MnLargeSymbols’164 MnLargeSymbols’171

# On the finiteness of certain factorization invariants

Laura Cossu  (L. Cossu) Department of Mathematics and Scientific Computing, University of Graz | Heinrichstrasse 36/III, 8010 Graz, Austria<EMAIL_ADDRESS>https://sites.google.com/view/laura-cossu and  Salvatore Tringali  (S. Tringali) School of Mathematical Sciences, Hebei Normal University | Shijiazhuang, Hebei province, 050024 China<EMAIL_ADDRESS>http://imsc.uni-graz.at/tringali

###### Abstract.

Let $H$ be a monoid, $\mathscr{F}(X)$ be the free monoid on a set $X$, and
$\pi_{H}$ be the unique extension of the identity map on $H$ to a monoid
homomorphism $\mathscr{F}(H)\to\allowbreak H$. Given $A\subseteq H$, an
$A$-word $\mathfrak{z}$ (i.e., an element of $\mathscr{F}(A)$) is minimal if
$\pi_{H}(\mathfrak{z})\neq\allowbreak\pi_{H}(\mathfrak{z}^{\prime})$ for every
permutation $\mathfrak{z}^{\prime}$ of a proper subword of $\mathfrak{z}$. The
minimal $A$-elasticity of $H$ is then the supremum of all rational numbers
$m/n$ with $m,n\in\allowbreak\mathbb{N}^{+}$ such that there exist minimal
$A$-words $\mathfrak{a}$ and $\mathfrak{b}$ of length $m$ and $n$, resp., with
$\pi_{H}(\mathfrak{a})=\pi_{H}(\mathfrak{b})$.

Among other things, we show that if $H$ is commutative and $A$ is finite, then
the minimal $A$-elasticity of $H$ is finite. This yields a non-trivial
generalization of the finiteness part of a classical theorem of Anderson et
al. from the case where $H$ is cancellative, commutative, and finitely
generated (f.g.) modulo units and $A$ is the set $\mathscr{A}(H)$ of its
atoms. We also check that commutativity is somewhat essential here, by proving
the existence of an atomic, cancellative, f.g. monoid with trivial group of
units whose minimal $\mathscr{A}(H)$-elasticity is infinite.

###### Key words and phrases:

Atoms; elasticity; [minimal] factorizations; irreducibles; monoids; preorders;
sets of lengths.

###### 2020 Mathematics Subject Classification:

Primary 20M13, 08A50, 20M05, 13A05. Secondary 20M14

##  1\. Introduction

Let $H$ be a (multiplicatively written) monoid, e.g., the multiplicative
monoid of a (unital, associative) ring. An irreducible of $H$ is an element
$a\in H$ that is neither a divisor of the identity $1_{H}$ nor a product of
two other elements $x,y\in H$ each of which is neither a divisor of $1_{H}$
nor of $a$. An atom of $H$ is, on the other hand, a non-unit that cannot be
expressed as the product of two non-units. The existence itself of an atom
implies that any divisor of $1_{H}$ is a unit [12, Lemma 2.2]; hence, every
atom is an irreducible, and the converse is true when the monoid is, e.g.,
commutative and cancellative (see Sect. 5).

We define the (classical) elasticity of $H$ as the supremum (with respect to
the standard ordering of the non-negative real numbers) of the set of rational
numbers of the form $m/n$ with $m,n\in\mathbb{N}^{+}$ (the positive integers)
such that $a_{1}\cdots a_{m}=b_{1}\cdots b_{n}$ for some irreducibles
$a_{1},\ldots,a_{m},b_{1},\ldots,b_{n}\in H$. Moreover, we say a monoid is
atomic if every non-unit has an atomic factorization, that is, the element
factors as a (finite) product of atoms.

Since the late 1980s, elasticity has received wide attention in the literature
(see Anderson’s survey [3] for an overview of results prior to 1997, [17,
Theorems 6.2 and 7.2] for some of the strongest finiteness criteria so far
available in the cancellative commutative setting, and [5, 7, 14, 19, 24, 25]
for a non-exhaustive list of recent contributions). Introduced by Valenza [22]
in his study of factorization in number rings, the notion was made popular by
Zaks [23] who used it as a measure of the deviation of an atomic monoid from
the condition of half-factoriality (a monoid is half-factorial if it is atomic
and any two atomic factorizations of the same element have the same _length_ ,
i.e., the same number of factors). Most notably, it is a classical result in
the arithmetic theory of monoids (and rings) that the elasticity of a
cancellative commutative monoid with finitely many non-associated atoms is a
rational number and hence finite. (Two elements $u,v\in H$ are associated if
$u$ divides $v$ (namely, $v\in HuH$) and $v$ divides $u$.) This was first
proved by Anderson et al. in [2, Theorem 7] and is herein referred to as
_Anderson et al.’s theorem_. The result has been later extended to unit-
cancellative commutative monoids with finitely many non-associated atoms by
Fan et al. [11, Proposition 3.4(1)], where the monoid $H$ is called unit-
cancellative if $yx\neq x\neq xy$ for all $x,y\in H$ such that $y$ is a non-
unit (obviously, a cancellative monoid is unit-cancellative).

In this article, we adopt a new point of view set forth in [4, 10] and obtain
a non-trivial generalization of the _finiteness_ part of Fan et al.’s result
(and hence of Anderson et al.’s theorem) to any commutative monoid that is
finitely generated up to units (Corollaries 3.3 and 3.4). The proof relies on
_Dickson’s lemma_ (see, e.g., Theorem 9.18 in [8]), which is the one and only
aspect in common with previous results in the same vein. A critical feature of
our approach is the use of _minimal_ factorizations to counter the blowup of
factorization lengths and related invariants that is typical of a non-unit-
cancellative or non-commutative setup (Examples 3.12 and 4.2). The conclusion
then becomes an immediate consequence of an essentially combinatorial theorem
(Theorem 2.3) that makes no reference to atoms, irreducibles, etc.

The price we pay is that we cannot say anything about the rationality of the
invariants we introduce along the way to generalize the classical elasticity
(Sect. 5). What we gain is that, in the very spirit of a series of recent
papers by the same authors [20, 9, 10, 21], many of our results are no longer
phrased in the language of _monoids and irreducibles_ (whose scope is, in a
sense, too narrow) but rather in the more abstract language of _monoids and
preorders_ (see Sect. 3 for details). Loosely speaking, this allows us to use
basically any sort of elements as “building blocks” in the factorization
process (Proposition 3.10).

Other than that, we prove a couple or so of finiteness results on length sets
and their unions in a (commutative or non-commutative) monoid with finitely
many irreducibles (Definition 3.5, Propositions 3.6 and 3.7, and Corollary
3.9), and we show by way of example that our main theorem (Theorem 2.3) fails
in the non-commutative setting (Sect. 4).

### Notation.

Through the paper, $\mathbb{N}$ is the set of non-negative integers, and for
$a,b\in\mathbb{N}\cup\\{\infty\\}$ we let $\llbracket
a,b\rrbracket:=\allowbreak\\{x\in\mathbb{N}\colon a\leq x\leq b\\}$ be the
discrete interval from $a$ to $b$.

A preorder on a set $S$ is a reflexive and transitive binary relation on $S$.
Given a preorder $\preceq$ on $S$, we write $x\prec y$ to mean that $x\preceq
y$ and $y\not\preceq x$. (This convention also applies to the symbols
$\sqsubseteq$ and $\sqsubset$, which we will likewise employ for preorders.)
We say that $\preceq$ is artinian if, for every $\preceq$-non-increasing
sequence $(x_{k})_{k\geq 0}$ in $S$, we have $x_{k}\preceq x_{k+1}$ for all
but finitely many $k\in\mathbb{N}$.

We refer to [16] for generalities on monoids. In particular, we denote by
$\mathscr{F}(X)$ the free monoid on a set $X$ and refer to the elements of
$\mathscr{F}(X)$ as $X$-words. We use the symbols $\ast_{X}$ and
$\varepsilon_{X}$, resp., for the operation and the identity of
$\mathscr{F}(X)$. We take $\|\mathfrak{u}\|_{X}$ to be the (word) length of an
$X$-word $\mathfrak{u}$, and for each $i\in\llbracket
1,\|\mathfrak{u}\|_{X}\rrbracket$ we let $\mathfrak{u}[i]$ be the
$i^{\text{th}}$ letter of $\mathfrak{u}$. An $X$-word $\mathfrak{v}$ is then a
(scattered) subword of $\mathfrak{u}$ if there is a strictly increasing
function $\sigma\colon\llbracket 1,\|\mathfrak{v}\|_{X}\rrbracket\to\llbracket
1,\|\mathfrak{u}\|_{X}\rrbracket$ such that
$\mathfrak{u}[\sigma(i)]=\mathfrak{v}[i]$ for each $i\in\llbracket
1,\allowbreak\|\mathfrak{v}\|_{X}\rrbracket$. When there is no serious risk of
ambiguity, we drop the subscript “$X$” from the above notation.

We use $H^{\times}$ for the group of units (or invertible elements) of a
monoid $H$ and $\langle X\rangle_{H}$ for the submonoid of $H$ generated by a
set $X\subseteq H$. We call $H$ reduced if its only unit is the identity
$1_{H}$ and we write $\pi_{H}$ for the unique extension of the identity map on
$H$ to a monoid homomorphism $\mathscr{F}(H)\to H$.

##  2\. Elasticity

By the definition given in the introduction, the (classical) elasticity of a
monoid $H$ is the supremum of the set of all rational numbers of the form
$\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|$ as $\mathfrak{a}$ and $\mathfrak{b}$
range over the non-empty $\mathscr{I}(H)$-words with
$\pi_{H}(\mathfrak{a})=\pi_{H}(\mathfrak{b})$, where $\mathscr{I}(H)$ is the
set of irreducibles of $H$. We aim to generalize this idea.

######  Definition 2.1.

Given a set $X$, we denote by $\sqsubseteq_{X}$ the binary relation on the
free monoid $\mathscr{F}(X)$ defined by
$\mathfrak{a}\sqsubseteq_{X}\mathfrak{b}$, for some $X$-words $\mathfrak{a}$
and $\mathfrak{b}$, if and only if $\mathfrak{a}$ is a permuted subword of
$\mathfrak{b}$, i.e., there is an injective function $\sigma\colon\llbracket
1,\|\mathfrak{a}\|\rrbracket\to\llbracket 1,\|\mathfrak{b}\|\rrbracket$ such
that $\mathfrak{a}[i]=\mathfrak{b}[\sigma(i)]$.

Since the composition of two injections is still an injection, it is immediate
that the relation $\sqsubseteq_{X}$ in Definition 2.1 is a preorder on
$\mathscr{F}(X)$. More precisely, $\sqsubseteq_{X}$ is an artinian preorder,
because $\mathfrak{a}\sqsubset_{X}\mathfrak{b}$ implies
$\|\mathfrak{a}\|<\|\mathfrak{b}\|$. This makes it natural to talk about
$\sqsubseteq_{X}$-minimality in $\mathscr{F}(X)$. Moreover, the pair
$(\mathscr{F}(X),\allowbreak\sqsubseteq_{X})$ is a _strongly positive monoid_
in the sense of [10, Definition 2.3], meaning that (i)
$\varepsilon\sqsubseteq_{X}\mathfrak{a}$ for every $X$-word $\mathfrak{a}$ and
(ii) $\mathfrak{a}\sqsubseteq_{X}\mathfrak{b}$ yields
$\mathfrak{u}\ast\mathfrak{a}\ast\mathfrak{v}\sqsubseteq_{X}\mathfrak{u}\ast\mathfrak{b}\ast\mathfrak{v}$
for all $\mathfrak{u},\mathfrak{v}\in\mathscr{F}(X)$, with the latter
inequality being strict whenever so is the former.

######  Definition 2.2.

Given a monoid $H$ and a set $A\subseteq H$, an $A$-word $\mathfrak{z}$ is
minimal if
$\pi_{H}(\mathfrak{z}^{\prime})\neq\allowbreak\pi_{H}(\mathfrak{z})$ whenever
$\mathfrak{z}^{\prime}\sqsubset_{H}\mathfrak{z}$. The minimal $A$-elasticity
$\varrho_{A}^{\mathsf{m}}(H)$ of $H$ is then the supremum of the set of all
rational numbers of the form $m/n$ with $m,n\in\mathbb{N}^{+}$ such that there
exist minimal $A$-words $\mathfrak{a}$ and $\mathfrak{b}$ with
$\|\mathfrak{a}\|=\allowbreak m$, $\|\mathfrak{b}\|=n$, and
$\pi_{H}(\mathfrak{a})=\pi_{H}(\mathfrak{b})$, where it is understood that
$\sup\varnothing:=0$.

With these preliminaries in place, we are just ready for the main theorem of
the paper. In the proof, we will make use of Dickson’s lemma [8, Theorem
9.18], stating that every non-empty subset of $\mathbb{N}^{\times n}$
($n\in\allowbreak\mathbb{N}^{+}$) has at least one minimal element with
respect to the product order $\leq_{n}$ induced on $\mathbb{N}^{\times n}$ by
the standard order on $\mathbb{N}$, meaning that
$\mathfrak{u}\leq_{n}\mathfrak{v}$, for some
$\mathfrak{u},\allowbreak\mathfrak{v}\in\mathbb{N}^{\times n}$, if and only if
$\mathfrak{u}[i]\leq\mathfrak{v}[i]$ for each $i\in\allowbreak\llbracket
1,n\rrbracket$ (here, we regard $\mathfrak{u}$ and $\mathfrak{v}$ as
$\mathbb{N}$-words of length $n$).

######  Theorem 2.3.

The minimal $A$-elasticity of a commutative monoid $H$ is finite for every
finite $A\subseteq H$.

###### Proof.

Suppose $A$ is a finite subset of $H$ and put $s:=|A|\in\mathbb{N}$. We may
assume $s\neq 0$, or else $A$ is empty and the conclusion is trivial because
the only minimal $A$-word is then the empty word $\varepsilon$. Accordingly,
let $a_{1},\ldots,a_{s}$ be an enumeration of $A$, and for each
$t\in\llbracket 1,s\rrbracket$ denote by $\mathsf{v}_{t}$ the function
$\mathscr{F}(A)\to\allowbreak\mathbb{N}$ that maps an $A$-word $\mathfrak{a}$
to its $a_{t}$-adic valuation, i.e., to the number of indices
$i\in\allowbreak\llbracket 1,\allowbreak\|\mathfrak{a}\|\rrbracket$ such that
$a_{t}=\mathfrak{a}[i]$. It is trivial that

$\|\mathfrak{a}\|=\mathsf{v}_{1}(\mathfrak{a})+\cdots+\mathsf{v}_{s}(\mathfrak{a}),\qquad\text{for
all }\mathfrak{a}\in\mathscr{F}(A).$ (1)

Let $S$ be the set of all triples $(\mathfrak{a},\mathfrak{b},\mathfrak{z})$
of $A$-words with $(\mathfrak{a},\mathfrak{b})\neq(\varepsilon,\varepsilon)$
and
$\pi_{H}(\mathfrak{a}\ast\mathfrak{z})=\pi_{H}(\mathfrak{b}\ast\mathfrak{z})$
such that $\mathfrak{a}\ast\mathfrak{z}$ is a minimal $A$-word (the definition
of $S$ is intentionally “asymmetric”, insomuch as we do not require that also
$\mathfrak{b}\ast\mathfrak{z}$ is a minimal $A$-word).

If $(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in S$, then
$\mathfrak{b}\neq\varepsilon$. Otherwise, $\mathfrak{a}$ would be a non-empty
$A$-word; and since $(\mathscr{F}(A),\allowbreak\sqsubseteq_{H})$ is a
strongly positive monoid (see the comments after Definition 2.1), we would
find that $\mathfrak{z}$ is a proper subword of $\mathfrak{a}\ast\mathfrak{z}$
with $\pi_{H}(\mathfrak{a}\ast\mathfrak{z})=\pi_{H}(\mathfrak{z})$,
contradicting that $\mathfrak{a}\ast\mathfrak{z}$ is a minimal $A$-word. So,
it makes sense to define

$\varrho^{\ast}:=\sup\bigl{\\{}\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|\colon(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in
S\\}.$

It is clear that, if $(\mathfrak{a},\mathfrak{b})$ is a pair of minimal
$A$-words with $\pi_{H}(\mathfrak{a})=\allowbreak\pi_{H}(\mathfrak{b})$ and
$\mathfrak{b}\neq\allowbreak\varepsilon$, then
$(\mathfrak{a},\mathfrak{b},\varepsilon)\in S$; and there is at least one such
pair, since we can take $\mathfrak{a}=\mathfrak{b}=a_{1}$. It follows that
$\varrho_{A}^{\mathsf{m}}(H)\leq\varrho^{\ast}$. Consequently, in order to
prove that $\varrho_{A}^{\mathsf{m}}(H)$ is finite, it suffices to check that
$\varrho^{\ast}<\infty$.

For, let $\leq_{3s}$ be the product order induced on $\mathbb{N}^{\times 3s}$
by the standard order on $\mathbb{N}$ and $f$ be the function

$S\to\mathbb{N}^{\times
3s}\colon(\mathfrak{a},\mathfrak{b},\mathfrak{z})\mapsto(\mathsf{v}_{1}(\mathfrak{a}),\ldots,\mathsf{v}_{s}(\mathfrak{a}),\mathsf{v}_{1}(\mathfrak{b}),\ldots,\mathsf{v}_{s}(\mathfrak{b}),\mathsf{v}_{1}(\mathfrak{z}),\ldots,\mathsf{v}_{s}(\mathfrak{z})).$

Since $A$ is non-empty, $f(S)$ is a non-empty subset of $\mathbb{N}^{\times
3s}$. Hence we gather from Dickson’s lemma that the set $T$ of
$\leq_{3s}$-minimal elements of $f(S)$ is finite and non-empty. We thus see
that

$R:=\bigl{\\{}\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|\colon(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in
f^{-1}(T)\bigr{\\}}$

is a non-empty finite subset of $\mathbb{Q}_{\geq 0}$. In particular, it is
straightforward from Eq. (1) that

$\begin{split}R&=\left\\{\frac{\mathsf{v}_{1}(\mathfrak{a})+\cdots+\mathsf{v}_{s}(\mathfrak{a})}{\mathsf{v}_{1}(\mathfrak{b})+\cdots+\mathsf{v}_{s}(\mathfrak{b})}\colon(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in
f^{-1}(T)\right\\}\\\
&=\left\\{\frac{n_{1}+\cdots+n_{s}}{d_{1}+\cdots+d_{s}}\colon(n_{1},\ldots,n_{s},d_{1},\ldots,d_{s},v_{1},\ldots,v_{s})\in
T\right\\},\end{split}$

so making it evident that $1\leq|R|\leq|T|<\infty$. As a result, $R$ has a
maximum element $r\in\mathbb{Q}_{\geq 0}$, and it is obvious that
$r\leq\rho^{\ast}$. We claim $r=\rho^{\ast}$ (of course, this will be enough
to show that $\rho^{\ast}<\infty$).

Assume to the contrary that $r<\rho^{\ast}$. The set $\bar{S}$ of all triples
$(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in S$ with
$r<\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|$ is then a non-empty subset of
$S\smallsetminus f^{-1}(T)$; in particular, if
$(\mathfrak{a},\mathfrak{b},\mathfrak{z})\in f^{-1}(T)$, then
$\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|\leq r$. Let $\preceq_{S}$ be the binary
relation on $S$ defined by
$(\mathfrak{a},\mathfrak{b},\mathfrak{z})\preceq_{S}(\mathfrak{a}^{\prime},\mathfrak{b}^{\prime},\mathfrak{z}^{\prime})$,
for some
$(\mathfrak{a},\mathfrak{b},\mathfrak{z}),(\mathfrak{a}^{\prime},\mathfrak{b}^{\prime},\mathfrak{z}^{\prime})\in
S$, if and only if

$\text{(i)
}\|\mathfrak{a}\|+\|\mathfrak{b}\|<\|\mathfrak{a}^{\prime}\|+\|\mathfrak{b}^{\prime}\|,\quad\text{or}\quad\text{(ii)
}\|\mathfrak{a}\|+\|\mathfrak{b}\|=\|\mathfrak{a}^{\prime}\|+\|\mathfrak{b}^{\prime}\|\text{
and }\|\mathfrak{z}\|\leq\|\mathfrak{z}^{\prime}\|.$

It is routine to verify that $\preceq_{S}$ is an artinian preorder on $S$.
Since $\bar{S}$ is a non-empty subset of $S$, we thus obtain from the well-
foundedness of artinian preorders (see, e.g., Remark 3.9(3) in [20]) that
$\bar{S}$ has a $\preceq_{S}$-minimal element
$\mathfrak{p}=(\mathfrak{a},\mathfrak{b},\mathfrak{z})$. By construction,
$\mathfrak{p}$ is in $S$ but not in $f^{-1}(T)$. In consequence, there exists
$\mathfrak{p}_{1}=\allowbreak(\mathfrak{a}_{1},\allowbreak\mathfrak{b}_{1},\mathfrak{z}_{1})\in
S$ such that
$f(\mathfrak{a}_{1},\mathfrak{b}_{1},\mathfrak{z}_{1})<_{3s}f(\mathfrak{a},\mathfrak{b},\mathfrak{z})$,
which means that

$\text{(j) }\mathfrak{a}_{1}\sqsubseteq_{H}\mathfrak{a},\
\mathfrak{b}_{1}\sqsubseteq_{H}\mathfrak{b},\text{ and
}\mathfrak{z}_{1}\sqsubseteq_{H}\mathfrak{z},\quad\text{and}\quad\text{(jj) at
least one of these inequalities is strict}.$ (2)

Denote by $\approx_{H}$ the relation of $\sqsubseteq_{H}$-equivalence on
$\mathscr{F}(H)$, and for all $i\in\llbracket 1,s\rrbracket$ and
$\mathfrak{u},\mathfrak{v}\in\mathscr{F}(H)$ set
$\Delta_{i}(\mathfrak{u},\allowbreak\mathfrak{v}):=\allowbreak|\mathsf{v}_{i}(\mathfrak{u})-\allowbreak\mathsf{v}_{i}(\mathfrak{v})|$.
We distinguish two cases depending on whether
$\mathfrak{b}_{1}\approx_{H}\mathfrak{b}$ or
$\mathfrak{b}_{1}\sqsubset_{H}\mathfrak{b}$ (by the last displayed equation,
there are no other possibilities). In each case, we will reach a
contradiction, thus finishing the proof of the theorem.

Case 1: $\mathfrak{b}_{1}\approx_{H}\mathfrak{b}$. If
$\mathfrak{a}_{1}\approx_{H}\mathfrak{a}$, then
$r<\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|=\|\mathfrak{b}_{1}\|^{-1}\|\mathfrak{a}_{1}\|$
(two $H$-words are $\sqsubseteq_{H}$-equivalent only if they have the same
length) and, by the second item of Eq. (2),
$\mathfrak{z}_{1}\sqsubset_{H}\allowbreak\mathfrak{z}$; it follows that
$\bar{S}\ni\allowbreak(\mathfrak{a}_{1},\allowbreak\mathfrak{b}_{1},\mathfrak{z}_{1})\prec_{S}\mathfrak{p}$,
which contradicts the $\preceq_{S}$-minimality of $\mathfrak{p}$ in $\bar{S}$.
Therefore, we gather from the first item of Eq. (2) that
$\mathfrak{a}_{1}\sqsubset_{H}\mathfrak{a}$. Since $H$ is commutative and the
equivalence $\mathfrak{b}_{1}\approx_{H}\mathfrak{b}$ translates to
$\mathfrak{b}_{1}$ and $\mathfrak{b}$ being the same $A$-word up to a
permutation of their letters, it is then seen that

$\pi_{H}(\mathfrak{a}_{1}\ast\mathfrak{z})=\pi_{H}(\mathfrak{a}_{1}\ast\mathfrak{z}_{1}\ast\bar{\mathfrak{z}}_{1})=\pi_{H}(\mathfrak{a}_{1}\ast\mathfrak{z}_{1})\kern
1.00006pt\pi_{H}(\bar{\mathfrak{z}}_{1})=\pi_{H}(\mathfrak{b}\ast\mathfrak{z}_{1})\kern
1.00006pt\pi_{H}(\bar{\mathfrak{z}}_{1})=\pi_{H}(\mathfrak{b}\ast\mathfrak{z})=\pi_{H}(\mathfrak{a}\ast\mathfrak{z}),$

where $\bar{\mathfrak{z}}_{1}$ is the $A$-word
$a_{1}^{\ast\Delta_{1}(\mathfrak{z},\mathfrak{z}_{1})}\ast\cdots\ast
a_{s}^{\ast\Delta_{s}(\mathfrak{z},\mathfrak{z}_{1})}$. Considering that
$\mathfrak{a}_{1}\ast\mathfrak{z}\sqsubset_{H}\mathfrak{a}\ast\mathfrak{z}$,
this is however in contradiction with the fact that
$\mathfrak{a}\ast\mathfrak{z}$ is a minimal $A$-word.

Case 2: $\mathfrak{b}_{1}\sqsubset_{H}\mathfrak{b}$. Let $\mathfrak{a}_{2}$
and $\mathfrak{b}_{2}$ be, resp., the $A$-words
$a_{1}^{\ast\Delta_{1}(\mathfrak{a},\mathfrak{a}_{1})}\ast\cdots\ast
a_{s}^{\ast\Delta_{s}(\mathfrak{a},\mathfrak{a}_{1})}$ and
$a_{1}^{\ast\Delta_{1}(\mathfrak{b},\mathfrak{b}_{1})}\ast\cdots\ast
a_{s}^{\ast\Delta_{s}(\mathfrak{b},\mathfrak{b}_{1})}$, and set
$\mathfrak{z}_{2}:=\mathfrak{a}_{1}\ast\mathfrak{z}$. We have
$\|\mathfrak{b}_{1}\|<\|\mathfrak{b}\|$ and hence
$\mathfrak{b}_{2}\neq\varepsilon$. Moreover,
$\mathfrak{a}_{2}\ast\mathfrak{z}_{2}=\mathfrak{a}_{2}\ast\mathfrak{a}_{1}\ast\mathfrak{z}\approx_{H}\allowbreak\mathfrak{a}\ast\allowbreak\mathfrak{z}$
and
$\pi_{H}(\mathfrak{a}_{2}\ast\mathfrak{z}_{2})=\allowbreak\pi_{H}(\mathfrak{a}\ast\mathfrak{z})$,
since $H$ is commutative and
$\mathfrak{a}_{2}\ast\mathfrak{a}_{1}=\mathfrak{a}$ up to a permutation of the
letters. This shows that $\mathfrak{a}_{2}\ast\mathfrak{z}_{2}$ is a minimal
$A$-word. In addition,

$\begin{split}\pi_{H}(\mathfrak{b}_{2}\ast\mathfrak{z}_{2})=\pi_{H}(\mathfrak{b}_{2}\ast\mathfrak{a}_{1}\ast\mathfrak{z})&=\pi_{H}(\mathfrak{b}_{2}\ast\bar{\mathfrak{z}}_{1})\kern
1.00006pt\pi_{H}(\mathfrak{a}_{1}\ast\mathfrak{z}_{1})=\pi_{H}(\mathfrak{b}_{2}\ast\bar{\mathfrak{z}}_{1})\kern
1.00006pt\pi_{H}(\mathfrak{b}_{1}\ast\mathfrak{z}_{1})\\\
&=\pi_{H}(\mathfrak{b}_{1}\ast\mathfrak{b}_{2}\ast\mathfrak{z}_{1}\ast\bar{\mathfrak{z}}_{1})=\pi_{H}(\mathfrak{b}\ast\mathfrak{z})=\pi_{H}(\mathfrak{a}\ast\mathfrak{z})=\pi_{H}(\mathfrak{a}_{2}\ast\mathfrak{z}_{2}),\end{split}$

where $\bar{\mathfrak{z}}_{1}$ is the same as in Case 1. It then follows that
$(\mathfrak{a}_{1},\mathfrak{b}_{1},\mathfrak{z}_{1})$ and
$(\mathfrak{a}_{2},\mathfrak{b}_{2},\mathfrak{z}_{2})$ are both in $S$, and

$\|\mathfrak{a}_{i}\|+\|\mathfrak{b}_{i}\|\leq\|\mathfrak{a}\|+\|\mathfrak{b}_{i}\|<\|\mathfrak{a}\|+\|\mathfrak{b}\|\qquad(i=1,2).$

So, we get from the $\preceq_{S}$-minimality of $\mathfrak{p}$ in $\bar{S}$
that $\|\mathfrak{b}_{i}\|^{-1}\|\mathfrak{a}_{i}\|\leq r$, and hence

$r<\frac{\|\mathfrak{a}\|}{\|\mathfrak{b}\|}=\frac{\|\mathfrak{a}_{1}\|+\|\mathfrak{a}_{2}\|}{\|\mathfrak{b}_{1}\|+\|\mathfrak{b}_{2}\|}\leq\max\left\\{\frac{\|\mathfrak{a}_{1}\|}{\|\mathfrak{b}_{1}\|},\frac{\|\mathfrak{a}_{2}\|}{\|\mathfrak{b}_{2}\|}\right\\}\leq
r,$

which is however absurd. (For the second inequality in the last display, see,
e.g., [6, Lemma 1.41].) ∎

It is not clear if, given a commutative monoid $H$ and a finite set
$A\subseteq H$, the $A$-elasticity of $H$ is not only finite (as guaranteed by
Theorem 2.3), but also rational (cf. Sect. 5).

##  3\. Factorizations and [unions of] length sets

Let $\mathcal{H}=(H,\preceq)$ be a premon, i.e., a monoid $H$ paired with a
preorder $\preceq$ on its underlying set (note that, in general, we require no
compatibility between the operation in $H$ and the preorder). In particular,
we write $\mid_{H}$ for the divisibility preorder on $H$ (that is,
$u\mid_{H}v$ if and only if $v\in H$ and $u\in HvH$) and $H^{\rm div}$ for the
divisibility premon $(H,\mid_{H})$ of $H$.

Two elements $u,v\in H$ are $\preceq$-equivalent if $u\preceq v\preceq u$, and
$u$ is a $\preceq$-unit (of $H$) if it is $\preceq$-equivalent to the identity
$1_{H}$ (otherwise, $u$ is a $\preceq$-non-unit). A $\preceq$-non-unit
$a\in\allowbreak H$ is then a $\preceq$-irreducible if $a\neq xy$ for all
$\preceq$-non-units $x,y\in H$ with $x\prec a$ and $y\prec a$. We use
$\mathcal{H}^{\times}$ for the set of $\preceq$-units, and
$\mathscr{I}(\mathcal{H})$ for the set of $\preceq$-irreducibles of the monoid
$H$, which we also refer to as the irreducibles of the premon $\mathcal{H}$.
These notions were first considered in [20, Definition 3.6] and further
studied in [9, 10, 21].

Following [10, Sect. 3], we denote by $\sqsubseteq_{\mathcal{H}}$ the
shuffling preorder induced by $\preceq$, that is, the preorder on
$\mathscr{F}(H)$ defined by
$\mathfrak{a}\sqsubseteq_{\mathcal{H}}\mathfrak{b}$, for some $H$-words
$\mathfrak{a}$ and $\mathfrak{b}$, if and only if there is an injective
function $\sigma\colon\llbracket
1,\allowbreak\|\mathfrak{a}\|\rrbracket\to\llbracket
1,\|\mathfrak{b}\|\rrbracket$ such that
$\mathfrak{a}[i]\preceq\mathfrak{b}[\sigma(i)]\preceq\mathfrak{a}[i]$ for
every $i\in\llbracket 1,\|\mathfrak{a}\|\rrbracket$. It turns out that,
similar to the case of the preorder $\sqsubseteq_{H}$ introduced in Definition
2.1, $\sqsubseteq_{\mathcal{H}}$ is artinian.

Accordingly, we let a $\preceq$-factorization of an element $x\in H$ be an
$\mathscr{I}(\mathcal{H})$-word $\mathfrak{a}\in\allowbreak\pi_{H}^{-1}(x)$
and we set
$\mathcal{Z}_{\mathcal{H}}(x):=\allowbreak\pi_{H}^{-1}(x)\cap\allowbreak\mathscr{F}(\mathscr{I}(\mathcal{H}))$.
A minimal $\preceq$-factorization of $x$ is then a
$\sqsubseteq_{\mathcal{H}}$-minimal word in $\mathcal{Z}_{\mathcal{H}}(x)$,
i.e., an $\mathscr{I}(\mathcal{H})$-word
$\mathfrak{a}\in\allowbreak\pi_{H}^{-1}(x)$ such that there is no
$\mathscr{I}(\mathcal{H})$-word $\mathfrak{b}\in\allowbreak\pi_{H}^{-1}(x)$
with $\mathfrak{b}\sqsubset_{\mathcal{H}}\mathfrak{a}$. We denote the set of
minimal $\preceq$-factorizations of $x$ by $\mathcal{Z}_{\mathcal{H}}^{\sf
m}(x)$. It follows from the artinianity of $\sqsubseteq_{\mathcal{H}}$ that
$x$ has a $\preceq$-factorization if and only if it has a minimal
$\preceq$-factorization (see Remark 3.3(1) in [10]).

######  Definition 3.1.

(1) Given a premon $\mathcal{H}=(H,\preceq)$, the $\preceq$-elasticity
$\varrho_{\mathcal{H}}(x)$ (resp., the minimal $\preceq$-elasticity
$\varrho_{\mathcal{H}}^{\mathsf{m}}(x)$) of an element $x\in H$ is the
supremum of $\|\mathfrak{b}\|^{-1}\|\mathfrak{a}\|$ as $\mathfrak{a}$ and
$\mathfrak{b}$ range over the non-empty $\preceq$-factorizations (resp., the
non-empty minimal $\preceq$-factorizations) of $x$. (It is understood that
$\sup\varnothing:=0$.)

(2) The elasticity $\varrho(\mathcal{H})$ (resp., the minimal elasticity
$\varrho^{\mathsf{m}}(\mathcal{H})$) of the premon $\mathcal{H}$ is then the
supremum of $\varrho_{\mathcal{H}}(x)$ (resp., of
$\varrho_{\mathcal{H}}^{\mathsf{m}}(x)$) as $x$ ranges over the $\preceq$-non-
units of $H$. In particular, we let the elasticity (resp., the minimal
elasticity) of the monoid $H$ be the elasticity (resp., the minimal
elasticity) of $H^{\rm div}$.

In the notation of Definition 3.1,
$\varrho_{\mathcal{H}}(x)=\varrho_{\mathcal{H}}^{\mathsf{m}}(x)=0$ for every
$x\in H$ whose set of $\preceq$-factorizations is empty. Also observe that
$\varrho_{\mathcal{H}}^{\mathsf{m}}(1_{H})=0$, since the only minimal
$\preceq$-factorization of the identity $1_{H}$ is the empty word. Lastly,
note that the elasticity of the monoid $H$ is nothing different from what we
called the _classical elasticity_ of $H$ in Sect. 1.

######  Theorem 3.2.

The minimal elasticity of a commutative premon with finitely many irreducibles
is finite.

###### Proof.

Let $\mathcal{H}=(H,\preceq)$ be a commutative premon with finitely many
irreducibles and denote by $\varrho_{\rm irr}^{\mathsf{m}}(H)$ the minimal
$\mathscr{I}(\mathcal{H})$-elasticity $\varrho_{\rm irr}^{\mathsf{m}}(H)$ of
$H$. Since $\mathscr{I}(\mathcal{H})$ is a finite set (by hypothesis), we have
from Theorem 2.3 that $\varrho_{\rm irr}^{\mathsf{m}}(H)<\infty$. This
suffices to finish the proof, as it is immediate that a minimal
$\preceq$-factorization of a $\preceq$-non-unit is a minimal
$\mathscr{I}(\mathcal{H})$-word in the sense of Definition 2.2 and hence
$\varrho^{\mathsf{m}}(\mathcal{H})\leq\varrho_{\rm irr}^{\mathsf{m}}(H)$. ∎

As mentioned in Sect. 1, it was proved by Fan et al. in [11, Proposition
3.4(1)] that, if $H$ is a commutative unit-cancellative monoid $H$ such that
the quotient $H/H^{\times}$ is finitely generated (i.e., $H$ is finitely
generated modulo units), then the classical elasticity of $H$ is rational (and
hence finite). In the next corollaries, we show how Theorem 2.3 can be used to
recover the _finiteness_ part of Fan et al.’s result.

######  Corollary 3.3.

If a commutative monoid $H$ is finitely generated modulo units, then its
minimal elasticity is finite.

###### Proof.

By the commutativity of $H$, an element $u\in H$ divides the identity $1_{H}$
if and only if $u$ is a unit. Consequently, an irreducible of $H$ is a non-
unit $a\in H$ that does not factor as a product of two non-units each of which
is not divisible by $a$; and the irreducibles of the quotient monoid
$H/H^{\times}$ are nothing but the cosets of the irreducibles of $H$. It
follows that $H$ and $H/H^{\times}$ have the same minimal elasticity. On the
other hand, since $H$ is finitely generated modulo units if and only if
$H/H^{\times}$ is finitely generated, we may assume without loss of generality
that $H$ is reduced. By [10, Example 2.5(1) and Theorem 4.7], the set of
irreducibles of $H$ is then finite; and by Theorem 2.3, this suffices to
complete the proof. ∎

######  Corollary 3.4.

If a commutative, unit-cancellative monoid is finitely generated modulo units,
then its (classical) elasticity is finite.

###### Proof.

This follows from Corollary 3.3 when considering that, in a commutative and
unit-cancellative monoid $H$, every irreducible (i.e., every
$\mid_{H}$-irreducible) is an atom [20, Corollary 4.4] and every minimal
$\mid_{H}$-factorization is a $\mid_{H}$-factorization [4, Proposition
4.7(v)]. ∎

The following definitions generalize sets of lengths and related invariants
from the classical theory of factorization, see [4, Sects. 2.3 and 4.1] and
[10, Definition 3.2].

######  Definition 3.5.

(1) Given a premon $\mathcal{H}=(H,\preceq)$ and an element $x\in H$, we let
$\mathsf{L}_{\mathcal{H}}(x):={\bigl{\\{}\|\mathfrak{a}\|_{H}:\mathfrak{a}\in\mathcal{Z}_{\mathcal{H}}(x)\bigr{\\}}}\subseteq\mathbb{N}\quad\text{and}\quad\mathsf{L}_{\mathcal{H}}^{\sf
m}(x):={\bigl{\\{}\|\mathfrak{a}\|_{H}:\mathfrak{a}\in\mathcal{Z}_{\mathcal{H}}^{\sf
m}(x)\bigr{\\}}}\subseteq\mathbb{N}$ be, resp., the length set and the minimal
length set of $x$ (relative to $\mathcal{H}$). Accordingly, we refer to
$\mathscr{L}(\mathcal{H}):=\bigl{\\{}\mathsf{L}_{\mathcal{H}}(x)\colon x\in
H\smallsetminus\mathcal{H}^{\times}\bigr{\\}}\quad\text{and}\quad\mathscr{L}^{\sf
m}(\mathcal{H}):=\bigl{\\{}\mathsf{L}_{\mathcal{H}}^{\sf m}(x)\colon x\in
H\smallsetminus\mathcal{H}^{\times}\bigr{\\}},$ resp., as the system of length
sets and the system of minimal length sets of $\mathcal{H}$; and given
$k\in\mathbb{N}$, we call  $\mathscr{U}_{k}(\mathcal{H}):=\bigcup\kern
1.00006pt\\{L\in\mathscr{L}(\mathcal{H})\colon k\in
L\\}\qquad\text{and}\qquad\mathscr{U}_{k}^{\mathsf{m}}(\mathcal{H}):=\bigcup\kern
1.00006pt\\{L\in\mathscr{L}^{\mathsf{m}}(\mathcal{H})\colon k\in L\\}$ resp.,
the union of length sets and the union of minimal length sets containing $k$.

(2) In particular, we take the system of length sets (resp., the system of
minimal length sets) of the monoid $H$ to be the system of length sets (resp.,
of minimal length sets) of the divisibility premon $H^{\rm div}$ of $H$, and
we write $\mathscr{L}(H)$ for $\mathscr{L}(H^{\rm div})$ and $\mathscr{L}^{\sf
m}(H)$ for $\mathscr{L}^{\sf m}(H^{\rm div})$. The same goes with [minimal]
length sets and unions of [minimal] length sets.

Note that, in the notation of Definition 3.5, the sets
$\mathscr{U}_{0}(\mathcal{H})$ and $\mathscr{U}_{0}^{\sf m}(\mathcal{H})$ are
both empty, because there is no $\preceq$-non-unit whose length set contains
$0$ (in fact, $0$ is only a $\preceq$-length of the identity $1_{H}$).

######  Proposition 3.6.

The following hold for a premon $\mathcal{H}=(H,\preceq)$:

  1. (i)

$\varrho(\mathcal{H})\leq 1$ (resp., $\varrho^{\mathsf{m}}(\mathcal{H})\leq
1$) if and only if $|\mathsf{L}_{\mathcal{H}}(x)|\leq 1$ (resp.,
$|\mathsf{L}_{\mathcal{H}}^{\sf m}(x)|\leq 1$) for each $x\in
H\smallsetminus\mathcal{H}^{\times}$.

  2. (ii)

If $\varrho(\mathcal{H})$ (resp., $\varrho^{\mathsf{m}}(\mathcal{H})$) is
finite, then $\mathscr{U}_{k}(\mathcal{H})$ (resp., $\mathscr{U}_{k}^{\sf
m}(\mathcal{H})$) is finite for every $k\in\mathbb{N}$.

###### Proof.

We focus on length sets and leave the corresponding statements on _minimal_
length sets to the reader (the proofs are pretty much the same in either
case).

(i) If $|\mathsf{L}_{\mathcal{H}}(x_{0})|\leq 1$ for every $\preceq$-non-unit
$x\in H$, then $\varrho_{\mathcal{H}}(x)$ is either $0$ or $1$, and hence
$\varrho(\mathcal{H})\leq 1$. As for the converse, assume
$\varrho(\mathcal{H})\leq 1$ and suppose by way of contradiction that there is
a $\preceq$-non-unit $x_{0}\in\allowbreak H$ such that
$|\mathsf{L}_{\mathcal{H}}(x_{0})|\geq 2$. There exist
$k_{1},k_{2}\in\mathsf{L}_{\mathcal{H}}(x_{0})$ with $1\leq k_{1}<k_{2}$,
implying that
$\varrho(\mathcal{H})\geq\allowbreak\varrho_{\mathcal{H}}(x_{0})\geq\allowbreak
k_{2}/k_{1}>1$, which is absurd.

(ii) Let $\varrho(\mathcal{H})$ be finite and suppose for a contradiction that
$|\mathscr{U}_{k}(\mathcal{H})|=\infty$ for some $k\in\mathbb{N}$. Since
$\mathscr{U}_{0}(\mathcal{H})$ is empty, $k$ is then a _positive_ integer and
there is a sequence
$(\mathfrak{a}_{1},\mathfrak{b}_{1}),\allowbreak(\mathfrak{a}_{2},\mathfrak{b}_{2}),\ldots$
of pairs of $\mathscr{I}(\mathcal{H})$-words such that, for every
$i\in\mathbb{N}^{+}$, $\mathfrak{a}_{i}$ and $\mathfrak{b}_{i}$ are
$\preceq$-factorizations of the same $\preceq$-non-unit $x_{i}\in H$ and
$k=\allowbreak\|\mathfrak{a}_{i}\|\leq\allowbreak\|\mathfrak{b}_{i}\|<\|\mathfrak{b}_{i+1}\|$.
This, however, contradicts the finiteness of $\varrho(\mathcal{H})$. ∎

######  Proposition 3.7.

Let $\mathcal{H}=(H,\preceq)$ be a premon. Then either there is a bound
$M\in\mathbb{N}$ such that $\|\mathfrak{a}\|\leq M$ for every minimal
$\preceq$-factorization $\mathfrak{a}$, or for each $k\in\mathbb{N}$ there is
a $\preceq$-minimal factorization of length $k$.

###### Proof.

Suppose that there is an integer $k\geq 1$ such that the set of minimal
$\preceq$-factorizations of length $k$ is empty. We claim that there is no
minimal $\preceq$-factorization of length $n\geq k$, and we proceed to prove
it by induction on $n$. If $n=k$, the assertion is obvious. Otherwise, let
$\mathfrak{a}=a_{1}\ast\cdots\ast a_{n}$ be an $\mathscr{I}(\mathcal{H})$-word
of length $n\geq k+1$ and assume inductively that there is no minimal
$\preceq$-factorization of length $n-1$. In particular, this means that
$\mathfrak{a}^{\prime}:=a_{1}\ast\cdots\ast a_{n-1}$ is not a minimal
$\preceq$-factorization, viz., there exists an $\mathscr{I}(\mathcal{H})$-word
$\mathfrak{b}$ with $\mathfrak{b}\sqsubset_{\mathcal{H}}\mathfrak{a}^{\prime}$
and $\pi_{H}(\mathfrak{b})=\pi_{H}(\mathfrak{a}^{\prime})$. Since
$(\mathscr{F}(\mathscr{I}(\mathcal{H})),\sqsubseteq_{\mathcal{H}})$ is a
strongly positive monoid (as noted in the second paragraph of this section),
it follows that $\mathfrak{b}\ast a_{n}\sqsubset_{\mathcal{H}}\mathfrak{a}$;
and since $\pi_{H}(\mathfrak{b}\ast a_{n})=\allowbreak\pi_{H}(\mathfrak{a})$,
we conclude that $\mathfrak{a}$ is not a minimal $\preceq$-factorization
either. ∎

In [13, Proposition 2 and Corollary 1], Geroldinger and Lettl proved that the
unions of length sets (of the divisibility premon) of a cancellative,
commutative, finitely generated monoid $H$ are all finite. This result is
generalized to a non-commutative, non-cancellative context by Corollary 3.9
below.

######  Theorem 3.8.

Let $\mathcal{H}=(H,\preceq)$ be a premon and suppose there is a finite set
$A\subseteq\mathscr{I}(\mathcal{H})$ such that every $\preceq$-irreducible is
$\preceq$-equivalent to an element of $A$. Then the minimal length sets of
$\mathcal{H}$ are all finite.

###### Proof.

Denote by $\sim_{\mathcal{H}}$ the relation of $\preceq$-equivalence on $H$
and by $\approx_{\mathcal{H}}$ the relation of
$\sqsubseteq_{\mathcal{H}}$-equivalence on $\mathscr{F}(H)$, and assume
without loss of generality that the elements of $A$ are pairwise
$\preceq$-inequivalent. Next, let $\\{q_{1},\ldots,q_{s}\\}$ be an enumeration
of the elements of $A$, where $s:=|A|\in\mathbb{N}^{+}$; and for each
$i\in\llbracket 1,s\rrbracket$, let
$\mathsf{v}_{i}\colon\mathscr{F}(\mathscr{I}(\mathcal{H}))\to\mathbb{N}$ be
the function that maps the empty word to $0$ and a non-empty
$\mathscr{I}(\mathcal{H})$-word $a_{1}\ast\allowbreak\cdots\ast a_{n}$ of
length $n$ to the number of indices $j\in\llbracket 1,n\rrbracket$ such that
$a_{j}\sim_{\mathcal{H}}q_{i}$.

By assumption, we are given that, for every $a\in\mathscr{I}(\mathcal{H})$,
there is a unique $i\in\llbracket 1,s\rrbracket$ such that
$a\sim_{\mathcal{H}}q_{i}$. Moreover, $q_{i}\sim_{\mathcal{H}}q_{j}$, for some
$i,j\in\llbracket 1,s\rrbracket$, if and only if $i=j$. It is then readily
seen that

$\|\mathfrak{a}\|=\mathsf{v}_{1}(\mathfrak{a})+\cdots+\mathsf{v}_{s}(\mathfrak{a}),\qquad\text{for
all }\mathfrak{a}\in\mathscr{F}(\mathscr{I}(\mathcal{H})).$ (3)

Now, fix $x\in H$. We need to show that the minimal length set
$\mathsf{L}_{\mathcal{H}}^{\sf m}(x)$ of $x$ is finite. If
$\mathsf{L}_{\mathcal{H}}^{\sf m}(x)$ is empty, the conclusion is obvious. So,
suppose that $x$ has at least one minimal $\preceq$-factorization, and let
$\leq_{s}$ be the product order induced on $\mathbb{N}^{\times s}$ by the
standard order on $\mathbb{N}$ and $f$ be the function

$\mathsf{Z}_{\mathcal{H}}^{\sf m}(x)\to\mathbb{N}^{\times
s}\colon\mathfrak{a}\mapsto(\mathsf{v}_{1}(\mathfrak{a}),\ldots,\mathsf{v}_{s}(\mathfrak{a})).$

Since $f(\mathsf{Z}_{\mathcal{H}}^{\sf m}(x))$ is a non-empty subset of
$\mathbb{N}^{\times s}$, we get from Dickson’s lemma that the set
$\mathcal{M}$ of $\leq_{s}$-minimal elements of
$f(\mathsf{Z}_{\mathcal{H}}^{\sf m}(x))$ is finite and non-empty. We claim
that $f(\mathfrak{a})\in\mathcal{M}$ for every
$\mathfrak{a}\in\mathsf{Z}_{\mathcal{H}}^{\sf m}(x)$; note that this will
finish the proof, as it implies by Eq. (3) that

$|\mathsf{L}_{\mathcal{H}}^{\sf
m}(x)|=|\\{\mathsf{v}_{1}(\mathfrak{a})+\cdots+\mathsf{v}_{s}(\mathfrak{a})\colon\mathfrak{a}\in\mathsf{Z}_{\mathcal{H}}^{\sf
m}(x)\\}|=|\\{k_{1}+\cdots+k_{s}\colon(k_{1},\ldots,k_{s})\in\mathcal{M}\\}|\leq|\mathcal{M}|<\infty.$

For the claim, suppose to the contrary that there exists
$\mathfrak{a}\in\mathsf{Z}_{\mathcal{H}}^{\sf m}(x)$ with
$f(\mathfrak{a})\notin\mathcal{M}$. Then $f(\mathfrak{b})<_{s}f(\mathfrak{a})$
for some $\mathfrak{b}\in f^{-1}(\mathcal{M})$, meaning that
$\mathsf{v}_{i}(\mathfrak{b})\leq\mathsf{v}_{i}(\mathfrak{a})$ for each
$i\in\llbracket 1,s\rrbracket$ and at least one of these inequalities is
strict. By the definition of $\sqsubseteq_{\mathcal{H}}$, it then follows that

$\mathfrak{b}\approx_{\mathcal{H}}q_{1}^{\mathsf{v}_{1}(\mathfrak{b})}\ast\cdots\ast
q_{s}^{\mathsf{v}_{s}(\mathfrak{b})}\sqsubseteq_{\mathcal{H}}q_{1}^{\mathsf{v}_{1}(\mathfrak{a})}\ast\cdots\ast
q_{s}^{\mathsf{v}_{s}(\mathfrak{a})}\approx_{\mathcal{H}}\mathfrak{a},$

which, in turn, yields $\mathfrak{b}\sqsubset_{\mathcal{H}}\mathfrak{a}$,
because two $\mathscr{I}(\mathcal{H})$-words are
$\sqsubseteq_{\mathcal{H}}$-equivalent only if they have the same length and,
by Eq. (3) and the above, $\|\mathfrak{b}\|<\|\mathfrak{a}\|$. We have thus
reached a contradiction, because $\mathfrak{a}$ and $\mathfrak{b}$ are both
minimal $\preceq$-factorizations of $x$. ∎

######  Corollary 3.9.

In a premon $\mathcal{H}=(H,\preceq)$ with finitely many
$\preceq$-irreducibles, unions of minimal length sets are all finite.

###### Proof.

Assume for a contradiction that
$|\mathscr{U}^{\mathsf{m}}_{k}(\mathcal{H})|=\infty$ for some
$k\in\mathbb{N}$. There is then a sequence
$(\mathfrak{a}_{1},\mathfrak{b}_{1}),\allowbreak(\mathfrak{a}_{2},\mathfrak{b}_{2}),\ldots$
of pairs of $\mathscr{I}(\mathcal{H})$-words such that, for each
$i\in\mathbb{N}^{+}$, $\mathfrak{a}_{i}$ and $\mathfrak{b}_{i}$ are minimal
$\preceq$-factorizations of one same element $x_{i}\in
H\smallsetminus\mathcal{H}^{\times}$ with
$k=\|\mathfrak{a}_{i}\|\leq\|\mathfrak{b}_{i}\|<\|\mathfrak{b}_{i+1}\|$.
However, the set of $\mathscr{I}(\mathcal{H})$-words of length $k$ is finite,
since the basis $\mathscr{I}(\mathcal{H})$ is finite. So, we can find a
(strictly) increasing sequence $i_{1},i_{2},\ldots$ of positive integers with
$\mathfrak{a}_{i_{1}}=\mathfrak{a}_{i_{j}}$ for all $j\in\mathbb{N}^{+}$,
implying that the minimal length set of $x_{i_{1}}$ is unbounded (as it
contains the increasing sequence
$\|\mathfrak{b}_{i_{1}}\|,\|\mathfrak{b}_{i_{2}}\|,\ldots$) and hence
contradicting Theorem 3.8. ∎

The next proposition shows that, under mild conditions on the premon
$\mathcal{H}$ (see Remark 3.11) and up to a suitable modification of
Definitions 3.1 and 3.5, the results of the present section can be extended to
the case when the factors used in the factorization process are taken from an
arbitrary set $A\subseteq\mathscr{I}(\mathcal{H})$.

######  Proposition 3.10.

Let $\mathcal{H}=(H,\preceq)$ be a premon in which the product of any two
$\preceq$-non-units is a $\preceq$-non-unit, and let $A$ be a set of
$\preceq$-irreducibles. There then exists a preorder $\preceq_{A}$ on $H$ such
that each of the following conditions is satisfied:

  1. (i)

$u\in H$ is a $\preceq_{A}$-non-unit if and only if $u\in\langle
A\rangle_{H}\smallsetminus\mathcal{H}^{\times}$.

  2. (ii)

$a\in H$ is a $\preceq_{A}$-irreducible if and only if $a\in A$.

  3. (iii)

$a\preceq b\preceq a$, for some $a,b\in A$, if and only if
$a\preceq_{A}b\preceq_{A}a$.

###### Proof.

Set $S:=\langle A\rangle_{H}\smallsetminus\mathcal{H}^{\times}$ and let $\phi$
be the function $H\to\mathbb{N}$ that maps an element $x\in S$ to the smallest
integer $n\geq 1$ such that $x\in A^{n}$ and an element in $H\smallsetminus S$
to $0$. We define a binary relation $\preceq_{A}$ on $H$ by taking
$x\preceq_{A}y$ if and only if one of the following conditions is satisfied:

$(1)\ x,y\in H\smallsetminus S,\quad(2)\ x,y\in S\text{ and
}\phi(x)<\phi(y),\quad\text{or}\quad(3)\ x,y\in S,\ \phi(x)=\phi(y),\text{ and
}x\preceq y.$

It is routine to check that $\preceq_{A}$ is a preorder, so we focus below on
proving that $\preceq_{A}$ satisfies (i)–(iii).

(i) It is clear that $u\in H$ is a $\preceq_{A}$-unit if and only if $u\notin
S$ (by the fact that $1_{H}\notin S$). Consequently, an element of $H$ is a
$\preceq_{A}$-non-unit if and only if it belongs to $S$.

(ii) Fix $a\in A$ and assume for a contradiction that $a$ is not a
$\preceq_{A}$-irreducible. Then, $a=bc$ for some $b,c\in S$ (i.e.,
$\preceq_{A}$-non-units) such that $b\prec_{A}a$ and $c\prec_{A}a$. Since
$\phi(a)=1$, it must be $\phi(b)=\phi(c)=1$ and $b\prec a$ and $c\prec a$.
Since $b,c\notin\mathcal{H}^{\times}$, we contradict that $a$ is a
$\preceq$-irreducible and we prove that every $a\in A$ is a
$\preceq_{A}$-irreducible. On the other hand, let $a\in H$ be a
$\preceq_{A}$-irreducible and assume for a contradiction that $a\notin A$.
Since $a$ is a $\preceq_{A}$-non-unit, then $a\in S$ and there exist $n\geq 2$
elements $a_{1},\dots,a_{n}\in A$ such that $a=a_{1}\cdots a_{n}$. Thus
$a=bc$, with $b:=a_{1}$ and $c:=a_{2}\cdots a_{n}$. But this contradicts that
$a$ is a $\preceq_{A}$-irreducible, since $b$ and $c$ are both in $S$ by the
fact that a non-empty product of $\preceq$-non-units is still a $\preceq$-non-
unit.

(iii) Pick $a,b\in A$. It is clear that $\phi(a)=\phi(b)=1$. Since $A$ is a
set of $\preceq$-irreducibles and a $\preceq$-irreducible is a $\preceq$-non-
unit, $a$ and $b$ are both elements of $S$. Therefore, the definition itself
of the preorder $\preceq_{A}$ implies that $a\preceq_{A}b\preceq_{A}a$ if and
only if $a\preceq b\preceq a$. ∎

######  Remark 3.11.

(1) Let $H$ be a monoid. If $x,y\in H$ and $xy\mid_{H}1_{H}$ (i.e., $xy$ is a
$\mid_{H}$-unit), then also $x$ and $y$ divide $1_{H}$; that is, the product
of any two $\mid_{H}$-non-units is still a $\mid_{H}$-non-unit and Proposition
3.10 applies to the divisibility premon $(H,\mid_{H})$ of $H$. The same is
true of the premons $(H,\vdash_{H})$ and $(H,\dashv_{H})$, where $\vdash_{H}$
and $\dashv_{H}$ are, resp., the “divides from the left” and the “divides from
the right” preorders on $H$ (i.e., $x\vdash_{H}y$ if and only if $x\in H$ and
$y\in xH$, and similarly for $\dashv_{H}$).

(2) Following [10, Definition 2.3], let a weakly positive monoid be a premon
$\mathcal{H}=(H,\preceq)$ such that $1_{H}\preceq\allowbreak x$ and
$uxv\preceq x\preceq yxz$ for all $x,y,z\in H$ and
$u,v\in\mathcal{H}^{\times}$. By [10, Remark 2.4(3)] , the product of two
$\preceq$-non-units is then a $\preceq$-non-unit. That is, Proposition 3.10
applies also to weakly positive monoids.

We conclude with a couple of examples that show how the paradigm of _minimal_
factorizations (as opposite to “ordinary factorizations”) can indeed mitigate
the effects of blowup phenomena that would otherwise affect the invariants
considered through this section, making them lose most of their significance.

######  Example 3.12.

(1) Let $H$ be the multiplicative monoid of the integers modulo $p^{n}$, where
$p\in\mathbb{N}$ is a prime and $n$ is an integer $\geq 2$. By [10, Example
3.4], $H$ is an atomic monoid and the atoms (resp., the units) of $H$ are
precisely the $\mid_{H}$-irreducibles (resp., the $\mid_{H}$-units). In
addition, every non-zero non-unit of $H$ has an essentially unique atomic
factorization, with “essentially unique” meaning that any two atomic
factorizations of the same element are equivalent with respect to the
shuffling preorder induced by the divisibility preorder $\mid_{H}$. On the
other hand, the residue class of $0$ modulo $p^{n}$ has an essentially unique
minimal $\mid_{H}$-factorization (of length $n$), but atomic factorizations of
any length $\geq n$. It follows that the (classical) elasticity of $H$ is
$\infty$; the minimal elasticity is $1$; and for every $k\in\mathbb{N}^{+}$,
we have  $\mathscr{U}_{k}(H)=\left\\{\begin{array}[]{ll}\\{k\\}&\text{if
}1\leq k<n,\\\ \llbracket k,\infty\rrbracket&\text{if }k\geq
n\end{array}\right.\qquad\text{and}\qquad\mathscr{U}_{k}^{\sf
m}(H)=\left\\{\begin{array}[]{ll}\\{k\\}&\text{if }1\leq k\leq n,\\\
\varnothing&\text{if }k>n.\end{array}\right.$

(2) Fix an integer $n\geq 2$ and let $\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n})$ be the reduced power monoid of the additive group
$\mathbb{Z}_{n}$ of integers modulo $n$, i.e., the (additively written) monoid
obtained by endowing the subsets of $\mathbb{Z}_{n}$ containing the zero
element $[0]_{n}\in\mathbb{Z}_{n}$ with the operation of setwise addition
$(X,Y)\mapsto\\{x+y\colon x\in X,\kern 1.00006pty\in Y\\}$.
By [20, Proposition 4.11(ii)], every $X\in\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n})$ factors as a product of irreducibles. On the other
hand, it is obvious that $\\{[0]_{n},x\\}$ is an irreducible of
$\mathcal{P}_{{\rm fin},0}(\mathbb{Z}_{n})$ for every non-zero
$x\in\mathbb{Z}_{n}$. Consequently, we get from the _proof_ of [4, Proposition
4.12(i) and Lemma 5.5] (reworked in terms of irreducibles) that every minimal
factorization in $\mathcal{P}_{{\rm fin},0}(\mathbb{Z}_{n})$ has length
smaller than $n$ and the interval $\llbracket 2,n-1\rrbracket$ is a minimal
length set of $\mathcal{P}_{{\rm fin},0}(\mathbb{Z}_{n})$. Since
$\mathbb{Z}_{n}$ is a proper idempotent of $\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n})$, it follows that, for every integer $k\geq 2$,
$\mathscr{U}_{k}(\mathcal{P}_{{\rm fin},0}(\mathbb{Z}_{n}))=\mathbb{N}_{\geq
2}\qquad\text{and}\qquad\mathscr{U}_{k}^{\sf m}(\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n}))=\left\\{\begin{array}[]{ll}\llbracket
2,n-1\rrbracket&\text{if }2\leq k<n,\\\ \varnothing&\text{if }k\geq
n.\end{array}\right.$ Note also that $\mathscr{U}_{1}(\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n}))=\\{1\\}$ when $n$ is odd, since every irreducible of
$H$ is then an atom [20, Theorem 4.12]; and $\mathscr{U}_{1}(\mathcal{P}_{{\rm
fin},0}(\mathbb{Z}_{n}))=\mathbb{N}^{+}$ when $n$ is even, because in this
latter case the set $\\{[0]_{n},[n/2]_{n}\\}$ is an idempotent irreducible of
$\mathcal{P}_{{\rm fin},0}(\mathbb{Z}_{n})$.

##  4\. An example

Given a set $X$ and a binary relation $R$ on the free monoid $\mathscr{F}(X)$,
we define $R^{\sharp}$ as the smallest monoid congruence on $\mathscr{F}(X)$
containing $R$. This means that $\mathfrak{u}\equiv\mathfrak{v}\bmod
R^{\sharp}$ if and only if there are
$\mathfrak{z}_{0},\mathfrak{z}_{1},\allowbreak\ldots,\mathfrak{z}_{n}\in\mathscr{F}(X)$
with $\mathfrak{z}_{0}=\mathfrak{u}$ and $\mathfrak{z}_{n}=\mathfrak{v}$ such
that, for each $i\in\llbracket 0,n-1\rrbracket$, there exist $X$-words
$\mathfrak{p}_{i}$, $\mathfrak{q}_{i}$, $\mathfrak{q}_{i}^{\prime}$, and
$\mathfrak{r}_{i}$ with the following properties:

(i) either $\mathfrak{q}_{i}=\mathfrak{q}_{i}^{\prime}$, or
$\mathfrak{q}_{i}\mathbin{R}\mathfrak{q}_{i}^{\prime}$, or
$\mathfrak{q}_{i}^{\prime}\mathbin{R}\mathfrak{q}_{i}$; (ii)
$\mathfrak{z}_{i}=\mathfrak{p}_{i}\ast\mathfrak{q}_{i}\ast\mathfrak{r}_{i}$
and
$\mathfrak{z}_{i+1}=\mathfrak{p}_{i}\ast\mathfrak{q}_{i}^{\prime}\ast\mathfrak{r}_{i}$.

We denote by $\mathrm{Mon}\langle X\mid R\rangle$ the monoid obtained by
taking the quotient of $\mathscr{F}(X)$ by the congruence $R^{\sharp}$; we
write $\mathrm{Mon}\langle X\mid R\rangle$ multiplicatively and call it a
(monoid) presentation. We refer to the elements of $X$ as the generators of
the presentation, to each pair $(\mathfrak{q},\mathfrak{q}^{\prime})\in R$ as
a defining relation, and to each $X$-word in a defining relation as a defining
word. If there is no danger of confusion, we systematically identify an
$X$-word $\mathfrak{z}$ with its equivalence class in $\mathrm{Mon}\langle
X\mid R\rangle$.

The left graph of a presentation $\mathrm{Mon}\langle X\mid R\rangle$ is the
undirected multigraph with vertex set $X$ and an edge from $y$ to $z$ for each
pair $(y\ast\mathfrak{u},z\ast\mathfrak{v})\in R$ with $y,z\in X$ and
$\mathfrak{u},\mathfrak{v}\in\mathscr{F}(X)$; this results in a loop when
$y=z$, and in multiple (or parallel) edges between $y$ and $z$ if there are
two or more defining relations of the form
$(y\ast\mathfrak{u},z\ast\mathfrak{v})$. The right graph is defined
analogously, using the right-most (instead of left-most) letters of each word
from a defining relation. The left and the right graphs of a presentation were
first considered by Adian [1], whence we refer to them as the Adian graphs of
$\mathrm{Mon}\langle X\mid R\rangle$.

On the other hand, a piece of a presentation $\mathrm{Mon}\langle X\mid
R\rangle$ is a non-empty $X$-word $\mathfrak{u}$ for which there exist
$\mathfrak{p},\mathfrak{p}^{\prime},\mathfrak{q},\mathfrak{q}^{\prime}\in\mathscr{F}(X)$
with $\mathfrak{p}\neq\mathfrak{p}^{\prime}$ or
$\mathfrak{q}\neq\mathfrak{q}^{\prime}$ such that
$\mathfrak{v}:=\mathfrak{p}\ast\mathfrak{u}\ast\mathfrak{q}$ and
$\mathfrak{v}^{\prime}:=\mathfrak{p}^{\prime}\ast\mathfrak{u}\ast\mathfrak{q}^{\prime}$
are defining words; in particular, it is not required that
$\mathfrak{v}\neq\mathfrak{v}^{\prime}$ or
$(\mathfrak{v},\mathfrak{v}^{\prime})\in R$. The notion was first conceived in
the study of _group_ presentations and later extended by Kashintsev to
semigroups (see [18, Section 1]).

Following [18], we say that a monoid $H$ is of class $K_{p}^{q}$, for some
$p,q\in\mathbb{N}^{+}$, if $H$ is isomorphic to a monoid presentation
$\textup{Mon}\langle X\mid R\rangle$ with finitely many generators such that
(i) no defining word can be expressed as the concatenation in $\mathscr{F}(X)$
of less than $p$ pieces and (ii) the Adian graphs of the presentation have
both girth $\geq q$ (we recall that the girth of an undirected multigraph $G$
is the shortest length of a cycle in $G$, with the understanding that the
girth of a cycle-free multigraph is $\infty$). Our interest in these
definitions is linked to the following result, first proved by Guba in [15,
Theorem 1] and hence referred to as _Guba’s_ (_embedding_) _theorem_ (note
that, in [15], there is a typo in the very definition of a piece, as discussed
on MathOverflow at https://mathoverflow.net/questions/353340/).

######  Theorem 4.1.

Every monoid of class $K_{3}^{2}$ embeds into a group (and hence is
cancellative).

Guba’s theorem will come in handy in the next example, which shows that
Theorem 2.3 does not carry over to the non-commutative setting in any obvious
way.

######  Example 4.2.

Let $H$ be the presentation $\textup{Mon}\langle A\mid R\rangle$, where $A$ is
the $3$-element set $\\{a,b,c\\}$ and $R$ is the set
$\\{(\mathfrak{s}_{n},\mathfrak{t}_{n})\colon
n=2,3,\ldots\\}\subseteq\mathscr{F}(A)\times\mathscr{F}(A)$, with

$\mathfrak{s}_{n}:=c\ast a^{\ast n}\ast b^{\ast 2^{n}}\ast a^{\ast n}\ast
c\quad\textup{and}\quad\mathfrak{t}_{n}:=a\ast c^{\ast n}\ast b^{\ast n}\ast
c^{\ast n}\ast a.$

It is routine to check that $H$ is a $3$-generated monoid with trivial group
of units whose $\mid_{H}$-irreducibles are $\underline{a}$, $\underline{b}$,
and $\underline{c}$, where we write $\underline{\mathfrak{u}}$ for the
$R^{\sharp}$-congruence class (in $H$) of an $A$-word $\mathfrak{u}$. Each of
$\underline{a}$, $\underline{b}$, and $\underline{c}$ is on the other hand an
atom, because the defining words in $R$ have all length $\geq 2$. Then, $H$ is
an atomic monoid and every $\mid_{H}$-factorization is an atomic factorization
and vice versa. In addition, it is immediate that, for each
$k\in\mathbb{N}^{+}$, none of the $A$-words

$c\ast a^{\ast k}\ast b,\quad b\ast a^{\ast k}\ast c,\quad a\ast c^{\ast
k}\ast b,\quad\textup{or}\quad b\ast c^{\ast k}\ast a$ (4)

is a piece of $\textup{Mon}\langle A\mid R\rangle$; if, e.g., a defining word
factors as $\mathfrak{p}\ast(a\ast c^{\ast k}\ast b)\ast\mathfrak{q}$ for some
$A$-words $\mathfrak{p}$ and $\mathfrak{q}$, then necessarily
$\mathfrak{p}=\varepsilon_{A}$ and $\mathfrak{q}=b^{\ast(k-1)}\ast c^{\ast
k}\ast a$ (and the other cases are similar). Since a non-empty $A$-word
$\mathfrak{s}:=x_{1}\ast\cdots\ast x_{l}$ is a piece of $\textup{Mon}\langle
A\mid R\rangle$ only if so is any subword of the form $x_{i}\ast\cdots\ast
x_{j}$ with $1\leq i\leq j\leq l$, it follows that the support
$\\{x_{1},\ldots,x_{l}\\}$ of $\mathfrak{s}$ is a _proper_ subset of $A$. As a
matter of fact, a piece is in the first place a subword of a defining word, so
that the shortest pieces with support $A$ (if there were any) would be those
listed in Eq. (4).

It follows that no defining word of $\textup{Mon}\langle A\mid R\rangle$ is
the concatenation in $\mathscr{F}(A)$ of less than three pieces (note that
each defining word is palindromic and its support is $A$). Consequently, $H$
is a monoid of class $K_{3}^{2}$, which implies, by Guba’s theorem, that $H$
is cancellative. So, it remains to see that, for each $n\in\mathbb{N}^{+}$,
there are minimal atomic factorizations $\mathfrak{a}_{n}$ and
$\mathfrak{b}_{n}$ of an element $x_{n}\in H$ such that
$\|\mathfrak{a}_{n}\|\geq n\kern 1.00006pt\|\mathfrak{b}_{n}\|$.

For, fix an integer $n\geq 2$. The $H$-words
$\mathfrak{a}_{n}:=\underline{c}\ast\underline{a}^{\ast
n}\ast\underline{b}^{\ast 2^{n}}\ast\underline{a}^{\ast n}\ast\underline{c}$
and $\mathfrak{b}_{n}:=\underline{a}\ast\underline{c}^{\ast
n}\ast\underline{b}^{\ast n}\ast\underline{c}^{\ast n}\ast\underline{a}$ are
non-empty atomic factorizations of the same element, with the further property
that $\|\mathfrak{b}_{n}\|^{-1}\|\mathfrak{a}_{n}\|\to\allowbreak\infty$ as
$n\to\infty$. Therefore, it suffices to check that $\mathfrak{a}_{n}$ and
$\mathfrak{b}_{n}$ are minimal atomic factorizations.

Assume to the contrary that $\mathfrak{a}_{n}$ is not a minimal atomic
factorization (the minimality of $\mathfrak{b}_{n}$ can be proved in a similar
fashion). Then $\mathfrak{s}_{n}\equiv\allowbreak\mathfrak{u}\bmod R^{\sharp}$
for some permutation $\mathfrak{u}$ of a _proper_ subword of
$\mathfrak{s}_{n}$; that is, there exist a smallest $k\in\mathbb{N}^{+}$ and
$A$-words
$\mathfrak{z}_{0},\mathfrak{z}_{1},\allowbreak\ldots,\mathfrak{z}_{k}$ with
$\mathfrak{z}_{0}=\allowbreak\mathfrak{s}_{n}$ and
$\mathfrak{z}_{k}=\mathfrak{u}$ such that, for each $i\in\llbracket
0,k-1\rrbracket$, there are an integer $m_{i}\geq 2$ and
$\mathfrak{p}_{i},\mathfrak{r}_{i}\in\mathscr{F}(A)$ with

(i)
$\mathfrak{z}_{i}=\mathfrak{p}_{i}\ast\mathfrak{s}_{m_{i}}\ast\mathfrak{r}_{i}$
and
$\mathfrak{z}_{i+1}=\mathfrak{p}_{i}\ast\mathfrak{t}_{m_{i}}\ast\mathfrak{r}_{i}$,
or (ii)
$\mathfrak{z}_{i}=\mathfrak{p}_{i}\ast\mathfrak{t}_{m_{i}}\ast\mathfrak{r}_{i}$
and
$\mathfrak{z}_{i+1}=\mathfrak{p}_{i}\ast\mathfrak{s}_{m_{i}}\ast\mathfrak{r}_{i}$.

However, it is readily seen (by induction on $i$) that this can only happen if
$\mathfrak{z}_{i}\in\\{\mathfrak{s}_{n},\mathfrak{t}_{n}\\}$ for every
$i\in\llbracket 0,k\rrbracket$, because, for an integer $m\geq 2$ with $m\neq
n$, neither $\mathfrak{s}_{m}$ nor $\mathfrak{t}_{m}$ is a divisor of
$\mathfrak{s}_{n}$ (resp., of $\mathfrak{t}_{n}$) in $\mathscr{F}(A)$. It
follows that $\mathfrak{u}=\mathfrak{t}_{n}$, which is impossible since
neither $\mathfrak{s}_{n}$ nor $\mathfrak{t}_{n}$ is a proper subword of
$\mathfrak{s}_{n}$.

##  5\. Prospects for future research

By Theorem 3.2, the minimal elasticity $\rho^{\rm m}(H)$ of (the divisibility
premon of) a cancellative, commutative, reduced monoid $H$ with finitely many
irreducibles is finite. In view of [11, Proposition 3.4(1)], we are led to ask
if $\rho^{\rm m}(H)$ is, in fact, a rational number (cf. the last lines of
Sect. 2).

On the other hand, we know from Sect. 4 that the minimal elasticity of an
atomic, cancellative, finitely generated, reduced monoid need not be finite.
Is this still the case with an acyclic, finitely generated, and reduced
monoid? Here, we say a monoid $H$ is acyclic if $uxv\neq x$ for all $u,v,x\in
H$ such that $u$ or $v$ is a non-unit. Acyclic monoids were introduced in [20,
Definition 4.2] and further studied in [10]; from an arithmetical point of
view, they provide a better alternative to cancellativity in the non-
commutative setting. For instance, we get from [20, Corollary 4.4] that, in an
acyclic monoid, irreducibles and atoms are the same thing, which generalizes
an observation made in the first lines of Sect. 1. It is also worth noting
that every acyclic monoid is unit-cancellative; the two notions (of acyclicity
and unit-cancellativity) coincide in the commutative setting, but not in
general [20, Example 4.8].

## Acknowledgments

L.C. was supported by the European Union’s Horizon 2020 research and
innovation programme under the Marie Skłodowska-Curie grant agreement No.
101021791, and by the FWF (Austrian Science Fund) project No. P33499-N. The
Marie Skłodowska-Curie grant also financed S.T.’s visit at University of Graz
in summer-fall 2022, during which the paper was written. The authors are both
members of the Gruppo Nazionale per le Strutture Algebriche, Geometriche e le
loro Applicazioni (GNSAGA) of the Italian Mathematics Research Institute
(INdAM).

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