Failure of the length-set structure theorem
for weakly C-monoids
Henry Zweiman
September 9, 2026
Abstract
We construct a reduced weakly C-monoid with factorial complete integral closure, nonempty conductor, and distance set exactly which does not satisfy the Structure Theorem for Sets of Lengths. The monoid is the union of a free rank-two boundary and a principal conductor ideal in a free abelian monoid of rank four. We prove its Mori property by a six-element colon witness and determine all of its length sets. An explicit sequence of elements has length sets consisting of two residue classes modulo three and one additional endpoint. Every almost arithmetical multiprogression representation of the th set, with difference and bound , satisfies . This gives a negative answer to Problem 20 of Geroldinger, Kim, and Loper for weakly C-monoids in Kainrath’s stated sense.
1. Introduction and main result
The Structure Theorem for Sets of Lengths describes all length sets of a monoid by almost arithmetical multiprogressions with a common bound and differences from a finite set. It holds for C-monoids and for tame monoids; the latter result is due to Geroldinger and Kainrath [2]. Geroldinger, Kim, and Loper ask whether it holds for every weakly C-monoid in the sense of Kainrath [3, Problem 20]. We give a counterexample.
All monoids below are commutative and cancellative, with identity. The notation denotes the free abelian monoid on . Put
Theorem 1.1.
The monoid in (1) is a reduced Mori weakly C-monoid with
It does not satisfy the Structure Theorem for Sets of Lengths. More precisely, for every the element
has length set
If is an almost arithmetical multiprogression with difference and bound , then
Intervals in this paper contain integers only. The counterexample concerns the full definition of weakly C-monoid in [4, Definition 5.1]. That definition does not require local tameness or the stronger condition (C) introduced in the same chapter. We make no assertion that a version restricted by either additional hypothesis fails.
2. Definitions and the Mori verification
For , write
where is the quotient group. An integral divisorial ideal is an ideal with . The monoid is Mori, or -Noetherian, if these ideals satisfy the ascending chain condition. Its complete integral closure is
We recall the part of the weak-C definition that must be checked. For a factorial monoid , two elements are -equivalent if exactly when for every ; write for the equivalence class. A submonoid is weakly C in [4, Definition 5.1] if:
is Mori, , and is saturated and cofinal;
there are an equivalence relation on with finitely many classes and an integer such that, whenever all lie in one class, some satisfies
In particular, the second condition is automatic for a finite ambient prime set: take equality for and .
Define the weight of to be , also for integral exponents in the quotient group.
Lemma 2.1.
For every nonempty , there is with and . Consequently is Mori.
Proof.
Choose at most four elements attaining the minimum of each of the four ambient prime valuations on . Write and choose one of these witnesses, , with . If contains an element with -valuation , add one. If the weights of the parts of the elements at -valuation are not all equal, add one whose weight differs from that of . Call the resulting set .
Suppose . The valuation minima imply . Put . If , every product belongs to . The case is impossible, since then.
Suppose . The witness at valuation , if present, would give a product of -valuation one. Thus there are no such elements. At valuation , membership requires the sum of the two weights to be zero. The weight witness either makes this impossible or ensures that the same equation holds for all elements of at this valuation. It already holds for . All other elements give products of -valuation at least two. Hence , as required.
Thus for a finite subset of each nonempty . To see ACC directly, take the union of an ascending chain of integral divisorial ideals. Its finite witness lies in one member of the chain, whose divisorial closure therefore contains the union. The chain stabilizes.
Proposition 2.2.
The monoid is reduced and weakly C, with , , and conductor .
Proof.
The set in (1) is a submonoid, and its only unit is . The elements show that its quotient group contains all four ambient primes. Every element is almost integral over , since for all . An element of with a negative ambient valuation cannot be almost integral over . This proves .
The ideal is in the conductor. No boundary element is in the conductor, since multiplying it by leaves . Thus the conductor is exactly . The closure embedding is the identity on , hence saturated and cofinal. Lemma 2.1 and the equality partition of the four primes now verify both weak-C conditions.
3. All factorization lengths
For an atomic monoid and , let be the set of numbers of atoms in factorizations of , with . In our monoid these sets are nonempty and finite: the sum of the four coordinates decreases in any proper factorization and bounds every factorization length. For a finite set , its distances are the differences between successive elements in increasing order; is their union over .
Call a monomial in boundary-free if it has no nonunit divisor from in the ambient monoid.
Lemma 3.1.
The atoms of are and precisely the elements
Proof.
The boundary is free abelian on by (4), and a boundary factorization cannot use positive -degrees. At -degree two or three, a proper factorization must remove a boundary atom. The displayed conditions exclude exactly these divisors. At -degree at least four, split off to obtain two nonunits in .
For , define
Notice that .
Proposition 3.2.
The length sets of are given by the following formulas.
If , then .
If or , then
If , then
Moreover .
Proof.
The first assertion follows from the freeness of . A factorization at -degree two or three uses exactly one positive- atom. If it contains copies of and copies of , its remaining exponents are
They must be nonnegative and boundary-free. After copies of , maximality with respect to forces
If , the residual -exponent is less than two, so maximality with respect to follows. If , it requires either or . Together with feasibility these conditions give exactly the range in (5). This proves (7).
For , the capacity no longer binds and
These values decrease successively by one or two. If the first value is capacity-limited and there is a next value, write with . The first difference is
Otherwise the same floor formula applies from the beginning. Thus is nonincreasing and the maximum number of boundary atoms whose product divides is . A maximum packing is maximal, and every smaller packing size is obtained by taking a subset of one of maximum size.
Now fix . If a factorization uses positive- atoms, then
Every such is possible, since is a sum of integers each equal to two or three. For any packing of boundary atoms, split its remaining monomial into its -part and its -part. Both are boundary-free. Adjoin factors equal to , and assign the -degrees just described. This gives every length in . The reverse inclusion follows because the boundary factors are a packing. Taking the union over consecutive gives (8).
The first-layer sets have distances at most two, the higher-layer sets are intervals, and boundary length sets are singletons. Finally, (7) at gives , exhibiting both distances.
4. An endpoint that prevents uniform structure
We use the standard AAMP definition from [2, Section 5]. A nonempty finite set is an almost arithmetical multiprogression of difference , period , and bound if
where , , , and . Equivalently there is a core interval with endpoints in , with both tails of width at most , on which the permitted periodic residue pattern is complete. The whole of lies in that same periodic pattern. This last containment, including the tails, is essential.
The Structure Theorem requires a single bound and a finite nonempty set of permitted differences for all length sets. Bounds and differences are unchanged by translating a finite set.
Lemma 4.1.
For , let be the set in (2). Every AAMP representation of with difference and bound satisfies (3).
Proof.
Let be its core interval. Since and , we have
Put . Suppose that contains at least consecutive integers. On , membership in is the pattern accepting residues zero and one modulo three. It is also invariant under translation by whenever both points lie in the core. Comparing three consecutive pairs in proves : a nonzero translation modulo three does not preserve the subset .
The endpoint belongs to . By the containment in (9), its residue modulo is permitted throughout the core. Since contains at least consecutive integers, it contains an integer in that residue class. This integer is two modulo three, hence is missing from . This is a contradiction.
Therefore . At most the endpoint lies in , giving . We conclude that
The structural assertions follow from Proposition 2.2, and the distance set from Proposition 3.2. At , the cutoffs are , . The first packing length is ; all others are
Separating odd and even gives (2). Lemma 4.1 proves (3). If the Structure Theorem held with common bound and differences bounded by , then (3) would give for every , which is impossible.
5. Scope and further questions
The example separates finite distances from uniform length structure inside the class of weakly C-monoids, even with factorial complete integral closure and a conductor principal in that closure. Its boundary is itself factorial, and every length set at -degree at least four is an interval. Thus the obstruction is confined to the first conductor layers and to the requirement that the AAMP tails respect the period of the core.
The union of a lattice boundary and a conductor ideal is related to the non-locally-tame example of Geroldinger and Hassler [1, Example 6.11]. The particular two-generator boundary in (1) permits an explicit incompatible endpoint. The result does not contradict the theorem for tame monoids [2]. Nor does it settle the length-structure problem under the stronger condition (C) discussed by Kainrath [4]. Identifying additional algebraic hypotheses that eliminate these endpoint obstructions remains a natural next question.
The accompanying standard-library Python script checks the atom characterization by direct decompositions, compares the full length formulas with independent factorizations in a finite box, and checks the ray formula and obstruction inequality in finite ranges. The proofs above, including the Mori property and the unbounded obstruction, do not depend on computation.
References
[1] A. Geroldinger and W. Hassler, Arithmetic of Mori domains and monoids, Journal of Algebra 319 (2008), 3419–3463. doi:10.1016/j.jalgebra.2007.11.025.
[2] A. Geroldinger and F. Kainrath, On the arithmetic of tame monoids with applications to Krull monoids and Mori domains, Journal of Pure and Applied Algebra 214 (2010), 2199–2218. doi:10.1016/j.jpaa.2010.02.023.
[3] A. Geroldinger, H. Kim, and K. A. Loper, On Long-Term Problems in Multiplicative Ideal Theory and Factorization Theory, arXiv:2502.21020v3, February 28, 2026.
[4] F. Kainrath, Arithmetic of Mori domains and monoids: the global case, in S. T. Chapman, M. Fontana, A. Geroldinger, and B. Olberding (eds.), Multiplicative Ideal Theory and Factorization Theory, Springer Proceedings in Mathematics & Statistics 170, Springer, 2016, 183–218. doi:10.1007/978-3-319-38855-7_8.