Monoids Determined by a Homogenous Linear Diophantine

Equation and the Half-Factorial Property

Scott T. Chapman∗

Trinity UniversityDepartment of Mathematics

715 Stadium DriveSan Antonio, Texas 78212-7200

Ulrich KrauseUniversitat Bremen

Fachbereich Mathematik/InformatikD-28334 Bremen, Germany

Eberhard OeljeklausUniversitat Bremen

Fachbereich Mathematik/InformatikD-28334 Bremen, Germany

Abstract

We study submonoids M of Nk consisting of the solutions of a homogenous lineardiophantine equation with integer coefficients. M is a Krull monoid and we computeits divisor class group in terms of the coefficients of the diophantine equation. Thisleads to a characterization of such submonoids M which are factorial. We concentrateon determining conditions under which M is half-factorial and offer not only criteriabased on the reduction of M to a simplier monoid, but also a geometric argument.

1 Introduction

Let D be an integral domain in which each nonzero nonunit of D can be factored intoa finite product of irreducible elements (such an integral domain is called atomic). Thequestion of when D is a unique factorization domain (or factorial domain) has been aprinciple question in commutative algebra. Recently, the following generalization of thefactorial condition has appeared in the literature: an atomic integral domain D is called ahalf-factorial domain (or HFD) if whenever x1, . . . , xn, y1, . . . , ym are irreducible elementsof D with x1 · · ·xn = y1 · · · ym then n = m. If D is a ring of integers in a finite algebraicextension of the rationals, then it is well known that D is a HFD if and only if the classnumber of D is less than or equal to two (see [6]). More generally, various conditions whichforce a Krull domain to be a HFD have been studied in [4][7][8][9][11][12][17]. The question

∗The first author gratefully acknowledges support under grants from the Deutscher Akademischer Aus-tausch Dienst and the Trinity University Faculty Development Committee.

1

of whether or not D is a HFD is purely dependent on the structure of the multiplicativemonoid D∗ = D\{0}; hence, it is of interest to consider this condition for atomic monoids.

In this paper, we consider the half–factorial property for monoids of the form

M(a1, . . . , an) = {(r1, . . . , rn) ∈ Nn |n∑

j=1

rjaj = 0}

where a1, . . . , an ∈ Z and the semigroup operation is given by (componentwise) addition. Inspite of the fact that the nonnegative integer solutions of a linear Diophantine equation asabove are hard to determine (cf. [20]) we adopt the strategy of finding structural propertiesof the monoid of solutions M(a1, . . . , an). These monoids are Krull monoids (as definedin Section 2). We define and explicitly compute their class groups and provide a simplecriterion for when such a monoid is factorial. Factoriality means that the solutions innonnegative integers are in a unique manner given as the sums of primary solutions, thelatter being easily computed from the coefficients of the linear Diophantine equation. InSections 3 and 4, we center our discussion on special monoids i.e., M(a1, . . . , an) where1 ≤ a1 < a2 < · · · an−1 < −an = β · lcm(a1, . . . , an−1) for some β ∈ N, and determinevarious conditions under which these monoids are half-factorial (see Lemmas 3.1 and 3.2,Corollary 3.3 and Theorem 4.1). In Section 5, we consider the more general situation wherean may not be a multiple of lcm(a1, . . . , an−1) and give necessary and sufficient conditionsfor M(a1, . . . , an) to be half-factorial in terms of a related special monoid.

In terms of nonnegative solutions of a linear Diophantine equation half–factoriality forthe monoids considered in section 5 means that each irreducible solution is in a uniquemanner given as a convex combination over the rational numbers of primary solutions.Since every solution is a sum of irreducible solutions one could in principle obtain anysolution from the simple primary solutions. The case of half–factoriality, however, is moreintriguing than that of factoriality since not every rational convex combination of primarysolutions is irreducible and it seems to be difficult to determine those which are irreducible.

2 Definitions and Basic Results

Let S be a commutative and cancellative multiplicative monoid. For simplicity we assumethe group of units to consist of 1 only. We denote the quotient group of S by Q(S) andthe set of irreducible elements of S by I(S). Let 5 be the divisibility relation (or quasi-ordering) on S induced by the following: x 5 y in Q(S) if and only if xz = y for somez ∈S. An irreducibly x ∈ S is prime whenever x ≤ yz implies x ≤ y or x ≤ z and it isprimary whenever x ≤ yz implies x ≤ y or x ≤ zk for some k. Let F be a nonempty familyof homomorphims f 6≡ 0 of Q(S) into Z such that for each x ∈ Q(S) the set

F(x) = {f ∈ F | f(x) 6= 0}

is finite. An element of F is called a state on S. If the monoid S is defined by F , i.e.,S= {x ∈Q(S) | f(x) ≥ 0 ∀ f ∈ F}, then S is a Krull monoid. A state f on S is

2

essential if for any x, y ∈ Q(S), there exists z ∈ Q(S) such that x 5 z, y 5 z andf(z) = max{f(x), f(y)}. Let I denote the set of essential states f which are normalized,i.e. f(Q(S)= Z. Consider the map

ϕ : S→ Z(I)+

defined byϕ(x)(f) = f(x)

where Z+ = N = {0, 1, 2, . . . , } and Z(I)+ is the direct sum of | I | copies of Z+. For a Krull

monoid S this map satisfies the following two properties:

1. for all x, y ∈S, x 5 y if and only if ϕ(x) ≤ ϕ(y) (here ≤ denotes the pointwiseordering),

2. each element of Z(I)+ is the minimum of finitely many elements for the set ϕ(S).

As described above, the essential states of a Krull monoid yield a divisor theory for S (see[18], [15]). The factor monoid

Z(I)+ /ϕ(S) ∼= Z(I)/ϕ(Q(S)),

where ϕ is extended in the obvious way to Q(S), is a group known as the divisor class group

of S and denoted Cl(S). The unit vectors ei in Z(I)+ , i.e. ei(j) = δij , are the prime divisors.

[x] denotes the class in Cl (S) generated by x ∈ Z(I). In this respect, some well–knownfacts are:

1. An integral domain D is a Krull domain if and only if D∗ = D\{0} is a Krull monoid[16].

2. If D is a Krull domain then the divisor class group of the Krull monoid D∗ is merelythe divisor class group of D [16].

3. A monoid S is a factorial monoid, i.e. each element of S is a unique product ofirreducible elements, if and only if it is a Krull monoid with trivial divisor class group[15].

As indicated already in the Introduction, a monoid S is half–factorial if wheneverx1, . . . , xn, y1, . . . , ym are irreducible elements of S with x1, . . . , xn = y1 . . . ym then n = m.

In the following we will be concerned with additive submonoids M of Nn. The twofactoriality properties then may be rephrased as follows. M is half–factorial (resp. factorial)

if whenever x1, . . . , xk are pairwise different irreducible elements of M withk∑

i=1nixi =

0, ni ∈ Z, thenn∑

i=1ni = 0 (resp. ni = 0 for all i).

3

With respect to the additive monoids

M(a1, . . . , an) = {(r1, . . . , rn) ∈ Nn |n∑

j=1

rjaj = 0}

where a1, . . . , an ∈ Z as defined earlier, we can assume without loss that ai 6= 0 for all i,not all ai are of equal sign and gcd (a1, . . . , an) = 1.

While these monoids represent basic structures, surprizingly little seems to be knownabout them (see [20] for a discussion). If x = (x1, . . . , xn) ∈Q(M(a1, . . . , an)) and πi(x) =xi is the ith projection of Q (M) into Z then

M(a1, . . . , an) = {x ∈ Q(S) | πi(x) ≥ 0 ∀ i ∈ {1, . . . , n}}

andM(a1, . . . , an) is an additive Krull monoid defined by the projections {πi}ni=1. The next

Lemma will be quite useful for what follows.

Lemma 2.1. (i) The projection πi of M(a1, . . . , an) is not essential if and only ifaiaj ≤ 0 for all j 6= i and aiaj < 0 for at least two different j.

(ii) The mapping α:M(a1, . . . , an) −→M(a′1, . . . , a′n), (r1, . . . , rn) 7−→

(r1c1

, . . . , rncn

)where

ci = gcd{aj | i 6= j = 1, . . . , n} and a′j =aj

Πcii6=j

for 1 ≤ i, j ≤ n

is a monoid isomorphism.

(iii) Up to the above isomorphism of the solution monoid we may assume that the equationa1r1 + . . . + anrn = 0 is normalized, i.e. gcd{aj | i 6= j = 1, . . . , n} = 1 for all1 ≤ i ≤ n. The projections πi for a normalized equation are all normalized.

Proof. (i) we take from [17, p. 682, Example 2].

(ii) As already remarked we may assume gcd{a1, . . . , an} = 1. This implies gcd{ci, ck} = 1for i 6= k and, hence

∏i 6=j

ci | aj for all j. For (r1, . . . , rn) ∈ M(a1, . . . , an) we must

have that ci | airi which together with ci | aj for i 6= j implies that ci | ri. Because

of a1r1 + . . . + anrn =n∏

i=1ci

(a′1

r1c1

+ . . . + a′nrncn

)the mapping α:M(a1, . . . , an) −→

M(a′1, . . . , a′n) is well–defined. Obviously, α is bijective and a monoid homomorphism.

(iii) For the coefficients a′j defined in (ii) we have gcd{a′j | i 6= j = 1, . . . , n} | gcd{

aj

ci|

i 6= j = 1, . . . , n}

and gcd{a′j | i 6= j = 1, . . . , n} = 1 by definition of ci. Let a1r1 +

. . .+anrn = 0 be normalized and πi the ith projection of Q(M) = Q(M(a1, . . . , an)

)

into Z. gcd{aj | i 6= j = 1, . . . , n} = 1 implies that∑j 6=i

ajxj = 1 with xj ∈ Z

4

and, hence, for xi ∈ Z arbitrary∑j 6=i

aj(−aixjxi) + aixi = 0. That is, πi(L) = Z

where L is the set of solutions (y1, . . . , yn) of a1y1 + . . . + anyn = 0 with yi ∈ Z.Lets see that L = Q(M). Obviously, Q(M) ⊂ L. By general assumptions, the setsI+ = {i | 1 ≤ i ≤ n, ai > 0} and I− = {j | 1 ≤ j ≤ n, aj < 0} are non–empty andI+ ∪ I− = {1, . . . , n}. By xi = − ∑

j∈I−aj for all i ∈ I+ and xj =

∑i∈I+

ai for all j ∈ I−

a particular solutionn∑

i=1aixi = 0 is defined with components in the strictly positive

integers. For y ∈ L there exist some k ∈ N such that kx + y ∈ M(a1, . . . , an) and,hence, y = (kx + y)− kx ∈ Q(M).

By the above Lemma for every projection πi of M(a1, . . . , an) 1ci

πi is normalized. Forn = 2 both projections are essential and the solution monoid is isomorphic to Z+, thus thiscase we don’t need to consider. For n ≥ 3 there may be exactly one ai < 0, in which casethe πj with i 6= j are the essential projections or, if n ≥ 4, there are at least two negativecoefficients and at least two positive coefficients, in which case all projections are essential.These two cases turn out to be quite different from each other and, hence, we will capturethe case 1 by the monoid

M(a1, . . . , an; b) ={

(r1, . . . , rn+1) ∈ Nn+1∣∣∣

n∑

j=1

rjaj = rn+1b}

where a1 . . . , an, b ∈ N+ = {m ∈ N | m > 0}and case 2 by

M(a1, . . . , an; b1, . . . , bk) ={

(r1, . . . , rn+k) ∈ Nn+k∣∣∣

n∑

j=1

rjaj =k∑

i=1

rn+ibi

}

where a1, . . . , an, b1, . . . , bk ∈ N+ and n ≥ 2, k ≥ 2.The monoidsM(a1, . . . , an) show quite distinctive features among Krull monoids as the

following proposition indicates.

Proposition 2.2. 1. Cl (M(a1, . . . , an)) has only finitely many divisor classes contain-ing prime divisors. Moreover, each class containing a prime divisor contains finitelymany.

2. M(a1, . . . , an) is not realizable as the multiplicative monoid D∗ of a Krull domain D.

3. M(a1, . . . , an) contains finitely many irreducible elements.

4. In M(a1, . . . , an) the primary elements are precisely the elements q for which ϕ(q) =kei, k = ord [ei] < ∞ for some i.

5

In the particular case of M(a1, . . . , am; b) the primary elements are given by

q =b

gcd(ai, b)ei +

ai

gcd(ai, b)em+1 for i = 1, 2, . . . ,m,

the ei being the unit vectors in Zm+1.

5. If an element of M(a1, . . . , an) is a sum of primary elements then these primaryelements are unique.

Proof. 1 follows directly from the definition of the divisor class group and 3 from a wellknown theorem of Dickson [10, Theorem 9.18]. The assertion 2 follows from a fundamentalresult of Skula [19] for Krull domains, namely if D a Krull domain with divisor class groupG and P is the set of prime divisors of D∗ then for every finite subset E ⊆ P we have

G = 〈{[p] | p ∈ P\E}〉.

Hence a Krull domain D has either a divisor class which contains infinitely many primedivisors, or infinitely many divisor classes which contain prime divisors. This gives assertion2.

To prove 4, suppose ϕ(q) = kei with k = ord [ei] < +∞ (with respect to the classgroup). q needs to be irreducible. If q ≤ x + y for x, y ∈ M(a1, . . . , an) and q � x thenkei = ϕ(q) ≤ ϕ(x) + ϕ(y) and kei = ϕ(q) � ϕ(x). Therefore, ei ≤ ϕ(y) and ϕ(q) = kei ≤ϕ(ky) which implies q ≤ ky. Conversely, suppose q to be primary and let ϕ(q) =

∑i∈I

qiei, I

finite, qi ≥ 1. By the definition of a divisor theory there exist for each i ∈ I elementsxij ∈ M(a1, . . . , an) such that ei = min

{ϕ(xij) | j ∈ Ii

}with Ii finite (“min” taken

componentwise). It follows that ϕ(q) ≤ c∑i∈I

ei ≤ c∑i∈I

ϕ(xij(i)) for some c ≥ 1 and j(i)

taken arbitrarily from Ii. This implies q ≤ c∑i∈I

xij(i) for all j(i) ∈ Ii. Since q is primary it

follows inductiviely that there exist i0 ∈ I and integers kj ≥ 1 such that q ≤ kjxi0j for allj ∈ Ii0 . For k′ = maxj kj , therefore, we have that

ϕ(q) ≤ k′min{

ϕ(xi0j) | j ∈ Ii0

}= k′ei0 .

That is, ϕ(q) = kei0 for some k ≤ k′ and, because q is irreducible, we must have ord[ei0 ] = k < +∞.

Concerning the particular monoid M(a1, . . . , am; b) let q be primary, that is ϕ(q) = kei

with k = ord [ei] < ∞. The element x = bgcd(ai,b)

ei + aigcd(ai,b)

em+1 is in M(a1, . . . , an; b)and it is obviously irreducible. By definition of ϕ, ϕ(x) = lei for some l ∈ N. k = ord [ei]implies k ≤ l and, hence, ϕ(q) ≤ ϕ(x). Irreducibility of x implies q = x.

Finally, concerning assertion 5 suppose∑i∈I

qi =∑j∈J

pj for primary elements qi, pj . By

applying ϕ and using property 4 above we obtain a bijection α: I −→ J such that qi = pα(i)

for all i ∈ I.

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We now compute the divisor class groups of the M(a1, . . . , an).

Theorem 2.3. 1. Consider the monoid M(a1, . . . , an; b) as in case one above whereci = gcd (b ∪ {aj}j 6=i) and c =

∏ni=1 ci. Then

Cl(M(a1, . . . , an; b)) = Zb/c.

2. Consider the monoid M(a1, . . . , an; b1, . . . , bk) as in case two above. Then

Cl(M(a1, . . . , an; b1, . . . , bk)) = Z.

Proof. For a monoid M = M(a1, . . . , an) of general type by Lemma 2.1 we have an iso-morphism α:M −→M′ where M′ = M(a′1, . . . , a

′n) is the normalized monoid. For ϕ and

ϕ′ defining the divisor theory of M and M′, respectively from the definition of α we havethat ϕ = ϕ′ ◦ α. Therefore, the two class groups Cl(M) and Cl(M′) are isomorphic and itsuffices to consider a normalized monoid.

To prove assertion 1, we may assume that M = M(a1, . . . , an; b) is normalized, that isci = 1 for 1 ≤ i ≤ n and c = 1. Choose integers y1, . . . , yn such that a1y1 + . . . + anyn = 1.The order of (y1, . . . , yn) in Cl(M) is clearly b. Choose for each i a ti ∈ N such thatki = yi + bti > 0. Then for all i ei − ki(y1, . . . , yn) is a solution of the normalized equationin integers and, hence, ki[(y1, . . . , yn)] = [ei] for all 1 ≤ i ≤ n. Thus, Cl(M) is cyclic oforder b.

Concerning assertion 2 we also stick to the case of a normalized monoid,M =M(a1, . . . , an; b1, . . . , bk). Let t = n + k and f :Zt −→ Z,

f(x1, . . . , xt) = a1x1 + . . . + anxn − b1xn+1 − . . .− bkxn+k.

Because the greatest common divisor of the ai and bj is 1 f must be surjective and, hence,

Zt/ ker f ∼= imf = Z.

Obviously, Q(M) = ker f . Since all projections of M are essential ϕ must be the identityand we obtain

Cl(M) = Zt/ϕ(Q(M)) = Zt/ ker f ∼= Z.

Corollary 2.4. For a monoid M =M(a1, . . . , am) the following statements are equivalent:

(i) M is factorial

(ii) M is of the form M =M(a1, . . . , an; b) and b = c =n∏

i=1ci.

(iii) M is of the form M =M(a1, . . . , an; b) and each element of M is a (unique) sum of

elementsb

gcd(ai, b)ei +

ai

gcd(ai, b)en+1 in M, i = 1, . . . , n.

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(iv) M is isomorphic to Zn+ (with addition).

Proof. By Theorem 2.3 and the general fact 3 statements (i) and (ii) are equivalent.(ii) =⇒ (iii): Since by Theorem 2.3 Cl(M) must be trivial we have that ord [ei] = 1 for1 ≤ i ≤ n. By Proposition 2.2 (4) the primary elements qi of M are given by ϕ(qi) = ei for

all i. If x ∈ M then ϕ(x) =n∑

i=1xiei =

n∑i=1

xiϕ(qi) = ϕ( n∑

i=1xiqi

)with xi ∈ N and, hence,

x =n∑

i=1xiqi. The result follows from assertions 4 and 5 of Proposition 2.2.

(iii) =⇒ (iv): By the uniqueness of the representation in (iii).(iv) =⇒ (i): The irreducible elements in the additive monoid Zn

+ are the ei, 1 ≤ i ≤ n,the representation by which is unique.

As the Corollary indicates, primary elements play their role. (In [20] so called “com-pletely fundamental” elements are considered which turn out to be equivalent to primaryelements.) If there are plenty primary elements in the sense that they generate the monoid,the monoid must be factorial according to the Corollary. Whereas in general primary el-ements need not be prime elements, this holds in factorial monoids. There exist simplemonoids which do not have primary elements at all. By Theorem 2.3 in conjunction withProposition 2.2 (4) the monoidsM(a1, . . . , an; b1, . . . , bk) (not that n, k ≥ 2) cannot possessprimary elements. As we shall see later on, half–factorial monoids of type M(a1, . . . , an; b)can be characterized by the property that there are enough primary elements in the sensethat they generate the monoids in a convex manner.

As is the case with a Krull domain, the divisor class group can be used to computethe complete set of irreducible elements of M(a1, . . . , an). Let G be an abelian group. Asequence {g1, . . . , gn} of not necessarily distinct nonzero elements of G is called a minimal-zero sequence if g1 + · · · + gn = 0 and no proper subsequence of {g1, . . . , gn} sums tozero. By [11, Propositon 1], the irreducible elements of M(a1, . . . , an) are in one-to-onecorrespondence with the minimal-zero sequences of Cl(M(a1, . . . , an)) consisting of thedivisor classes containing a prime divisor.

The following examples illustrate some of our definitions and results.

Examples 2.5. 1. Nonnegative solutions of linear Diophantine equations play a role invarious parts of algebra. The following equation we take from [14] to illustrate ourgeneral principles. The equation 2x1+5x2 = 3x3 is already in normalized form and forthe solution monoidM =M(2, 5; 3) it follows Cl(M) = Z3 by Theorem 2.3, implyingthat M cannot be factorial. According to Proposition 2.2 there are two primaryelements q1 = 3e1 + 2e3 = (3, 0, 2) and q2 = 3e2 + 5e3 = (0, 3, 5) corresponding toord [e1] = ord [e2] = 3. The minimal zero–sequences, beside 3[e1] = 3[e2] = 0 are[e1] + 2[e2] = 0 and 2[e1] + [e2] = 0. Thus, there are two more irreducible elements,x = (1, 2, 4) and y = (2, 1, 3),none of which is primary. The latter can be representedin a convex manner by 3x = q1 + 2q2, 3y = 2q1 + q2. Thus by this representationthe member of irreducible summands neither increases nor decreases and we conclude

8

that M must be half–factorial. The above representations also imply that M has noprime element because q1 ≤ 3x, q2 ≤ 3y but neither q1 ≤ x nor q2 ≤ y.

If we change the given equation a little bit to be x1 + 5x2 = 3x3 then the solutionmonoid M(1, 5; 3) is no longer half–factorial. By the same method as before weobtain as the irreducible elements q1 = (3, 0, 1), q2 = (0, 3, 5), x = (1, 1, 2) whereq1, q2 are primary but not prime and x is not primary. Because of the relationship3x = q1 + q2 M(1, 5; 3) is not half–factorial.

2. Consider the equation 2x1 + 3x2 + 4x3 = 6x4. Since c1 = c3 = 1, c2 = 2, c = 2normalization yields x1 + 3x2 + x3 = 3x4. The class group for both equations is Z3,henceM =M(1, 3, 1; 3) is not factorial. Proceeding as in example 1, we obtain forMas primary elements q1 = (3, 0, 0, 1), q2 = (0, 1, 0, 1), q3 = (0, 0, 3, 1) corresponding to3[e1] = 3[e3] = 0 and [e2] = 0. This shows that q2 is a prime element. Beside the above,the only minimal zero–sequences are [e1] + 2[e3] = 0 and 2[e1] + [e3] = 0 which yieldthe irreducible elements x = (1, 0, 2, 1) and y = (2, 0, 1, 1). Furthermore, 3x = q1+2q3

and 3y = 2q1 + q3 which shows that q1 and q3 are not prime and that M and, hence,M(2, 3, 4; 6) are half–factorial. Thus M has 5 irreducible elements consisting of oneprime element, 2 primary elements which are not prime and 2 irreducible elementswhich are not primary.

Although the class groups in the two examples are isomorphic and both have somerelations in common the monoids in the two examples are not isomorphic, e.g. theformer contains a prime but the latter does not.

For the remainder of this paper, we focus on the problem of determining when themonoid M(a1, . . . , an; b) is half-factorial. While there is a characterization of such half-factorial monoids in terms of the cross numbers of the minimal zero-sequences (see [17]or [7] for an analysis), we focus on conditions involving the coefficients a1, . . . , an, b. Theproblem of determining when the monoidM(a1, . . . , an; b1, . . . , bk) is half-factorial is closelyrelated to the work done in [4] and [5]. While we leave investigation of this problem to futurework, we provide an example which helped motivate our current research.

Example 2.6. In [17, Example 2)], the authors argue that the monoidM(5, 2; 5) is factorial,the monoidM(1, 1, 2) is half-factorial and provide the monoidM(1, 1; 1, 1) as an example ofa monoid where all projections are essential. The monoid M(1, 1; 1, 1) is also half-factorial.This can be seen using elementary linear algebra as follows. Using a matrix argument,it is easy to show that the only irreducible elements of M(1, 1; 1, 1) are u1 = (1, 0, 1, 0),u2 = (1, 0, 0, 1), u3 = (0, 1, 1, 0) and u4 = (0, 1, 0, 1). If

x1u1 + x2u2 + x3u3 + x4u4 = y1u1 + y2u2 + y3u3 + y4u4

then x1+x2 = y1+y2 and x3+x4 = y3+y4 and thus x1+x2+x3+x4 = y1+y2+y3+y4. Thisresult can also be seen using the divisor class group. Since [e1]− [e2] = 0 and [e3]− [e4] = 0in the class group, we have that M(1, 1; 1, 1) is a Krull monoid with divisor class group Z

9

such that all the prime divisors of M(1, 1; 1, 1) are contained in the divisor classes {−1, 1}.By [9, Theorem 4.1], M(1, 1; 1, 1) is half-factorial.

We can extend this argument as follows: let S= M(a1, . . . , an) where ai = ±1 for1 ≤ i ≤ n. If S is of the form M(a1, . . . , an; b) then S has trivial divisor class group and isfactorial. If S is of the form M(a1, . . . , an; b1, . . . , bk) then a class group argument as thatoffered above shows that S is half-factorial.

For n = 2 it is easy to characterize half–factoriality as the following simple result shows.

Proposition 2.7. Let S= M(a1, a2, b) and set a′1 = a1/c2, a′2 = a2/c1 and b′ = b/c. S ishalf-factorial if and only if either

1. b′ ≤ 2, or

2. b′ | (a′1 − a′2).

Proof. (⇐) If b′ ≤ 2 then the divisor class group of M(a1, a2, b) is trivial or Z2 and theresult follows by [17, Proposition 2]. If b′ | (a′1−a′2) then [e1] = [e2] and results follows from[8, Theorem 3.1]. (⇒) We use here another theorem of Skula [19]: if H is a Krull monoidwith divisor class group G and set of primes P then for every p0 ∈ P , we have

G = 〈{[p] | p ∈ P\{po}}〉.

If neither 1 or 2 of the Proposition holds, then [e1] 6= [e2] and by the Skula result both [e1]and [e2] lie in generators of the cyclic class group. M(a1, a2, b) is not half-factorial by [8,Theorem 3.8].

3 Special monoids

Let d1, . . . , dk, b be finitely many integers with gcd(d1, . . . , dk) = 1 and

1 ≤ d1 < · · · < dk < b = β · lcm(d1, . . . , dk) for some β ∈ N+: = N\{0}.

We call

S := S(d1, . . . , dk; b) := {(r1, . . . , rk+1) ∈ Nk+1|k∑

i=1

ridi = rk+1b}

a special monoid. S is an additive subsemigroup of Nk+1 and we are interested in criteria ford1, . . . , dk, b so that S is half– factorial. Considering factoriality, this one is easy to check.

For, by Corollary 2.4 a special monoid S is factorial iff b =k∏

i=1ci where ci = gcd{dj}j 6=i. As

an elementary calculation shows the latter condition is equivalent to bk−1 =k∏

i=1di. Thus,

10

a given special monoid S = S(d1, . . . , dk; b) is factorial iffk∏

i=1di = bk−1. It is much more

difficult to characterize half–factorial special monoids. We start with

Lemma 3.1. A special monoid S = S(d1, . . . , dk; b) is half–factorial if and only if rk+1 = 1for every irreducible element (r1, . . . , rk+1) ∈ S.

Proof: With βi := bdi∈ N, 1 ≤ i ≤ k, we get the irreducible elements

ai := (0, . . . , 0, βi, 0, . . . , 0, 1) ∈ S, 1 ≤ i ≤ k. Assume that S is half–factorial and thatc = (r1, . . . , rk+1) ∈ S is irreducible. The relation

b(c−k∑

i=1

ri

βiai) = 0 = bc−

k∑

i=1

ridiai

yields

rk+1b =k∑

i=1

ridi = b,

hence rk+1 = 1.To verify the other direction let cj := (rj1, . . . , rjk, 1) be irreducible elements in S,

1 ≤ j ≤ t, and assume that∑t

j=1 mjcj = 0 for some mj ∈ Z. Summing up the last compo-nents gives

∑tj=1 mj = 0. 2

We note a few elementary facts for a special monoid S = S(d1, . . . , dk; b):

1) If (r1, . . . , rk,m) ∈ S is irreducible then m = 1 or ridi < b for 1 ≤ i ≤ k. In particularm ≤ max{1, k − 1}.

2) From 1) and Lemma 3.1 it follows immediately that S is half–factorial for k ≤ 2.

3) If S(d1, . . . , dk; b) is half–factorial then S(d1, . . . , dk; rb) is half–factorial for every r ∈N+.

4) S = S(1, d2, d3; b) is half–factorial: Let (r1, r2, r3,m) ∈ S be irreducible and assumethat m = 2. Then

r1 + r2d2 = 2b− r3d3 > b

and0 < r′1 := r1 + r2d2 − b = r1 − (b− r2d2) < r1.

It follows that(r′1, 0, r3, 1) ∈ S, (r1 − r′1, r2, 0, 1) ∈ S,

and(r1, r2, r3,m) = (r1, r2, r3, 2) = (r′1, 0, r3, 1) + (r1 − r′1, r2, 0, 1),

contrary to the irreducibility of (r1, r2, r3,m). Therefore m = 1.

11

5) We shall see later (Corollary 4.5 of Theorem 4.1) that S(d1, d2, d3; b) is always halffactorial. But for k ≥ 3 there is a standard procedure of constructing special monoidsS(1, d1, . . . , dk; b) which are not half factorial:Let q1 > q2 > · · · > qk ≥ 2 be finitely many natural numbers with gcd(qi, qj) = 1 for

1 ≤ i < j ≤ k and a :=k∑

i=1

1qi≥ 1. Define b :=

k∏

i=1

qi and di := bqi

, 1 ≤ i ≤ k.

Proposition: The special monoid M = M(1, d1, . . . , dk; b) is not half factorial andcontains an irreducible element of the form (1, t1, . . . , tk,m) with m > a.

Proof: LetA := {(r1, . . . , rk) ∈ Nk | 0 ≤ ri ≤ qi − 1, 1 ≤ i ≤ k}

and define

φ : A → Z/bZ, φ(r1, . . . , rk) :=k∑

i=1

ridimod b.

By our assumptions it is immediate that φ is injective, hence bijective.

Take (t1, . . . , tk) ∈ A withk∑

i=1

tidi ≡ −1 mod b. Then 1 +k∑

i=1

tidi = mb for some

m ∈ N and x := (1, t1, . . . , tk,m) ∈ M . Note that ti ≥ 1 for 1 ≤ i ≤ k since ti = 0would lead to the contradiction qi|1.

We claim that x is irreducible in M and that m > a, thereby proving the theorem.

Assume that x = y + z, y, z ∈ M . Then y (or z) is of the form y = (0, r1, . . . , rk, r)0 ≤ ri ≤ qi − 1, 1 ≤ i ≤ k. But qi|ridi and therefore qi|ri, since gcd(qi, di) = 1. Itfollows that ri = 0 and y = 0. We have

m =1b

(1 +

k∑

i=1

tidi

)=

1b

(1 + b

k∑

i=1

tiqi

)

=1b

+k∑

i=1

tiqi≥ 1

b+

k∑

i=1

1qi

> a.

2

For k = 3 and q1 = 5, q2 = 3, q3 = 2 we obtain the special monoid S = S(1, 6, 10, 15; 30)which contains the irreducible element (1, 4, 2, 1, 2).

6) Let S = S(d1, . . . , dk; b) be a special monoid, k ≥ 3. If there existsi0 ∈ I := {1, . . . , k} with the property

q := gcd(di, dj) = gcd(d1, . . . , di0−1, di0+1, . . . , dk)

12

for all i, j ∈ J := I\{i0}, then S is half–factorial if and only ifSq := S(d1

q , . . . ,di0−1

q , di0 ,di0+1

q , . . . , dkq ; b

q ) is half–factorial.For a proof note that S→ Sq,

(r1, . . . , rk,m) 7→ (r1, . . . , ri0−1,ri0q , ri0+1, . . . , rk, m), is an isomorphism between S

and Sq. 2

The special monoid S(d1, . . . , dk; b) is half–factorial if the principal ideals Zdi ⊂ Z forma descending chain Zd1 ⊃ Zd2 ⊃ . . . ⊃ Zdk in Z:

Lemma 3.2. If di|dj for 1 ≤ i ≤ j ≤ k, then S = S(d1, . . . , dk; b) is half–factorial.

Proof: We proceed by induction on b and may therefore assume that d1 = 1. If di = γid2,2 ≤ i ≤ k, then 1 = γ2 < γ3 < · · · < γk; moreover γi|γj for 2 ≤ i, j ≤ k. Let (r1, . . . , rk,m) ∈S be irreducible,

r1 +k∑

i=2

ridi = mb = r1 + (k∑

i=2

riγi)d2.

It follows that r1 = ad2 for some a ∈ N and

(a + r2, r3, . . . , rk, mb

d2) ∈ S(1, γ2, . . . , γk;

b

d2) =: S.

By induction S is half–factorial, so there are natural numbers r1 ≤ a + r2, r′3 ≤r3, . . . , r

′k ≤ rk with

r1 +k∑

i=3

r′iγi =b

d2.

Then

r1d2 +k∑

i=3

r′idi = b

with r1d2 ≤ r1 + r2d2. Choosing r′1, r′2 ∈ N with r′1 ≤ r1, r′2 ≤ r2 and r1d2 = r′1 + r′2d2 we

finally get (r′1, r′2, . . . , r

′k, 1) = (r1, r2, . . . , rk,m). 2

Corollary 3.3. S = S(d1, . . . , dk; b) is half–factorial if b = pm for some prime number pand m ∈ N. 2

4 A sufficient condition for half–factoriality

In this section we prove the half factoriality of special monoids S(d1, . . . , dk; b) under theassumption that the numbers d1, . . . , dk are relatively small compared tob ∈ lcm(d1, . . . , dk) · N.

13

Theorem 4.1. Let S = S(d1, . . . , dk; b) be a special monoid with the property

(∗) max

{dk−1

k∑

i=2

di, dk

k−1∑

i=1

di

}< b +

k∑

i=1

di.

Then S is half–factorial.

Before starting the proof we note a few implications of the condition (∗).

We fix a special monoid S = S(d1, . . . , dk; b) and assume that k ≥ 3 and that d1 > 1 ifk = 3 (cf. elementary facts 2) and 4) above). Moreover we fix an element (r1, . . . , rk,m) ∈ Swith m ≥ 2.

Lemma 4.2. If (∗) is true, then

i0 := max{i ∈ {2, . . . , k} | ri ≥ dk−1},i1 := max{i ∈ {1, . . . , i0 − 1} | ri ≥ di0}

exist.

Proof: Since

r1d1 +(dk−1−1)k∑

i=2

di ≤ (b

d1−1)d1 +(dk−1−1)

k∑

i=2

di = b+dk−1

k∑

i=2

di−k∑

i=1

di < 2b ≤ mb,

it follows that i0 exists.

For i0 ≤ k − 1 we get

(di0 − 1)i0−1∑

i=1

di + ri0di0 + (dk−1 − 1)k∑

i=i0+1

di ≤ (dk−1 − 1)k∑

i=1i6=i0

di + (b

di0

− 1)di0

≤ (dk−1 − 1)k∑

i=2

di + b− di0 ≤ dk−1

k∑

i=2

di −k∑

i=1

di + b < mb

and for i0 = k:

(dk − 1)k−1∑

i=1

di + rkdk ≤ (dk − 1)k−1∑

i=1

di + (b

dk− 1)dk = dk

k−1∑

i=1

di −k∑

i=1

di + b < mb.

Therefore i1 also exists. 2

We define q := gcd(di0 , di1) ∈ N and γi, δi ∈ N with ri = γiq + δi, 0 ≤ δi < q, 1 ≤ i ≤ k.

14

Lemma 4.3. The following inequalities hold if (∗) is true:

α) If i0 ≥ 3, then dk−1di0 +k∑

i=i0+1

ridi < b

β) If i0 = 2, then (dk−1 − 2)d2 +k∑

i=3

ridi < b

γ) (di0

q− 1)di1 +

k∑

i=1

δidi < (m− 1)b + δi1di1 + δi0di0.

Proof: α) We assume at first that k = i0 ≥ 3. Then

dk−1dk + dk

k−2∑

i=1

di = dk

k−1∑

i=1

di < b +k∑

i=1

di,

dk−1dk < b + (1− dk)k−2∑

i=1

di + dk−1 + dk.

We have to show that

(dk − 1)k−2∑

i=1

di ≥ dk−1 + dk.

If k ≥ 4 or d1 ≥ 3 thenk−2∑

i=1

di ≥ 3 and

(dk − 1)k−2∑

i=1

di ≥ 3(dk − 1) = 2dk − 1 + (dk − 2) ≥ 2dk − 1 = (dk − 1) + dk ≥ dk−1 + dk.

If k = 3 and d1 = 2 we know by (∗) that

d2d3 + d22 < b + d1 + d2 + d3 = b + 2 + d2 + d3

d2d3 + 2d3 < b + d1 + d2 + d3 = b + 2 + d2 + d3

We have to show that d22 ≥ 2 + d2 + d3 or d3 ≥ 2 + d2. Assume that d3 = 1 + d2. Note

that d2 ≥ 3. Therefore4 ≤ (d2 − 1)2 = d2

2 − 2d2 + 1.

In particular d22 ≥ 2d2 + 3 = 2 + d2 + (d2 + 1) = 2 + d2 + d3 as desired. This proves α) in

case i0 = k.

Now we assume that k − 1 ≥ i0 ≥ 3. We get

15

dk−1di0 +k∑

i=i0+1

ridi ≤ dk−1di0 + (dk−1 − 1)k∑

i=i0+1

di = dk−1

k∑

i=i0

di −k∑

i=i0+1

di,

hence

dk−1di0 +k∑

i=i0+1

ridi < b +k∑

i=1

di −i0−1∑

i=2

didk−1 −k∑

i=i0+1

di = b +i0∑

i=1

di − dk−1

i0−1∑

i=2

di.

It suffices to show that

(dk−1 − 1)i0−1∑

i=2

di ≥ d1 + di0 .

We have

(dk−1 − 1)i0−1∑

i=2

di ≥ (dk−1 − 1)d2 ≥ 2(dk−1 − 1)

and2(dk−1 − 1) ≥ d1 + dk−1 ≥ d1 + di0

since dk−1 ≥ d3 ≥ 2 + d1. The proof of α) is finished.

β) Assume that i0 = 2. We have

(dk−1 − 2)d2 +k∑

i=3

ridi ≤ (dk−1 − 2)d2 + (dk−1 − 1)k∑

i=3

di

= dk−1

k∑

i=2

di − d2 −k∑

i=2

di < b +k∑

i=1

di − d2 −k∑

i=2

di

= b + d1 − d2 < b.

γ) We have

(di0

q− 1)di1 +

k∑

i=1

δidi ≤ (di0

q− 1)di1 + (q − 1)

k∑i=1

i6=i0,i6=i1

di + δi1di1 + δi0di0 .

Hence it suffices to show that

(di0

q− 1)di1 + (q − 1)

k∑i=1

i 6=i0,i6=i1

di < (m− 1)b.

This is true for q = 1 since

(di0 − 1)di1 < ri1di1 < b ≤ (m− 1)b.

16

Assume now that q ≥ 2. Then

(di0

q− 1)di1 + (q − 1)

k∑i=1

i6=i0,i 6=i1

di ≤ (di0

2− 1)di1 + (di1 − 1)

k∑i=1

i6=i0,i6=i1

di

= di1

k∑i=1i6=i1

di − di1di0

2−

k∑i=1i 6=i0

di ≤ dk−1

k∑

i=2

di −k∑

i=1i6=i0

di − di1di0

2

< b +k∑

i=1

di −k∑

i=1i6=i0

di − di1di0

2= b + di0 −

di1di0

2

= b + di0(1−di1

2) ≤ b ≤ (m− 1)b,

since di1 ≥ q ≥ 2. 2

Proof of Theorem 4.1: Assume that S = S(d1, . . . , dk; b) is not half–factorial andthat (r1, . . . , rk,m) ∈ S is irreducible with m ≥ 2. We know already that this implies k ≥ 3and d1 > 1 if k = 3. Our assumption will lead to a contradiction. Let i0, i1, q, γi, δi bedefined as above. We distinguish the cases i0 ≥ 3 and i0 = 2.

a) i0 ≥ 3:

By statement α) of lemma 4.3 we have

(m− 1)b = mb− b < mb− (dk−1di0 +k∑

i=i0+1

ridi) =k∑

i=1

ridi − (dk−1di0 +k∑

i=i0+1

ridi)

= ri1di1 + (ri0 − dk−1)di0 +i0−1∑i=1i6=i1

ridi = ri1di1 + (ri0 − dk−1)di0 +i0−1∑i=1i 6=i1

(γiq + δi)di

≤ ri1di1 + (ri0 − dk−1)di0 +i0−1∑i=1i6=i1

(γiq + δi)di +k∑

i=i0+1

δidi

= ri1di1 + (ri0 − dk−1)di0 + q

i0−1∑i=1i6=i1

γidi +k∑

i=1i6=i0,i 6=i1

δidi =: x,

and by statement γ) of Lemma 4.3:

x > (m− 1)b > (di0

q− 1)di1 +

k∑i=1

i6=i0,i6=i1

δidi.

17

Remember that ri1 >di0q − 1 and that ri0 − dk−1 ≥ 0. The two inequalities show that we

can find

r′i1 ∈ N, ri1 ≥ r′i1 ≥di0

q− 1, r′i0 ∈ N, r′i0 ≤ ri0 − dk−1,

γ′i ∈ N, γ′i ≤ γi, 1 ≤ i ≤ i0 − 1, i 6= i1,

such that

(m− 1)b ≥ y := r′i1di1 + r′i0di0 + q

i0−1∑i=1i6=i1

γ′idi +k∑

i=1i 6=i0,i6=i1

δidi > (m− 1)b− qdi0 .

Since q divides di1 and di0 it follows that q dividesk∑

i=1i6=i0,i 6=i1

δidi = mb − ri0di0 − ri1di1 − qk∑

i=1i 6=i0,i6=i1

γidi. In particular, q divides x, y and α :=

(m − 1)b − y < qdi0 . Note that α > 0 since α = 0 would contradict our assumption that(r1, . . . , rk,m) is irreducible.

There are s, t ∈ N, 0 ≤ t ≤ di0q − 1 with α = sdi0 − tdi1 . In particular

(r′i1 − t)di1 + (r′i0 + s)di0 +i0−1∑i=1i6=i1

(γ′iq + δi)di +k∑

i=i0+1

δidi = (m− 1)b.

It remains to show that r′i1 − t ≥ 0 and r′i0 + s ≤ ri0 .Whereas the first inequality is trivial we will verify the second inequality. From

qdi0 > α = sdi0 − tdi1 ≥ sdi0 − (di0

q− 1)di1 = di0(s−

di1

q) + di1

we haves− di1

q+

di1

di0

< q or s < q +di1

q− di1

di0

.

Since s ∈ N and di1q ∈ N it follows that

s ≤ q +di1

q− 1.

Thenr′i0 + s ≤ ri0 − dk−1 + q − 1 +

di1

q

and, since di1 ≤ dk−1, it suffices to show that

di1 + 1 ≥ q +di1

q.

18

But this equivalent to

di1(1−1q) ≥ q − 1.

Division by q leads todi1

q(1− 1

q) ≥ 1− 1

q.

This is true for q = 1. If q ≥ 2, this is equivalent to di1q ≥ 1, which is true because di1

q ∈ N+.This finishes the proof for i0 ≥ 3.

b) i0 = 2 :

Using Lemma 4.3, statement β), we get in the same way as above:

r1d1 + (r2 + 2− dk−1)d2 +k∑

i=3

δidi > (m− 1)b

(d2

q− 1)d1 +

k∑

i=3

δidi < (m− 1)b.

(**)

There are r′1, r′2 ∈ N with r1 ≥ r′1 ≥ d2

q − 1, r′2 ≤ r2 + 2− dk−1, such that

r′1d1 + r′2d2 +k∑

i=3

δidi = (m− 1)b− α,

where 0 < α < d2 and q|α.Let s, t ∈ N, 0 ≤ t ≤ d2

q − 1, with α = sd2 − td1. As above, it remains to show that

r′2 + s ≤ r2.

An easy calculation yields s ≤ d1q . Then

r′2 + s ≤ r2 + 2− dk−1 +d1

q

and it suffices to show thatdk−1 ≥ d1

q+ 2

ordk−1 − 1 ≥ d1

q+ 1.

If dk−1 ≥ 2 + d1, e.g. if k ≥ 4, it follows that

dk−1 − 1 ≥ d1 + 1 ≥ d1

q+ 1.

19

For k = 3 and dk−1 = d2 = 1 + d1 ≥ 3 we have q = 1 and r1 = ri1 ≥ di0 = d2 = d1 + 1 > d1.The inequalities (**) yield

r1d1 + (r2 + 1− d1)(d1 + 1) > b,

d21 < b.

In this special case, using the relation r1 > d1, we can find r′1, r′2 ∈ N with

2 ≤ d1 ≤ r′1 < r1

and r′2 ≤ r2 + 1− d1 ≤ r2, such that

b > y := r′1d1 + r′2(d1 + 1) > b− d1.

Again let s, t ∈ N, 0 ≤ t ≤ d1, with α := b− y = s(d1 + 1)− td1 ≤ d1− 1. Finally it sufficesto show that

s ≤ d1 − 1,

since this implies r′2 + s ≤ r2 + 1− d1 + s ≤ r2. The inequality

d1 − 1 ≥ α = s(d1 + 1)− td1 ≥ s(d1 + 1)− d21

yields

s ≤ d21 + d1 − 1d1 + 1

= d1 − 1d1 + 1

.

Hence s ≤ d1 − 1 since s ∈ N. This completes the proof of Theorem 4.1. 2

Corollary 4.4. The special monoid S = S(d1, . . . , dk; b) is half–factorial if k ≤ 2 or ifgcd(di, dj) = 1 for 1 ≤ i < j ≤ k.

Proof: We may assume that k ≥ 3 and that d1 ≥ 2 if k = 3 (cf. elementary facts 2)and 4)). Since b ∈ lcm(d1, . . . , dk) · N we may also assume (cf. elementary fact 3)) that

b =k∏

i=1

di.

It suffices to show that

1) dk−1

k∑

i=2

di ≤k∏

i=1

di,

2) dk

k−1∑

i=1

di ≤k∏

i=1

di.

ad 1):

20

i) k ≥ 5:d2 + · · ·+ dk ≤ (k − 1)dk ≤ (k − 2)(k − 3)dk ≤ dk−2dk−3dk

ii) k = 4:d2 + d3 + d4 ≤ 3d4 ≤ d2d4 if d2 ≥ 3.

For d2 = 2 we have d4 ≥ 2 + d3 since gcd(d4, 2) = gcd(d3, 2) = 1 and therefore

d2 + d3 + d4 = (2 + d3) + d4 ≤ 2d4 = d2d4.

iii) k = 3, d1 ≥ 2:d2 + d3 ≤ 2d3 ≤ d1d3.

ad 2)

i) k ≥ 5:

d1 + · · ·+ dk−1 ≤ (k − 1)dk−1 ≤ (k − 3)(k − 2)dk−1 ≤ dk−3dk−2dk−1.

ii) k = 4:d1 + d2 + d3 ≤ 3d3 ≤ d2d3 if d2 ≥ 3.

If d2 = 2, then d1 + d2 + d3 = 3 + d3 ≤ d3 + d3 = 2d3 = d2d3.

iii) k = 3, d1 ≥ 2:d1 + d2 ≤ 2d2 ≤ d1d2.

2

Corollary 4.5. A special monoid S = S(d1, d2, d3; b) is half–factorial.

Proof: By elementary fact 6) we may assume that gcd(di, dj) = 1 for 1 ≤ i < j ≤ 3. NowCorollary 4.4 applies. 2

5 The general situation

We want to characterize half–factorial monoids of the form

M :=M(a1, . . . , an; b) := {(r1, . . . , rn+1) ∈ Nn+1 |n∑

j=1

rjaj = rn+1b}

with a1, . . . , an, b ∈ N+. Let the monoidM =M(a1, . . . , an; b) be given, normalized accord-ing to Lemma 2.1, i.e. gcd{b∪{aj}j 6=i} = 1 for i = 1, . . . , n and gcd{aj | 1 ≤ j ≤ n} = 1 and

21

let aj = αjdj , b = βjdj , dj = gcd(aj , b) ∈ N, 1 ≤ j ≤ n. Let e1 := (1, 0, . . . , 0), . . . , en+1 :=(0, . . . , 0, 1) be the unit vectors in Qn+1.

By Proposition 2.2 the primary elements ofM are given by qi = βiei+αien+1, 1 ≤ i ≤ n.

For x =n+1∑

i=1

xiei let m(x) = lcm{ ord[ϕ(ei)]

gcd(ord[ϕ(ei)], xi)

∣∣∣xi > 0}

and mi = xim(x)

ord[ϕ(ei)]. By

using ϕ(qi) = ord [ϕ(ei)]ϕ(ei) from Proposition 2.2 we obtain

ϕ(m(x)x) =n∑

i=1

xim(x)ϕ(ei) =n∑

i=1

miϕ(qi) = ϕ( n∑

i=1

miqi

)

and, hence, m(x)x =n∑

i=1miqi. Thus, for each x ∈ M a certain multiple m(x)x splits

completely into primary elements, the number of which is n(x): =n∑

i=1mi. The rate z(x) =

n(x)m(x) of primary elements needed in this splitting per unit of x is also called the Zaks–Skulafunction (cf. [17], [19]).

In terms of splittings into primary elements half–factoriality can be characterized asfollows.

Proposition 5.1. The following statements for the monoid M = M(a1, . . . , an; b) areequivalent:

i) M is half–factorial.

ii) I(M) is contained in the simplex of convex combinations over Q of the primary ele-ments.

iii) The Zaks–Skula function is identically 1 on I(M).

iv) Every element (r1, . . . , rn+1) ∈ I(M) satisfies the equationn∑

j=1rjdj = b.

Proof. (i) ⇐⇒ (ii): If M is half–factorial and x ∈ I(M) then by the above considerations

m(x)x =n∑

i=1miqi with m(x) =

n∑i=1

mi. Therefore, x =n∑

i=1kiqi, ki = mi

m(x) ∈ Q, 0 ≤ ki

andn∑

i=1ki = 1. By Proposition 2.2 representation in terms of primary elements is unique

and, hence, the convex set generated by the qi over Q is a simplex. Conversely, from (ii)it follows that for any x ∈ I(M) a certain multiple of x splits into an equal number ofprimary elements. The uniqueness for the representation by primary elements yields half–factoriality.

(ii) ⇐⇒ (iii): z(x) = 1 means m(x)x =n∑

i=1miqi with m(x) =

n∑i=1

mi and, hence, arguments

of the previous step apply.

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(iii) ⇐⇒ (iv): We show that z(x) = 1b

n∑i=1

xidi for any x =n+1∑i=1

xiei in M. By definition

z(x) = n(x)m(x) =

∑i

mi

m(x)=

∑

i

xi

ord[ϕ(ei)]. Obviously, in Cl(M) ord [ϕ(ei)] = b

difor 1 ≤ i ≤ n

which gives the required formula.

Remarks:

1. Properties ii) and iv) of the above Proposition provide geometrical criteria for half–factoriality. To test the latter one has to check whether a certain simplex (over Q)or a certain affine subspace (over Z), both specified by the given data, contain allirreducible elements of the monoid.

2. The proof of the proposition also shows that the Zaks–Skula function, which is definedfor arbitrary Krull monoids with a torsion class group, can be explicitly calculated in

the case of a monoid M(a1, . . . , an; b) as z(x) = 1b

n∑i=1

xidi.

3. It is not difficult to see that for M =M(a1, . . . , an; b) one has that b = lcm {m(x) |x ∈ I(M)}. Thus, b is an upper bound for the possible values of m(x) for x ∈ I(M).Also, from a general result on Krull monoids [17, Theorem 1] it follows that z(x) forx ∈ I(M) is bounded from above by the so called cross number k

(Cl(M)

)of the

class group. A fortiori we have thatn∑

i=1xidi ≤ b k

(Cl(M)

)for all x ∈ I(M).

Proposition 5.1 we illustrate by coming back to Example 2.5.

Example 5.2. For M(2, 5; 3) the primary elements are q1 = (3, 0, 2) and q2 = (0, 3, 5).Since m(x) | b by Remark 3 above, for the non–primary irreducible elements x = (1, 2, 4)and y = (2, 1, 3) we must have m(x) = m(y) = 3. Indeed, 3x = q1 + 2q2 and 3y = 2q1 + q2

and, hence, all irreducible elements are contained in the one–dimensional simplex in Q3

spanned by q1 and q2. The affine subspace is given by x1 + x2 = 3 and also contains allirreducible elements. Concerning the slight variationM(1, 5; 3) the irreducible elements areq1 = (3, 0, 1), q2 = (0, 3, 5) and x = (1, 1, 2), where q1, q2 are primary and x is not. Again,m(x) = 3 and 3x = q1 + q2. Thus, x is not contained in the simplex spanned by q1 and q2.The affine subspace is the same as before containing q1 and q2 but not x.

Corollary 5.3. Let M =M(a1, . . . , an; b) be half–factorial and dj = gcd(aj , b), 1 ≤ j ≤ n.Then

1)∑n

j=1 rjdj ∈ b · N for every (r1, . . . , rn+1) ∈M2) aidj − ajdi ∈ bN for all i, j ∈ {1, . . . , n}.

Proof:

23

ad 1): The statement follows directly from Proposition 5.1.

ad 2): Take m ∈ N with mb− aj ∈ N for 1 ≤ j ≤ n. Then

aimb = ai(mb− aj) + aiaj for all i, j ∈ {1, . . . , n}.

Therefore by 1):

(mb− aj)di + aidj = mb− (aidj − ajdi) ∈ bN.

2

To simplify the notation, we fix the following setting which can always be arranged bya permutation of the coefficients: Let aj , b ∈ N+, 1 ≤ j ≤ n, dj := gcd(aj , b) ∈ N+ andk, s ∈ {1, . . . , n}, k ≤ s, with the property

1 ≤ d1 < d2 < · · · < dk , {di|1 ≤ i ≤ k} = {dj |1 ≤ j ≤ n},ai 6= aj , 1 ≤ i < j ≤ s , {ai|1 ≤ i ≤ s} = {aj |1 ≤ j ≤ n}.

We call

R :=M(a1, . . . , as; b) := {(r1, . . . , rs+1) ∈ Ns+1 |s∑

j=1

rjaj = rs+1b}

the reduction of M(a1, . . . , an; b).If q := gcd(a1, . . . , an, b) = gcd(a1, . . . , as, b) = gcd(d1, . . . , dk, b), thenM(a1, . . . , an; b) =

M(a1q , . . . , an

q ; bq ) and therefore we may (and do) assume that gcd(d1, . . . , dk, b) = 1. If

dk = b, ak = αb, then

φ :M(a1, . . . , an; b) −→ N×M(a1, . . . , ak−1, ak+1, . . . , an; b),φ(r1, . . . , rn+1) := (rk, r1, . . . , rk−1, rk+1, . . . , rn, rn+1 − αrk),

is an isomorphism of semigroups. In particular M(a1, . . . , an; b) is half–factorial if and onlyif M(a1, . . . , ak−1, ak+1, . . . , an; b) is half–factorial. Therefore we assume additionally thatdk < b. We note:

Lemma 5.4. The monoidM =M(a1, . . . , an; b) is half–factorial if and only if its reductionR =M(a1, . . . , as; b) is half–factorial.

For the proof we need an auxiliary statement:

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Lemma 5.5. Let T be an abelian semigroup and ϕ : T → N a semigroup morphism. Thesemigroup

H := {(n, z) |n ∈ N, z ∈ T, ϕ(z) ≥ n}is half–factorial if and only if T is half–factorial.

Proof: Let yj = (nj , zj) ∈ H be irreducible elements, 1 ≤ j ≤ t, with∑t

j=1 mjyj = 0,mj ∈ Z. In particular

∑tj=1 mjzj = 0. The elements zj are irreducible in T since otherwise a

decomposition zj = u+w, u,w ∈ T , u 6= zj 6= w, would lead to nj ≤ ϕ(zj) = ϕ(u)+ϕ(w) andto α, β ∈ N with α ≤ ϕ(u), β ≤ ϕ(w), α+β = nj . In particular yj = (nj , zj) = (α, u)+(β, w)would be reducible, contrary to our assumption.

If T is half–factorial then∑t

j=1 mjzj = 0 implies∑t

j=1 mj = 0. If T is not half–factorialand

∑tj=1 mjzj = 0 for some irreducible elements zj ∈ T ,

∑tj=1 mj 6= 0, then (0, zj) ∈ H is

irreducible, 1 ≤ j ≤ t, and∑t

j=1 mj(0, zj) = 0. 2

Proof of Lemma 5.4: We proceed by induction on n ≥ s. Assuming that n ≥ s + 1we choose j0 ∈ {1, . . . , s} with an = aj0 . Define M := M(a1, . . . , an−1; b) and ϕ : M→ N,ϕ(s1, . . . , sn−1, sn) := sj0 . Then

H := {(n, z) ∈ N× M, ϕ(z) ≥ n}

is a semigroup and is isomorphic to M via

φ :M→ H,φ(r1, . . . , rn+1) := (rn, r1, . . . , rj0−1, rj0 + rn, rj0+1, . . . , rn−1, rn+1),

with

φ−1(m, s1, . . . , sn) = (s1, . . . , sj0−1, sj0 −m, sj0+1, sn−1,m, sn).

This implies, together with Lemma 5.5:

M half factorial ⇐⇒ H half–factorial ⇐⇒M half–factorial.

Note that R = M(a1, . . . , as; b) is also the reduction of M = M(a1, . . . , an−1; b). The in-duction step finishes the proof. 2

Let M = M(a1, . . . , an; b) be a monoid and k ≤ s as above. We call the specialmonoid S = M(d1, . . . , dk; b) the special reduction of M. Since S is the reduction ofS =M(d1, . . . , dn; b) it follows that S is half–factorial if and only if S is half–factorial. Wecall M =M(a1, . . . , an; b) a regular monoid if

b|(aidj − ajdi) for all i, j ∈ {1, . . . , n}, dj := gcd(aj , b).

25

Lemma 5.6. A regular monoid M = M(a1, . . . , an; b) is isomorphic to the monoid S =M(d1, . . . , dn; b). The morphism

ϕ :M→ S, ϕ(r1, . . . , rn+1) := (r1, . . . , rn,1b

n∑

j=1

rjdj)

is bijective and ϕ−1(s1, . . . , sn+1) = (s1, . . . , sn, 1b

∑sjai).

Proof: It suffices to show that for all (r1, . . . , rn+1) ∈ Nn+1 it is true that

1b

n∑

j=1

rjdj ∈ N⇐⇒ 1b

∑rjaj ∈ N.

Let aj = αjdj , b = βjdj , kij := aidj−ajdi

b ∈ N, 1 ≤ i, j ≤ n. Note that b = lcm(β1, . . . , βn)because of our assumption gcd(d1, . . . , dn) = 1. For (r1, . . . , rn+1) ∈ Nn+1 and every i ∈{1, 2, . . . , n} we have

n∑

j=1

rjaj =n∑

j=1

rj(aidj − kijb

di) =

ai

di

n∑

j=1

rjdj − b

di

n∑

j=1

rjkij = αi

n∑

j=1

rjdj − βi ·n∑

j=1

rjkij .

If∑n

j=1 rjdj ∈ bN, then βi|∑n

j=1 rjaj for 1 ≤ i ≤ n und therefore∑n

j=1 rjaj ∈ bN.If

∑nj=1 rjaj ∈ bN, then βi|

∑nj=1 rjdj for 1 ≤ i ≤ n, since gcd(αi, βi) = 1. Hence∑n

j=1 rjdj ∈ bN. 2

A regular monoid is half–factorial if and only if its special reduction is half factorial:

Proposition 5.7. The following statements about a monoid M = M(a1, . . . , an; b) areequivalent:

i) M is half–factorial.

ii) M is regular and its special reduction S =M(d1, . . . , dk; b) is half–factorial.

Proof: If M is regular then it is isomorphic to S = M(d1, . . . , dn; b) (Lemma 5.6)Moreover S is half–factorial if and only if its reduction S is half–factorial (Lemma 5.4) 2

Remark: Proposition 5.7 does not hold when “half–factorial” is replaced by “factorial”.For example, M =M(1, 1; 2) is not factorial by Corollary 2.4. But M is obviously regularand its special reduction S = M(1; 2) is factorial. This example shows also that a half–factorial monoid may possess a factorial special reduction.

Finally, we collect our results and give a list of sufficient conditions for a monoid M =M(a1, . . . , an; b) to be half–factorial. The conditions are not independent from each other.

26

Theorem 5.8. Let M = M(a1, . . . , an; b) be a regular monoid with special reduction S =S(d1, . . . , dk; b). M is half–factorial if at least one of the following conditions is satisfied:

1) k ≤ 3

2) di|dj for 1 ≤ i, j ≤ k

3) b = pm or b = pq for prime numbers p, q and m ∈ N.

4) gcd(di, dj) = 1 for 1 ≤ i < j ≤ k

5) There exists i0 ∈ I := {1, . . . , k} with gcd(di0 , dj) = 1 andgcd(di, dj) = gcd(d1, . . . , di0−1, di0+1, . . . , dk) for all i, j ∈ J := I\{j0}, i 6= j.

6) max{dk−1

k∑

i=2

di, dk

k−1∑

i=1

di} < b +k∑

i=1

di.

2

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