C consists of the classes of G containing nonzero prime ideals;(2) the set C consists of the classes of G containing infinitely many nonzero prime ideals;(3) the set B consists of the classes of G containing finitely many nonzero prime ideals.Moreover, the number of prime ideals in each class contained in B can be specified arbitrarily. Further refinements of Claborn’s theorem in the direction of Theorem 6.1 were given by Grams [41],Michel and Steffan [43], and Skula [49]. We are in a position now to exhibit a Dedekind domain with apurely short (resp., long) irreducible but no purely long (resp., short) irreducibles.
Example 6.2.
In this example, we will produce a Dedekind domain in which there is a purely shortirreducible but no purely long irreducibles. Toward this end, consider a Dedekind domain D with classgroup Cl( D ) ∼ = Z / Z . Since Cl( D ) is finite and the class of Cl( D ) corresponding to 2 + 3 Z generatesCl( D ) as a monoid, one can invoke Theorem 6.1 to assume that the non-principal prime ideals of D distribute within Cl( D ) in the following way. There is a unique nonzero prime ideal P in the class ofCl( D ) corresponding to 1 + 3 Z and infinitely many nonzero prime ideals in the class corresponding to2 + 3 Z . Observe that we can separate non-prime irreducible principal ideals I of D into the followingthree types, according to their (unique) factorizations into prime ideals:(1) I = P , (2) I = P Q, or (3) I = Q Q Q , where Q, Q , Q , and Q are prime ideals in the class of Cl( D ) corresponding to 2+3 Z . By construction,there is only one irreducible principal ideal of type (1), namely, P .Take a ∈ A ( D ) such that ( a ) = P . Proving that a ∈ S ( D ) amounts to verifying that ( a ) is apurely short irreducible in the atomic monoid M consisting of all nonzero principal ideals of D . To doso, suppose that ( a ) appears in an irredundant factorization relation ( z , z ) ∈ ker π M , where z := ( P ) k m Y i =1 ( P Q i ) n Y j =1 ( Q j, Q j, Q j, ) and z := m ′ Y i =1 ( P Q ′ i ) n ′ Y j =1 ( Q ′ j, Q ′ j, Q ′ j, )for some k ∈ N , m, n, m ′ , n ′ ∈ N , and ideals Q i , Q ′ i , Q j,i , and Q ′ j,i in the class corresponding to2 + 3 Z . As D is Dedekind, comparing the numbers of copies of P in all the irreducibles of the idealfactorizations z and z , one obtains that 3 k = m ′ − m . Now comparing the numbers of copies of prime6 S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH ideals in the class 2 + 3 Z in all the irreducibles of the ideal factorizations z and z , one obtains that m ′ − m = 3( n − n ′ ), and so n − k = n ′ . As a result, | z | = k + m + n = (3 k + m ) + ( n − k ) − k = m ′ + n ′ − k < | z | . Therefore a ∈ S ( D ), as desired.Let us proceed to argue that D contains no purely long irreducibles. By the previous paragraph, L ( D ) contains no irreducibles generating ideals of type (1). Take a , a ∈ A ( D ) such that the ideals( a ) and ( a ) are of type (2) and type (3), respectively. Then take prime ideals Q , . . . , Q in the classof Cl( D ) corresponding to 2 + 3 Z such that the equalities ( a ) = P Q and ( a ) = Q Q Q hold, and Q / ∈ { Q , Q , Q , Q } . Now consider the ideal factorizations z := ( P )( Q Q Q ) ,z := ( P Q )( P Q )( P Q ) ,z := ( P )( P Q )( Q Q Q )( Q ) , and z := ( P Q ) ( Q Q Q ) . Notice that ( z , z ) ∈ ker π M is irredundant and satisfies | z | < | z | . Because Q Q Q appears in z ,it follows that a / ∈ L ( D ). On the other hand, ( z , z ) ∈ ker π M is also irredundant, and it satisfies | z | < | z | . Because P Q appears in z , one finds that a / ∈ L ( D ). As a result, no irreducible generatingan ideal of type (2) or type (3) is purely long, whence L ( D ) = ∅ .To complement Example 6.2, we proceed to construct a Dedekind domain having a purely longirreducible but no purely short irreducibles. Example 6.3.
Let D be a Dedekind domain with Cl( D ) ∼ = Z . Since Cl( D ) is countable and the set {± } generates Cl( D ) as a monoid, we can assume in light of Theorem 6.1 that the non-principal primeideals of D distribute within Cl( D ) as follows. There is a unique nonzero prime ideal, which we denoteby P , in the class of Cl( D ) corresponding to −
2; there is a unique nonzero prime ideal, which we denoteby Q , in the class of Cl( D ) corresponding to 2; and there are infinitely many nonzero prime ideals in eachof the classes corresponding to − − N and the prime ideals in the class corresponding to 1 by (annotated) M . Notice thatwe can separate non-prime irreducible principal ideals I of D into the following four types, according totheir (unique) factorizations into prime ideals:(1) I = P Q, (2) I = P N N , (3) I = QM M , or (4) I = N M, where
N, N , N belong to the class of Cl( D ) corresponding to − M, M , M belong to the classof Cl( D ) corresponding to 1.Take a ∈ A ( D ) such that ( a ) = P Q . As in Example 6.2, proving that a ∈ L ( D ) amounts to showingthat the principal ideal ( a ) is a long irreducible in the atomic monoid M consisting of all nonzeroprincipal ideals of D . To do this, suppose that ( a ) appears in an irredundant ideal factorization relation( z , z ) ∈ ker π M , and write z := ( P Q ) k m Y i =1 ( P N i, N i, ) n Y j =1 ( QM j, M j, ) t Y k =1 ( N k M k )and z := m ′ Y i =1 ( P N ′ i, N ′ i, ) n ′ Y j =1 ( QM ′ j, M ′ j, ) t ′ Y k =1 ( N ′ k M ′ k )for some k, m, n, t, m ′ , n ′ , t ′ ∈ N . Since D is Dedekind, after comparing the numbers of copies of theprime ideals P and Q that appear in all irreducibles of z and z , one obtains that m − m ′ = n − n ′ = − k . ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS − z and z , one obtains that t − t ′ = − m − m ′ ) = 2 k . As a result, | z | = k + m + n + t = ( m ′ + n ′ + t ′ ) + k + ( m − m ′ ) + ( n − n ′ ) + ( t − t ′ ) = | z | + k > | z | . Therefore we can conclude that a ∈ L ( D ).Finally, let us verify that D contains no purely short irreducibles. Since any irreducible generator of P Q is purely long, D contains no purely short irreducibles of type (1). Take a ∈ A ( D ) such that ( a )has type (2), and then take prime ideals N , N in the class of Cl( D ) corresponding to − a ) = P N N . In addition, take distinct prime ideals N and N in the class of Cl( D ) correspondingto − N , N / ∈ { N , N } . Finally, take distinct prime ideals M and M in the class of Cl( D )corresponding to 1. Consider the ideal factorizations z := ( P Q )( P N N )( M N )( M N ) and z := ( P N N )( P N N )( QM M ) . Observe that ( z , z ) ∈ ker π M is an irredundant factorization relation satisfying | z | > | z | . Since P N N appears in z , it follows that a / ∈ S ( D ). As a result, no irreducible generating an ideal oftype (2) can be purely short. In a similar manner, one can verify that no irreducible generating anideal of type (3) is purely short. Now let a be an irreducible of D such that ( a ) is a principal idealof type (4). Take N and M in the classes of Cl( D ) corresponding to − a ) = N M , and then consider the ideal factorizations z := ( P Q )( N M ) and z := ( P N )( QM ) . Notice that ( z , z ) ∈ ker π M is an irredundant factorization relation satisfying | z | > | z | . Since N M appears in z , it follows that a / ∈ S ( D ). Therefore none of the irreducibles generating ideals of type (4)is purely short. As a consequence, S ( D ) = ∅ .6.2. Integral Domains Do Not Satisfy the PLS Property.
In Section 5, we have proved that inthe class of torsion-free monoids with rank at most 2, being a proper length-factorial monoid and being aPLS monoid are equivalent notions. It was proved in [22] that an integral domain has the length-factorialproperty only if it is a unique factorization domain. In addition, being a proper length-factorial monoidimplies having both purely long and purely short irreducibles. This begs the tantalizing question as towhether there is an atomic integral domain with both purely long and a purely short irreducibles. TheDedekind domains constructed in the previous subsection do not satisfy this property, and this is not acoincidence.
Theorem 6.4.
Let R be an atomic domain. Then either L ( R ) = ∅ or S ( R ) = ∅ .Proof. Suppose, by way of contradiction, that R is an atomic domain such that both L ( R ) nor S ( R )are nonempty sets. We recall that by Corollary 4.4 both L ( R ) and S ( R ) are finite sets. Set ℓ := | L ( R ) | and s := | S ( R ) | , and then write L ( R ) =: { α , α , . . . , α ℓ } and S ( R ) =: { β , β , . . . , β s } . Take ρ ∈ ker π R to be an irredundant and unbalanced factorization relation. It follows from Proposi-tion 4.3 that each α i appears in the longer factorization component of ρ and each β j appears in theshorter factorization component of ρ . Therefore there exist factorizations z, z ′ ∈ Z ( R ) such that ρ = ( α a α a · · · α a ℓ ℓ z, β b β b · · · β b s s z ′ )for some a , . . . , a ℓ , b , . . . , b s ∈ N such that none of the α i ’s appears in z and none of the β j ’s appearsin z ′ . In addition, as the factorization is irredundant, none of the α i ’s appears in z ′ and none of the β j ’sappears in z . We now derive contradictions in the following three cases.8 S. T. CHAPMAN, J. COYKENDALL, F. GOTTI, AND W. W. SMITH
CASE 1: ℓ, s ≥
2. Consider the element(6.1) x := α a α a · · · α a ℓ ℓ ( α a − β b ) π R ( z ) ∈ R. We claim that x = 0. Because R contains no nonzero zero-divisors, verifying that x = 0 amounts toshowing that α a − β b = 0. Indeed, this must be the case: if ( α a , β b ) ∈ ker π R , then the fact that ℓ ≥ α to appear in the left factorization component of therelation ( α a , β b ). Hence both α a − β b and x belong to R • . On the other hand, it follows from (6.1)that β divides x in R . Thus, there exist w ∈ Z ( R ) and w ∈ Z R ( α a − β b ) such that β w ∈ Z R ( x )and w := α a α a · · · α a ℓ ℓ w z ∈ Z R ( x ). Now consider the factorization relation ( β w , w ) ∈ ker π R .CASE 1.1: β does not appear in w . Since β ∈ S ( R ), it follows that | w | > | β w | . Now theinclusion α ∈ L ( R ) implies that α must appear in w and, therefore, in w . Thus, α divides β b in R . As a result, there is a factorization relation ( α w ′ , β b ) ∈ ker π R for some w ′ ∈ Z ( R ). Clearly,( α w ′ , β b ) is a non-diagonal factorization relation. Since β ∈ S ( R ), th inequality | α w ′ | > | β b | musthold. Therefore ( α w ′ , β b ) is an unbalanced factorization relation in which β does not appear. Thiscontradicts that β ∈ S ( R ).CASE 1.2: β appears in w . Because β does not appear in z , we see that β must appear in w .This implies that β divides α a in R . Now we can follow an argument completely analogous to thatwe just used in CASE 1.1 to obtain the desired contradiction.CASE 2: { ℓ, s } = { , n } for some n ∈ N ≥ . Assume will first assume that ℓ = 1. Recall that ρ = ( α a z, β b β b · · · β b n n z ′ ). In this case, we also impose the condition that the exponent a is theminimum number of copies of the purely long irreducible α that can appear in any irredundant andunbalanced factorization relation in ker π R . Using notation similar to that of CASE 1, we now set(6.2) x := α a − ( α − β b ) π R ( z ) ∈ R. Notice that x = 0 as otherwise α = β b , which is clearly impossible. Since ℓ = 1, it followsfrom (6.2) that β divides x in R . Then one can take w ∈ Z ( R ) and w ∈ Z R ( α − β b ) suchthat ( β w , α a − w z ) ∈ ker π R . It is clear that β does not divide α − β b , whence β does not appearin α a − w z , which implies that | β w | < | α a − w z | . By the minimality of a , we see that α mustappear in w . Thus, α must divide β b in R . In this case, ( α w , β b ) ∈ ker π R for some w ∈ Z ( R ),which is a contradiction because β does not appear in β b . The case when ℓ > s = 1 followssimilarly.CASE 3: ℓ = s = 1. In this case, ρ = ( α a z, β b z ′ ). We assume that the exponent a satisfies thesame minimality condition that we imposed in CASE 2. Consider the element(6.3) x := α a − ( α − β ) π R ( z ) ∈ R. As α − β is nonzero and β divides x in R , there exist factorizations w ∈ Z ( R ) and w ∈ Z R ( α − β )such that ( β w , α a − w z ) ∈ ker π R . Since β does not divide α − β in R , it cannot appear in α a − w z and, therefore, | β w | < | α a − w z | . This, along with the minimality of a , implies that α appears in w . However, this contradicts that α does not divide α − β in R . (cid:3) As a consequence of Theorem 6.4, we rediscover the main result of [22].
Corollary 6.5. [22, Theorem 2.10]
Let R be an integral domain. Then R is a unique factorizationdomain if and only if R • is a length-factorial monoid.Proof. Clearly, if R is a unique factorization domain, then R • is a factorial monoid and, therefore, alength-factorial monoid. For the reverse implication, suppose that R • is a length-factorial monoid. ByTheorem 6.4, either L ( R ) is empty or S ( R ) is empty. Therefore R • is not a proper length-factorialmonoid, and so it is a factorial monoid. Hence R is a unique factorization domain. (cid:3) ENGTH-FACTORIALITY IN MONOIDS AND DOMAINS Acknowledgments
The authors would like to thank an anonymous referee for providing comments and suggestionsthat helped improve the quality of the present paper. During this collaboration, the third author wassupported by the NSF award DMS-1903069.
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