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# Sets of lengths in maximal orders in central simple algebras

Daniel Smertnig  Institut für Mathematik und Wissenschaftliches Rechnen
Karl-Franzens-Universität Graz
Heinrichstraße 36
8010 Graz, Austria<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{O}$ be a holomorphy ring in a global field $K$, and $R$ a
classical maximal $\mathcal{O}$-order in a central simple algebra over $K$. We
study sets of lengths of factorizations of cancellative elements of $R$ into
atoms (irreducibles). In a large majority of cases there exists a transfer
homomorphism to a monoid of zero-sum sequences over a ray class group of
$\mathcal{O}$, which implies that all the structural finiteness results for
sets of lengths—valid for commutative Krull monoids with finite class
group—hold also true for $R$. If $\mathcal{O}$ is the ring of algebraic
integers of a number field $K$, we prove that in the remaining cases no such
transfer homomorphism can exist and that several invariants dealing with sets
of lengths are infinite.

###### Key words and phrases:

sets of lengths, maximal orders, global fields, Brandt groupoid, divisorial
ideals, Krull monoids

###### 2010 Mathematics Subject Classification:

16H10, 16U30, 20M12, 20M13, 11R54

The author is supported by the Austrian Science Fund (FWF): W1230, Doctoral
Program “Discrete Mathematics”

##  1\. Introduction

Let $H$ be a (left- and right-) cancellative semigroup and $H^{\times}$ its
group of units. An element $u\in H\setminus H^{\times}$ is called
_irreducible_ (or an _atom_) if $u=ab$ with $a,b\in H$ implies that $a\in
H^{\times}$ or $b\in H^{\times}$. If $a\in H\setminus H^{\times}$, then
$l\in\mathbb{N}$ is a _length of $a$_ if there exist atoms
$u_{1},\ldots,u_{l}\in H$ with $a=u_{1}\cdot\ldots\cdot u_{l}$, and the _set
of lengths_ of $a$, written as $\mathsf{L}(a)$, consists of all such lengths.
If there is a non-unit $a\in H$ with $\lvert\mathsf{L}(a)\rvert>1$, say
$1<k<l\in\mathsf{L}(a)$, then for every $n\in\mathbb{N}$, we have
$\mathsf{L}(a^{n})\supset\\{\,kn+\nu(l-k)\mid\nu\in[0,n]\,\\}$, which shows
that sets of lengths become arbitrarily large. If $H$ is commutative and
satisfies the ACC on divisorial ideals, then all sets of lengths are finite
and non-empty.

Sets of lengths (and all invariants derived from them, such as the set of
distances) are among the most investigated invariants in factorization theory.
So far research has almost been entirely devoted to the commutative setting,
and it has focused on commutative noetherian domains, commutative Krull
monoids, numerical monoids, and others (cf. [1, 12, 27, 28, 26, 20, 7]).
Recall that a commutative noetherian domain is a Krull domain if and only if
the monoid of non-zero elements is a Krull monoid and this is the case if and
only if the domain is integrally closed. Suppose that $H$ is a Krull monoid
(so completely integrally closed and the ACC on divisorial two-sided ideals
holds true). Then the monoid of divisorial two-sided ideals is a free abelian
monoid. If $H$ is commutative (or at least normalizing), this gives rise to
the construction of a transfer homomorphism $\theta\colon
H\to\mathcal{B}(G_{P})$, where $\mathcal{B}(G_{P})$ is the monoid of zero-sum
sequences over a subset $G_{P}$ of the class group $G$ of $H$. Transfer
homomorphisms preserve sets of lengths, and if $G_{P}$ is finite, then
$\mathcal{B}(G_{P})$ is a finitely generated commutative Krull monoid, whose
sets of lengths can be studied with methods from combinatorial number theory.
This approach has lead to a large variety of structural results for sets of
lengths in commutative Krull monoids (see [27, 24] for an overview).

Only first hesitant steps were taken so far to study factorization properties
in a non-commutative setting (for example, quaternion orders are investigated
in [19, 18, 16]), semifirs in ([14, 15], semigroup algebras in [37]). The
present paper provides an in-depth study of sets of lengths in classical
maximal orders over holomorphy rings in global fields.

Let $\mathcal{O}$ be a commutative Krull domain with quotient field $K$, $A$ a
central simple algebra over $K$, $R$ a maximal order in $A$, and $R^{\bullet}$
the semigroup of cancellative elements (equivalently, $R$ is a PI Krull ring).
Any approach to study sets of lengths, which runs as described above and
involves divisorial two-sided ideals, is restricted to normalizing Krull
monoids ([25, Theorem 4.13]). For this reason we develop the theory of
divisorial one-sided ideals. In Section 3 we fix our terminology in the
setting of cancellative small categories. Following ideas of Asano and Murata
[5] and partly of Rehm [45, 46], we provide in Section 4 a factorization
theory of integral elements in arithmetical groupoids, and introduce an
abstract transfer homomorphism for a subcategory of such a groupoid (Section
4). In Section 5 the divisorial one-sided ideal theory of maximal orders in
quotient semigroups is given, and Section 5 establishes the relationship with
arithmetical groupoids. Section 5 is a main result in the abstract setting of
arithmetical maximal orders (Remarks 5.2 and 5.1 reveal how the well-known
transfer homomorphisms for normalizing Krull monoids fit into our abstract
theory). For maximal orders over commutative Krull domains, we see that all
sets of lengths are finite and non-empty (Section 5.2). In Section 6 we
demonstrate that classical maximal orders over holomorphy rings in global
fields fulfill the abstract assumptions of Section 5, which implies the
following structural finiteness results on sets of lengths.

######  Theorem .

Let $\mathcal{O}$ be a holomorphy ring in a global field $K$, $A$ a central
simple algebra over $K$, and $R$ a classical maximal $\mathcal{O}$-order of
$A$. Suppose that every stably free left $R$-ideal is free. Then there exists
a transfer homomorphism $\theta\colon
R^{\bullet}\to\mathcal{B}(\operatorname{\mathcal{C}}_{A}(\mathcal{O}))$, where

$\operatorname{\mathcal{C}}_{A}(\mathcal{O})=\mathcal{F}^{\times}(\mathcal{O})\,/\,\\{\,a\mathcal{O}\mid
a\in K^{\times},\,a_{v}>0\text{ for all archimedean places $v$ of $K$ where
$A$ is ramified.}\,\\}$

is a ray class group of $\mathcal{O}$, and
$\mathcal{B}(\operatorname{\mathcal{C}}_{A}(\mathcal{O}))$ is the monoid of
zero-sum sequences over $\operatorname{\mathcal{C}}_{A}(\mathcal{O})$. In
particular,

  1. 1.

The set of distances $\Delta(R^{\bullet})$ is a finite interval, and if it is
non-empty, then $\min\Delta(R^{\bullet})=1$.

  2. 2.

For every $k\in\mathbb{N}$, the union of sets of lengths containing $k$,
denoted by $\mathcal{U}_{k}(R^{\bullet})$, is a finite interval.

  3. 3.

There is an $M\in\mathbb{N}_{0}$ such that for every $a\in R^{\bullet}$ the
set of lengths $\mathsf{L}(a)$ is an AAMP with difference
$d\in\Delta(R^{\bullet})$ and bound $M$.

Thus, under the additional hypothesis that every stably free left $R$-ideal is
free, we obtain a transfer homomorphism to a monoid of zero-sum sequences over
a finite abelian group. Therefore, sets of lengths in $R$ are the same as sets
of lengths in a commutative Krull monoid with finite class group.

If $A$ satisfies the Eichler condition relative to $\mathcal{O}$, then every
stably free left $R$-ideal is free by Eichler’s Theorem. In particular, if $K$
is a number field and $\mathcal{O}$ is its ring of algebraic integers, then
$A$ satisfies the Eichler condition relative to $\mathcal{O}$ unless $A$ is a
totally definite quaternion algebra. Thus in this setting Section 1 covers the
large majority of cases, and the following complementary theorem shows that
the condition that every stably free left $R$-ideal is free is indeed
necessary.

######  Theorem .

Let $\mathcal{O}$ be the ring of algebraic integers in a number field $K$, $A$
a central simple algebra over $K$, and $R$ a classical maximal
$\mathcal{O}$-order of $A$. If there exists a stably free left $R$-ideal that
is not free, then there exists no transfer homomorphism $\theta\colon
R^{\bullet}\to\mathcal{B}(G_{P})$, where $G_{P}$ is any subset of an abelian
group. Moreover,

  1. 1.

$\Delta(R^{\bullet})=\mathbb{N}$.

  2. 2.

For every $k\geq 3$, we have $\mathbb{N}_{\geq
3}\subset\mathcal{U}_{k}(R^{\bullet})\subset\mathbb{N}_{\geq_{2}}$.

The proof of Section 1 is based on recent work of Kirschmer and Voight ([39,
40]), and will be given in Section 7. If $H$ is a commutative Krull monoid
with an infinite class group such that every class contains a prime divisor,
then Kainrath showed that every finite subset of $\mathbb{N}_{\geq 2}$ can be
realized as a set of lengths ([38], or [27, Section 7.4]), whence
$\Delta(H)=\mathbb{N}$ and $\mathcal{U}_{k}(H)=\mathbb{N}_{\geq 2}$ for all
$k\geq 2$. However, we explicitly show that in the above situation no transfer
homomorphism is possible, implying that the factorization of $R^{\bullet}$
cannot be modeled by a monoid of zero-sum sequences. A similar statement about
sets of lengths in the integer-valued polynomials, as well as the
impossibility of a transfer homomorphism to a monoid of zero-sum sequences,
was recently shown by Frisch [21].

##  2\. Preliminaries

Let $\mathbb{N}$ denote the set of positive integers and put
$\mathbb{N}_{0}=\\{\,0\,\\}\cup\mathbb{N}$. For integers $a,b\in\mathbb{Z}$,
let $[a,b]=\\{\,x\in\mathbb{Z}\mid a\leq x\leq b\,\\}$ denote the discrete
interval. All semigroups and rings are assumed to have an identity element,
and all homomorphisms respect the identity. By a factorization we always mean
a factorization of a cancellative element into irreducible elements (a formal
definition follows in Section 3). In order to study factorizations in
semigroups we will have to investigate their divisorial one-sided ideal
theory, in which the multiplication of ideals only gains sufficiently nice
properties if one considers it as a partial operation that is only defined for
certain pairs of ideals. This is the reason why we introduce our concepts in
the setting of groupoids and consider subcategories of these groupoids.

Throughout the paper there will be many statements that can be either
formulated “from the left” or “from the right”, and most of the time it is
obvious how the symmetric statement should look like. Therefore often just one
variant is formulated and it is left to the reader to fill in the symmetric
definition or statement if required.

###  2.1. Small categories as generalizations of semigroups

Let $H$ be a small category. In the sequel the objects of $H$ play no role,
and therefore we shall identify $H$ with the set of morphisms of $H$. We
denote by $H_{0}$ the set of identity morphisms (representing the objects of
the category). There are two maps $s,t\colon H\to H_{0}$ such that two
elements $a,b\in H$ are composable to a (uniquely determined) element $ab\in
H$ if and only if $t(a)=s(b)$. 111This choice of $t$ and $s$ is compatible
with the usual convention for groupoids, but unfortunately opposite to the
usual convention for categories. For $e,f\in H_{0}$ we set $H(e,f)=\\{\,a\in
H\mid s(a)=e,\,t(a)=f\,\\}$, $H(e)=H(e,e)$, $H(e,\cdot)=\bigcup_{f^{\prime}\in
H_{0}}H(e,f^{\prime})$ and $H(\cdot,f)=\bigcup_{e^{\prime}\in
H_{0}}H(e^{\prime},f)$. Note that an element $e\in H$ lies in $H_{0}$ if and
only if $s(e)=t(e)=e$, $ea=a$ for all $a\in H(e,\cdot)$ and $ae=a$ for all
$a\in H(\cdot,e)$.

A semigroup may be viewed as a category with a single object (corresponding to
its identity element), and elements of the semigroup as morphisms with source
and target this unique object. In this way the notion of a small category
generalizes the usual notion of a semigroup ($H$ is a semigroup if and only if
$\lvert H_{0}\rvert=1$). We will consider a semigroup to be a small category
in this sense whenever this is convenient, without explicitly stating this
anymore. For $A,B\subset H$ we write $AB=\\{\,ab\in H\mid a\in A,b\in B\text{
and }t(a)=s(b)\,\\}$ for the set of all possible products, and if $b\in H$,
then $Ab=A\\{b\\}$ and $bA=\\{b\\}A$.

An element $a\in H$ is called _left-cancellative_ if it is an epimorphism
($ab=ac$ implies $b=c$ for all $b,c\in H(t(a),\cdot)$), and it is called
_right-cancellative_ if it is a monomorphism ($ba=ca$ implies $b=c$ for all
$b,c\in H(\cdot,s(a))$), and _cancellative_ if it is both. The set of all
cancellative elements is denoted by $H^{\bullet}$, and $H$ is called
_cancellative_ if $H=H^{\bullet}$. The set of isomorphisms of $H$ will also be
called the _set of units_ , and we denote it by $H^{\times}$. A subcategory
$D\subset H$ is _wide_ if $D_{0}=H_{0}$.

In line with the multiplicative notation, if $H$ and $D$ are two small
categories, we call a functor $f\colon H\to D$ a homomorphism (of small
categories). Explicitly, a map $f\colon H\to D$ is a homomorphism if
$f(H_{0})\subset D_{0}$ and whenever $a,b\in H$ with $t(a)=s(b)$ then also
$f(a)\cdot f(b)$ is defined (i.e., $t(f(a))=s(f(b))$) and $f(ab)=f(a)f(b)$.

If $H$ is a commutative semigroup, and $D\subset H$ is a subsemigroup, then a
localization $D^{-1}H$ with an embedding $H\hookrightarrow D^{-1}H$ exists
whenever all elements of $D$ are cancellative, and in particular $H$ has a
group of fractions if and only if $H$ is cancellative. If $H$ is a non-
commutative semigroup and $D\subset H$, then a semigroup of right fractions
with respect to $D$, $HD^{-1}$, in which every element can be represented as a
fraction $ad^{-1}$ with $a\in H$, $d\in D$, together with an embedding
$H\hookrightarrow HD^{-1}$, exists if and only if $D$ is cancellative and $D$
satisfies the _right Ore condition_ , meaning $aD\cap dH\neq\emptyset$ for all
$a\in H$ and $d\in D$. For a semigroup of left fractions, $D^{-1}H$, one gets
the analogous _left Ore condition_ , and if $D$ satisfies both, the left and
the right Ore condition, then every semigroup of right fractions is a
semigroup of left fractions and conversely. In this case we write
$D^{-1}H=HD^{-1}$. If $H^{\bullet}$ satisfies the left and right Ore
condition, we also write
$\mathbf{q}(H)=H(H^{\bullet})^{-1}=({H^{\bullet}}^{-1})H$ for the
corresponding semigroup of fractions.

The notion of semigroups of fractions generalizes to categories of fractions
with analogous conditions ([23]). Let $H$ be a small category, and $D\subset
H^{\bullet}$ a subset of the cancellative elements. Then $D$ _admits a
calculus of right fractions_ if $D$ is a wide subcategory of $H$ and it
satisfies the right Ore condition, i.e., $aD\cap dH\neq\emptyset$ for all
$a\in H$ and $d\in D$ with $s(a)=s(d)$. In that case there exists a small
category $HD^{-1}$ with $(HD^{-1})_{0}=H_{0}$ and an embedding $j\colon H\to
HD^{-1}$ (i.e., $j$ is a faithful functor) with $j\mid
H_{0}=\operatorname{id}$ and such that every element of $HD^{-1}$ can be
represented in the form $j(a)j(d)^{-1}$ with $a\in H$, $d\in D$ and
$t(a)=t(d)$, $j(D)\subset H^{\times}$ and it is universal with respect to that
property, i.e., if $f\colon H\to S$ is any homomorphism with $f(D)\subset
S^{\times}$, then there exists a unique $D^{-1}f\colon HD^{-1}\to S$ such that
$D^{-1}f\circ j=f$. We can assume $H\subset HD^{-1}$ and take $j$ to be the
inclusion map, and we call $HD^{-1}$ the _category of right fractions_ of $H$
with respect to $D$. If $D$ also admits a _left calculus of fractions_ , then
$HD^{-1}$ is also a category of left fractions, and we write
$HD^{-1}=D^{-1}H$.

A _monoid_ is a cancellative semigroup satisfying the left and right Ore
condition (following the convention of [25]). Every monoid has a (left and
right) group of fractions which is unique up to unique isomorphism. A
semigroup $H$ is called _normalizing_ if $aH=Ha$ for all $a\in H$. It is
easily checked that a normalizing cancellative semigroup is already a
normalizing monoid.

Let $\mathcal{M}$ be a directed multigraph (i.e., a quiver). For every edge
$a$ of $\mathcal{M}$ we write $s(a)$ for the vertex that is its source and
$t(a)$ for the vertex that is its target. The _path category_ on
$\mathcal{M}$, denoted by $\mathcal{F}(\mathcal{M})$, is defined as follows:
It consists of all tuples $y=(e,a_{1},\ldots,a_{k},f)$ with
$k\in\mathbb{N}_{0}$, $e,f$ vertices of $\mathcal{M}$ and $a_{1},\ldots,a_{k}$
edges of $\mathcal{M}$ with either $k=0$ and $e=f$ or $k>0$, $s(a_{1})=e$,
$t(a_{i})=s(a_{i+1})$ for all $i\in[1,k-1]$ and $t(a_{k})=f$. The set of
identities $\mathcal{F}(\mathcal{M})_{0}$ is the set of all tuples with $k=0$,
and given any tuple $y$ as above, $s(y)=(e,e)$ and $t(y)=(f,f)$. Composition
is defined in the obvious manner by concatenating tuples and removing the two
vertices in the middle. We identify the set of vertices of $\mathcal{M}$ with
$\mathcal{F}(M)_{0}$ so that $(e,e)=e$. Every subset $M$ of a small category
$H$ will be viewed as a quiver, with vertices $\\{\,s(a)\mid a\in
M\,\\}\cup\\{\,t(a)\mid a\in M\,\\}$ and for each $a\in M$ a directed edge
(again called $a$) from $s(a)$ to $t(a)$.

###  2.2. Groupoids

A groupoid $G$ is a small category in which every element is a unit (i.e.,
every morphism is an isomorphism). If $e,f,e^{\prime},f^{\prime}\in G_{0}$ and
there exist $a\in G(e,f)$ and $b\in G(e^{\prime},f^{\prime})$, then

$\begin{cases}G(e,e^{\prime})&\to G(f,f^{\prime})\\\ x&\mapsto
a^{-1}xb\end{cases}$ (1)

is a bijection.

For all $e\in G_{0}$ the set $G(e)$ is a group, called the _vertex group_ or
_isotropy group_ of $G$ at $e$. If $f\in G_{0}$ and $a\in G(e,f)$, then,
taking $b=a$, the map in (1) is a group isomorphism from $G(e)$ to $G(f)$. If
$G(e)$ is abelian, it can be easily checked that this isomorphism does not
depend on the choice of $a$: If $a,a^{\prime}\in G(e,f)$, then

$a^{\prime}(a^{-1}xa)a^{\prime-1}=(a^{\prime}a^{-1})x(aa^{\prime-1})=(a^{\prime}a^{-1})(aa^{\prime-1})x=x.$

In particular, if $G$ is connected (meaning $G(e,e^{\prime})\neq\emptyset$ for
all $e,e^{\prime}\in G_{0}$) and one vertex group is abelian, then all vertex
groups are abelian, and they are canonically isomorphic.

In this case we define for $e\in G_{0}$ and $x\in G(e)$ the set
$(x)=\\{\,a^{-1}xa\mid a\in G(e,\cdot)\,\\}$, and the _universal vertex group_
as

$\mathbb{G}=\\{\,(x)\mid x\in G(e),\,e\in G_{0}\,\\}.$

$\mathbb{G}$ indeed has a natural abelian group structure: For every $e\in
G_{0}$ there is a bijection $j_{e}\colon G(e)\to\mathbb{G},x\mapsto(x)$
inducing the structure of an abelian group on $\mathbb{G}$, and because the
diagrams

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commute for every choice of $e,f\in G_{0}$ and $a\in G(e,f)$, this group
structure is independent of the choice of $e$, yielding a canonical group
isomorphism $j_{e}\colon G(e)\to\mathbb{G}$ for every $e\in G_{0}$. We will
use calligraphic letters to denote elements of $\mathbb{G}$. If
$\mathcal{X}\in\mathbb{G}$, then the unique representative of $\mathcal{X}$ in
$G(e)$, $j_{e}^{-1}(\mathcal{X})$, will be denoted by $\mathcal{X}_{e}$.

If $G$ is a groupoid, and $H\subset G$ is a subcategory, then $HH^{-1}$
denotes the set of all right fractions of elements of $H$. Furthermore,
$HH^{-1}\subset G$ is a subgroupoid if and only if $H$ satisfies the right Ore
condition.

###  2.3. Krull monoids and Krull rings

A monoid $H$ is called a _Krull monoid_ if it is completely integrally closed
(in other words, a maximal order) and satisfies the ACC on divisorial two-
sided ideals. A prime Goldie ring $R$ is a _Krull ring_ if it is completely
integrally closed and satisfies the ACC on divisorial two-sided ideals
(equivalently, its monoid $R^{\bullet}$ of cancellative elements is a Krull
monoid; see [25]). The theory of commutative Krull monoids is presented in
[32, 27]. The simplest examples of non-commutative Krull rings are classical
maximal orders in central simple algebras over Dedekind domains (see Section
5.2). We discuss monoids of zero-sum sequences.

Let $G=(G,0_{G},+)$ be an additively written abelian group, $G_{P}\subset G$ a
subset and let $\mathcal{F}_{ab}(G_{P})$ be the (multiplicatively written)
free abelian monoid with basis $G_{P}$. Elements $S\in\mathcal{F}_{ab}(G_{P})$
are called _sequences over $G_{P}$_, and are written in the form
$S=g_{1}\cdot\ldots\cdot g_{l}$ where $l\in\mathbb{N}_{0}$ and
$g_{1},\ldots,g_{l}\in G_{P}$. We denote by $\lvert S\rvert=l$ the _length_ of
$S$. Such a sequence $S$ is said to be a _zero-sum sequence_ if
$\sigma(S)=g_{1}+\ldots+g_{l}=0_{G}$. The submonoid

$\mathcal{B}(G_{P})=\\{\,S\in\mathcal{F}_{ab}(G_{P})\mid\sigma(S)=0\,\\}\;\subset\;\mathcal{F}_{ab}(G_{P})$

is called the _monoid of zero-sum sequences_ over $G_{P}$. It is a reduced
commutative Krull monoid, which is finitely generated whenever $G_{P}$ is
finite ([27, Theorem 3.4.2]). Moreover, every commutative Krull monoid
possesses a transfer homomorphism onto a monoid of zero-sum sequences, and
thus $\mathcal{B}(G_{P})$ provides a model for the factorization behavior of
commutative Krull monoids ([27, Section 3.4]).

##  3\. Arithmetical Invariants

In this section we introduce our main arithmetical invariants (rigid
factorizations, sets of lengths, sets of distances) and transfer homomorphisms
in the setting of cancellative small categories.

_Throughout this section, let $H$ be a cancellative small category._

$H$ is _reduced_ if $H^{\times}=H_{0}$. An element $u\in H\setminus
H^{\times}$ is an _atom_ (or _irreducible_) if $u=bc$ with $b,c\in H$ implies
$b\in H^{\times}$ or $c\in H^{\times}$. By $\mathcal{A}(H)$ we denote the set
of all atoms of $H$, and call $H$ _atomic_ if every $a\in H\setminus
H^{\times}$ can be written as a (finite) product of atoms. A left ideal of $H$
is a subset $I\subset H$ with $HI\subset I$, and a right ideal of $H$ is
defined similarly. A _principal left (right) ideal of $H$_ is a set of the
form $Ha$ ($aH$) for some $a\in H$. If $H$ is a commutative monoid, then $p\in
H\setminus H^{\times}$ is a _prime element_ if $p\mid ab$ implies $p\mid a$ or
$p\mid b$ for all $a,b\in H$.

######  Proposition .

If $H$ satisfies the ACC on principal left and right ideals, then $H$ is
atomic.

###### Proof.

We first note that if $a,b\in H$ then $aH=bH$ if and only if $a=b\varepsilon$
with $\varepsilon\in H^{\times}$, and similarly $Ha=Hb$ if and only if
$a=\varepsilon b$ with $\varepsilon\in H^{\times}$. [We only show the
statement for the right ideals. The non-trivial direction is showing that
$aH=bH$ implies $a=b\varepsilon$. Since $aH=bH$ implies $a=bx$ and $b=ay$ with
$x,y\in H$, we get $a=a(yx)$ and $b=b(xy)$. Since $H$ is cancellative, this
implies $xy=t(b)=s(x)$ and $yx=t(a)=s(y)$, hence $y=x^{-1}$ and therefore
$x,y\in H^{\times}$.]

  1. Claim A.

If $a\in H\setminus H^{\times}$, then there exist $u\in\mathcal{A}(H)$ and
$a_{0}\in H$ such that $a=ua_{0}$.

###### Proof of Claim A.

Assume the contrary. Then the set

$\Omega=\\{\,a^{\prime}H\mid\text{$a^{\prime}\in H\setminus H^{\times}$ such
that there are no $u\in\mathcal{A}(H)$, $a_{0}\in H$ with
$a^{\prime}=ua_{0}$}\,\\}$

is non-empty, and hence, using the ascending chain condition on the principal
right ideals, possesses a maximal element $aH$ with $a\in H\setminus
H^{\times}$. Then $a\not\in\mathcal{A}(H)$, and therefore $a=bc$ with $b,c\in
H\setminus H^{\times}$. But $aH\subsetneq bH$ since $c\not\in H^{\times}$, and
thus maximality of $aH$ in $\Omega$ implies $b=ub_{0}$ with
$u\in\mathcal{A}(H)$ and $b_{0}\in H$. But then $a=u(b_{0}c)$, a
contradiction. ∎

We proceed to show that every $a\in H\setminus H^{\times}$ is a product of
atoms. Again, assume that this is not the case. Then

$\Omega^{\prime}=\\{\,Ha^{\prime}\mid\text{$a^{\prime}\in H\setminus
H^{\times}$ such that $a^{\prime}$ is not a product of atoms}\,\\}$

is non-empty, and hence possesses a maximal element $Ha$ with $a\in H\setminus
H^{\times}$ (this time using the ascending chain condition on principal left
ideals). Again $a\not\in\mathcal{A}(H)$ as otherwise it would be a product of
atoms. By Claim A, $a=ua_{0}$ with $u\in\mathcal{A}(H)$ and $a_{0}\in H$.
Since $a\not\in\mathcal{A}(H)$, $a_{0}\not\in H^{\times}$. Moreover,
$Ha\subsetneq Ha_{0}$ since $u\not\in H^{\times}$ and therefore
$a_{0}=u_{1}\cdot\ldots\cdot u_{l}$ with $l\in\mathbb{N}$ and
$u_{1},\ldots,u_{l}\in\mathcal{A}(H)$. Thus $a=uu_{1}\cdot\ldots\cdot u_{l}$
is a product of atoms, a contradiction. ∎

The following definition provides a natural notion of an ordered factorization
(called a _rigid factorization_) in a cancellative small category. It is
modeled after a terminology by Cohn [14, 15].

Let $\mathcal{F}(\mathcal{A}(H))$ denote the path category on atoms of $H$. We
define

$H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))=\\{\,(\varepsilon,y)\in
H^{\times}\times\mathcal{F}(\mathcal{A}(H))\mid t(\varepsilon)=s(y)\,\\},$

and define an associative partial operation on
$H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))$ as follows: If
$(\varepsilon,y),(\varepsilon^{\prime},y^{\prime})\in
H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))$ with
$\varepsilon,\varepsilon^{\prime}\in H^{\times}$,

$y=(e,u_{1},u_{2},\ldots,u_{k},f)\in\mathcal{F}(\mathcal{A}(H))\;\text{ and
}\;y^{\prime}=(e^{\prime},v_{1},v_{2},\ldots,v_{l},f^{\prime})\in\mathcal{F}(\mathcal{A}(H)),$

then the operation is defined if $t(y)=s(\varepsilon^{\prime})$, and

$(\varepsilon,y)\cdot(\varepsilon^{\prime},y^{\prime})=(\varepsilon,(e,u_{1},\ldots,u_{k}\varepsilon^{\prime},v_{1},v_{2},\ldots,v_{l},f^{\prime}))\quad\text{if
$k>0$, }$

while
$(\varepsilon,y)\cdot(\varepsilon^{\prime},y^{\prime})=(\varepsilon\varepsilon^{\prime},y^{\prime})$
if $k=0$. In this way $H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))$ is
again a cancellative small category (with identities $\\{\,(e,(e,e))\mid e\in
H_{0}\\}$ that we identify with $H_{0}$ again,
$s(\varepsilon,y)=s(\varepsilon)$ and $t(\varepsilon,y)=t(y)$). We define a
congruence relation $\sim$ on it as follows: If
$(\varepsilon,y),(\varepsilon^{\prime},y^{\prime})\in
H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))$ with $y,y^{\prime}$ as
before, then $(\varepsilon,y)\sim(\varepsilon^{\prime},y^{\prime})$ if $k=l$,
$\varepsilon u_{1}\cdot\ldots\cdot
u_{k}=\varepsilon^{\prime}v_{1}\cdot\ldots\cdot v_{l}\in H$ and either $k=0$
or there exist $\delta_{2},\ldots,\delta_{k}\in H^{\times}$ and
$\delta_{k+1}=t(u_{k})$ such that

$\varepsilon^{\prime}v_{1}=\varepsilon
u_{1}\delta_{2}^{-1}\quad\text{and}\quad
v_{i}=\delta_{i}u_{i}\delta_{i+1}^{-1}\text{ for all $i\in[2,k]$}.$

######  Definition 3.1.

The _category of rigid factorizations of $H$_ is defined as

$\mathsf{Z}^{*}(H)=(H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H)))/\sim,$

For $z\in\mathsf{Z}^{*}(H)$ with
$z=[(\varepsilon,(e,u_{1},u_{2},\ldots,u_{k},f))]_{\sim}$ we write
$z=\varepsilon u_{1}*\ldots*u_{k}$ and the operation on $\mathsf{Z}^{*}(H)$ is
also denoted by $*$. The _length_ of $z$ is $\lvert z\rvert=k$. There is a
surjective homomorphism $\pi\colon\mathsf{Z}^{*}(H)\to H$, induced by
multiplying out the elements of the factorization in $H$, explicitly
$\pi(z)=\varepsilon u_{1}u_{2}\cdot\ldots\cdot u_{k}\in H$. For $a\in H$, we
define $\mathsf{Z}^{*}(a)=\mathsf{Z}_{H}^{*}(a)=\pi^{-1}(\\{a\\})$ to be the
_set of rigid factorization of $a$_.

To simplify the notation, we make the following conventions:

  * •

If, for a rigid factorization $z=\varepsilon
u_{1}*\ldots*u_{k}\in\mathsf{Z}^{*}(H)$, we have $k>0$ (i.e., $\pi(z)\not\in
H^{\times}$), then the unit $\varepsilon$ can be absorbed into the first
factor $u_{1}$ (replacing it by $\varepsilon u_{1}$), and we can essentially
just work in $\mathcal{F}(\mathcal{A}(H))/\sim$, with $\sim$ defined to match
the equivalence relation on $H^{\times}\times_{r}\mathcal{F}(\mathcal{A}(H))$.

  * •

If $H$ is reduced but $\lvert H_{0}\rvert>1$, we often still write
$s(u_{1})u_{1}*\ldots*u_{k}$ instead of the shorter $u_{1}*\ldots*u_{k}$, as
$k=0$ is allowed and in the path category there is a different empty path for
every $e\in H_{0}$.

######  Remark .

  1. 1.

If $H$ is reduced, then $\mathsf{Z}^{*}(H)=\mathcal{F}(\mathcal{A}(H))$.

If $H$ is not reduced, the $H^{\times}$ factor allows us to represent trivial
factorizations of units, and the equivalence relation $\sim$ allows us to deal
with trivial insertion of units. In the commutative setting these
technicalities can easily be avoided by identifying associated elements and
passing to the reduced monoid $H_{\text{red}}=\\{\,aH^{\times}\mid a\in
H\,\\}$. Unfortunately, associativity (left, right or two-sided) is in general
no congruence relation in the non-commutative case.

  2. 2.

If $H$ is a commutative monoid, then $\mathsf{Z}^{*}(H)\cong
H^{\times}\times\mathcal{F}(\mathcal{A}(H_{\text{red}}))$, where
$\mathcal{F}(\mathcal{A}(H_{\text{red}}))$ is the free monoid on
$\mathcal{A}(H_{\text{red}})$, while a _factorization_ in this setting is
usually defined as an element of the free abelian monoid
$\mathsf{Z}(H)=\mathcal{F}_{\text{ab}}(\mathcal{A}(H_{\text{red}}))$, implying
in particular that factorizations are unordered while rigid factorizations are
ordered. The homomorphism $\pi:\mathsf{Z}^{*}(H_{\text{red}})\to
H_{\text{red}}$ obviously factors through the multiplication homomorphism
$\mathsf{Z}(H_{\text{red}})\to H_{\text{red}}$, and the fibers consist of the
different permutations of a factorization.

In the following we will only be concerned with invariants related to the
lengths of factorizations, which may as well be defined using rigid
factorizations.

######  Definition 3.2.

Let $a\in H$.

  1. 1.

We call

$\mathsf{L}(a)=\mathsf{L}_{H}(a)=\\{\,\lvert z\rvert\in\mathbb{N}_{0}\mid
z\in\mathsf{Z}^{*}(a)\,\\}$

the _set of lengths of $a$_.

  2. 2.

The _system of sets of lengths of $H$_ is defined as
$\mathcal{L}(H)=\\{\,\mathsf{L}(a)\subset\mathbb{N}_{0}\mid a\in H\,\\}$.

  3. 3.

A positive integer $d\in\mathbb{N}$ is a _distance of $a$_ if there exists an
$l\in\mathsf{L}(a)$ such that $\\{\,l,l+d\,\\}\in\mathsf{L}(a)$ and
$\mathsf{L}(a)\cap[l+1,l+d-1]=\emptyset$. The _set of distances of $a$_ is the
set consisting of all such distances and is denoted by
$\Delta(a)=\Delta_{H}(a)$. The _set of distances of $H$_ is defined as

$\Delta(H)=\bigcup_{a\in H}\Delta(a).$

  4. 4.

We define
$\mathcal{U}_{k}(H)=\bigcup_{\begin{subarray}{c}L\in\mathcal{L}(H)\\\ k\in
L\end{subarray}}L$ for $k\in\mathbb{N}_{0}$.

  5. 5.

$H$ is _half-factorial_ if $\lvert\mathsf{L}(a)\rvert=1$ for all $a\in H$
(equivalently, $H$ is atomic and $\Delta(H)=\emptyset$).

We write $b\mid_{H}^{r}a$ if $a\in Hb$ and similarly $b\mid_{H}^{l}a$ if $a\in
bH$.

######  Definition & Lemma .

Let $H\subset D$ be subcategories of a groupoid. The following are equivalent:

  1. (a)

For all $a,b\in H$, $b\mid_{D}^{r}a$ implies $b\mid_{H}^{r}a$,

  2. (b)

$HH^{-1}\cap D=H$.

$H\subset D$ is called _right-saturated_ if these equivalent conditions are
fulfilled.

###### Proof.

(a) $\Rightarrow$ (b): Let $c=ab^{-1}$ with $a,b\in H$, $t(a)=t(b)$ and $c\in
D$. Then $cb=a$, i.e., $b\mid_{D}^{r}a$ and hence also $b\mid_{H}^{r}a$. Since
the left factor is uniquely determined as $c=ab^{-1}$, it follows that $c\in
H$.

(b) $\Rightarrow$ (a): Let $b\mid_{D}^{r}a$. There exists $c\in D$ with
$cb=a$, and thus $c=ab^{-1}$. Therefore $c\in HH^{-1}\cap D=H$, hence
$b\mid_{H}^{r}a$. ∎

######  Definition 3.3.

Let $B$ be a reduced cancellative small category. A homomorphism $\theta\colon
H\to B$ is called a _transfer homomorphism_ if it has the following
properties:

  1. (T1)

$B=\theta(H)$ and $\theta^{-1}(B_{0})=H^{\times}$.

  2. (T2)

If $a\in H$, $b_{1},b_{2}\in B$ and $\theta(a)=b_{1}b_{2}$, then there exist
$a_{1},a_{2}\in H$ such that $a=a_{1}a_{2}$, $\theta(a_{1})=b_{1}$ and
$\theta(a_{2})=b_{2}$.

The notion of a transfer homomorphism plays a central role in studying sets of
lengths. It is easily checked that the following still holds in our
generalized setting (cf. [27, §3.2] for the commutative case, [25, Proposition
6.4] for the non-commutative monoid case).

######  Proposition .

If $\theta\colon H\to B$ is a transfer homomorphism, then
$\mathsf{L}_{H}(a)=\mathsf{L}_{B}(\theta(a))$ for all $a\in H$ and hence all
invariants defined in terms of lengths coincide for $H$ and $B$. In
particular,

  * •

$\mathcal{L}(H)=\mathcal{L}(B)$,

  * •

$\mathcal{U}_{k}(H)=\mathcal{U}_{k}(B)$ for all $k\in\mathbb{N}_{0}$,

  * •

$\Delta_{H}(a)=\Delta_{B}(\theta(a))$ for all $a\in H$, and
$\Delta(H)=\Delta(B)$.

######  Proposition .

Let $H$ be a cancellative small category, $G$ a finite abelian group and
$\theta\colon H\to\mathcal{B}(G)$ a transfer homomorphism. Then $H$ is half-
factorial if and only if $\lvert G\rvert\leq 2$. If $\lvert G\rvert\geq 3$,
then we have

  1. 1.

$\Delta(H)$ is a finite interval, and if it is non-empty, then
$\min\Delta(H)=1$,

  2. 2.

for every $k\geq 2$, the set $\mathcal{U}_{k}(H)$ is a finite interval,

  3. 3.

there exists an $M\in\mathbb{N}_{0}$ such that for every $a\in H$ the set of
lengths $\mathsf{L}(a)$ is an almost arithmetical multiprogression (AAMP) with
difference $d\in\Delta(H)$ and bound $M$.

###### Proof.

By the previous lemma it is sufficient to show these statements for the monoid
of zero-sum sequences $\mathcal{B}(G)$ over a finite abelian group $G$.
$\mathcal{B}(G)$ is half-factorial if and only if $\lvert G\rvert\leq 2$ by
[27, Proposition 2.5.6]. The first statement is proven in [29], the second can
be found in [24, Theorem 3.1.3]. For the definition of AAMPs and a proof of 3
see [27, Chapter 4]. ∎

The description in 3. is sharp by a realization theorem of W.A. Schmid [49].

##  4\. Factorization of integral elements in arithmetical groupoids

In this section we introduce arithmetical groupoids and study the
factorization behavior of integral elements. In Section 5 we will see that the
divisorial fractional one-sided ideals of suitable semigroups form such
groupoids. Thus in non-commutative semigroups arithmetical groupoids
generalize the free abelian group of divisorial fractional two-sided ideals
familiar from the commutative setting (see Section 4 and Section 4). This
abstract approach to factorizations was first used by Asano and Murata in [5].
We follow their ideas and also those of Rehm in [45, 46], who studies
factorizations of ideals in rings in a different abstract framework. The
notation and terminology for lattices follows [30], a reference for l-groups
is [51]. Section 4 is the main result on factorizations of integral elements
in a lattice-ordered groupoid (due to Asano and Murata). We introduce an
abstract norm homomorphism $\eta$, and as the main result in this section, we
present a transfer homomorphism to a monoid of zero-sum sequences in Section
4.

######  Definition 4.1.

A lattice-ordered groupoid $(G,\leq)$ is a groupoid $G$ together with a
relation $\leq$ on $G$ such that for all $e,f\in G_{0}$

  1. 1.

$(G(e,\cdot),\,\leq\mid_{G(e,\cdot)})$ is a lattice (we write
$\wedge^{\prime}_{e}$ and $\vee^{\prime}_{e}$ for the meet and join),

  2. 2.

$(G(\cdot,f),\,\leq\mid_{G(\cdot,f)})$ is a lattice (we write
$\wedge^{\prime\prime}_{f}$ and $\vee^{\prime\prime}_{f}$ for the meet and
join),

  3. 3.

$(G(e,f),\,\leq\mid_{G(e,f)})$ is a sublattice of both $G(e,\cdot)$ and
$G(\cdot,f)$. Explicitly: For all $a,b\in G(e,f)$ it holds that
$a\wedge^{\prime}_{e}b=a\wedge^{\prime\prime}_{f}b\in G(e,f)$ and
$a\vee^{\prime}_{e}b=a\vee^{\prime\prime}_{f}b\in G(e,f)$.

If $a,b\in G$ and $s(a)=s(b)$ we write $a\wedge b=a\wedge^{\prime}_{s(a)}b$
and $a\vee b=a\vee^{\prime}_{s(a)}b$. If $t(a)=t(b)$ we write $a\wedge
b=a\wedge^{\prime\prime}_{t(a)}b$ and $a\vee b=a\vee^{\prime\prime}_{t(a)}b$.
By 3 this is unambiguous if $s(a)=s(b)$ and $t(a)=t(b)$ both hold. The
restriction of $\leq$ to any of $G(e,\cdot)$, $G(\cdot,f)$ or $G(e,f)$ will in
the following simply be denoted by $\leq$ again. (Keep in mind however that
$\leq$ need not be a partial order on the entire set $G$, and $\wedge$ and
$\vee$ do not represent meet and join operations on the entire set $G$ in the
order-theoretic sense.)

An element $a$ of a lattice-ordered groupoid is called _integral_ if $a\leq
s(a)$ and $a\leq t(a)$, and we write $G_{+}$ for the subset of all integral
elements of $G$.

######  Definition 4.2.

A lattice-ordered groupoid $G$ is called an _arithmetical groupoid_ if it has
the following properties for all $e,f\in G_{0}$:

  1. (P1)

For $a\in G$, $a\leq s(a)$ if and only if $a\leq t(a)$.

  2. (P2)

$G(e,\cdot)$ and $G(\cdot,f)$ are modular lattices.

  3. (P3)

If $a\leq b$ for $a,b\in G(e,\cdot)$ and $c\in G(\cdot,e)$, then $ca\leq cb$.
Analogously, if $a,b\in G(\cdot,f)$ and $c\in G(f,\cdot)$, then $ac\leq bc$.

  4. (P4)

For every non-empty subset $M\subset G(e,\cdot)\cap G_{+}$, $\sup(M)\in
G(e,\cdot)$ exists, and similarly for $M\subset G(\cdot,f)\cap G_{+}$. If
moreover $M\subset G(e,f)$ then $\sup_{G(e,\cdot)}(M)=\sup_{G(\cdot,f)}(M)$.

  5. (P5)

$G(e,f)$ contains an integral element.

  6. (P6)

$G(e,\cdot)$ and $G(\cdot,f)$ satisfy the ACC on integral elements.

_For the remainder of this section, let $G$ be an arithmetical groupoid._

P5 implies in particular $G(e,f)\neq\emptyset$ for all $e,f\in G_{0}$, i.e.,
$G$ is connected. If $e,e^{\prime}\in G_{0}$ and $c\in G(e^{\prime},e)$, then
$G(e,\cdot)\to G(e^{\prime},\cdot),x\mapsto cx$ is an order isomorphism by P3,
and similarly every $d\in G(f,f^{\prime})$ induces an order isomorphism from
$G(\cdot,f)$ to $G(\cdot,f^{\prime})$. P2 could therefore equivalently be
required for a single $e$ and a single $f\in G_{0}$. Moreover, since the map
$(G(e,\cdot),\leq)\to(G(\cdot,e),\geq),x\mapsto x^{-1}$ is also an order
isomorphism (Lemma 4.1) and the property of being modular is self-dual, it is
in fact sufficient that one of $G(e,\cdot)$ and $G(\cdot,e)$ is modular for
one $e\in G_{0}$.

Using P5 we also observe that it is sufficient to have the ACC on integral
elements on one $G(e,\cdot)$ and one $G(\cdot,f)$: If, say, $a_{1}\leq
a_{2}\leq a_{3}\leq\ldots$ is an ascending chain of integral elements in
$G(e^{\prime},\cdot)$ and $c\in G(e,e^{\prime})$ is integral, then $ca_{1}\leq
ca_{2}\leq ca_{3}\leq\ldots$ is an ascending chain of integral elements in
$G(e,\cdot)$ (Lemma 4.2), hence becomes stationary, and multiplying by
$c^{-1}$ from the left again shows that the original chain also becomes
stationary.

We summarize some basic properties that follow immediately from the
definitions.

######  Lemma .

Let $e,f\in G_{0}$.

  1. 1.

$a\leq x\Leftrightarrow a^{-1}\geq x^{-1}$ holds if either $a,x\in G(e,\cdot)$
or $a,x\in G(\cdot,f)$. In particular, for $a\in G$ the following are
equivalent: (a) $a\leq s(a)$; (b) $a\leq t(a)$; (c) $a^{-1}\geq s(a)$; (d)
$a^{-1}\geq t(a)$.

  2. 2.

Let $a\in G(e,f)$. If $x\in G(\cdot,e)$ and $y\in G(f,\cdot)$ are integral,
then $xa\leq a$ and $ay\leq a$.

  3. 3.

If $a\in G(e,f)$, $x\in G(\cdot,e)$ and $y\in G(f,\cdot)$, then

    1. (i)

$x(a\vee b)=xa\vee xb$ and $x(a\wedge b)=xa\wedge xb$ if $b\in G(e,\cdot)$,

    2. (ii)

$(a\vee b)y=ay\vee by$ and $(a\wedge b)y=ay\wedge by$ if $b\in G(\cdot,f)$.

  4. 4.

Let $\emptyset\neq M\subset G(e,\cdot)$ and $x\in G(\cdot,e)$. If
$\sup_{G(e,\cdot)}(M)$ exists, then also $\sup_{G(s(x),\cdot)}(xM)$ exists,
and $\sup(xM)=x\sup(M)$. Moreover, then also $\inf_{G(\cdot,e)}(M^{-1})$
exists and $\inf(M^{-1})=\sup(M)^{-1}$. Analogous statements hold for
$\emptyset\neq M\subset G(\cdot,f)$ and $x\in G(f,\cdot)$.

  5. 5.

$G(e,\cdot)$, $G(\cdot,f)$, $G(e,f)$ and in particular $G(e)$ are
conditionally complete as lattices.

  6. 6.

The set $G_{+}$ of all integral elements forms a reduced wide subcategory of
$G$, and $G=\mathbf{q}(G_{+})$ is the groupoid of (left and right) fractions
of this subcategory.

  7. 7.

For every $a\in G(e,f)$, there exist $b\in G(e)$ and $c\in G(f)$ with $b\leq
a$ and $c\leq a$.

###### Proof.

  1. 1.

Assume first $s(x)=s(a)$. By P3, $a\leq x$ if and only if $x^{-1}a\leq t(x)$.
By P1 this is equivalent to $x^{-1}a\leq t(a)$. Again by P3 this is equivalent
to $x^{-1}\leq a^{-1}$. The case $t(x)=t(a)$ is proven similarly.

(a) $\Leftrightarrow$ (b) and (c) $\Leftrightarrow$ (d) by P1. For (a)
$\Leftrightarrow$ (c) set $x=s(a)$.

  2. 2.

Since $x\leq t(x)=s(a)$, we have $xa\leq s(a)a=a$ by P3. Similarly, $by\leq
y$.

  3. 3.

We show (i), (ii) is similar. Since $a\leq a\vee b$ and $b\leq a\vee b$, P3
implies $xa\leq x(a\vee b)$ and $xb\leq x(a\vee b)$, thus $xa\vee xb\leq
x(a\vee b)$. Therefore

$a\vee b=(x^{-1}xa)\,\vee\,(x^{-1}xb)\leq x^{-1}(xa\vee xb),$

and multiplying by $x$ from the left gives $x(a\vee b)\leq xa\vee xb$. Dually,
$x(a\wedge b)=xa\wedge xb$.

  4. 4.

Let $c=\sup(M)$. Then for all $m\in M$, $xm\leq xc$, hence $xc$ is an upper
bound for $xM$. If $d\in G(s(x),\cdot)$ is another upper bound for $xM$, then
$m\leq x^{-1}d$ for all $m\in M$, hence $c\leq x^{-1}d$ and thus $xc\leq d$.
Therefore $xc=\sup(xM)$.

For $d\in G(e,\cdot)$ we have $m\leq d$ for all $m\in M$ if and only if
$m^{-1}\geq d^{-1}$ (in $G(\cdot,e)$), and $\inf(M^{-1})=\sup(M)^{-1}$
follows.

  5. 5.

We show the claim for $G(e,\cdot)$, for $G(\cdot,f)$ the proof is similar. Let
$\emptyset\neq M\subset G(e,\cdot)$ be bounded, say $x\leq m\leq y$ for some
$x,y\in G(e,\cdot)$ and all $m\in M$. Then $y^{-1}M\subset G(t(y),\cdot)$ is
integral, hence $\sup(y^{-1}M)$ exists by P4, and $\sup(M)=y\sup(y^{-1}M)$ by
4. Similarly, $M^{-1}x\subset G(\cdot,t(x))$ is integral, and therefore
$\sup(M^{-1}x)$ exists, implying
$\inf(M)=\sup(M^{-1})^{-1}=x\sup(M^{-1}x)^{-1}$.

The proof for $G(e,f)$ is similar but uses in addition
$\sup_{G(t(y),\cdot)}(y^{-1}M)=\sup_{G(\cdot,f)}(y^{-1}M)$ (from P4), to
ensure that the supremum lies in $G(e,f)$ again.

  6. 6.

By 2 and the fact that every $e\in G_{0}$ is integral by definition, $G_{+}$
forms a wide subcategory of $G$. If $a\in G_{+}\setminus G_{0}$, then
$a<s(a)$, thus $a^{-1}>s(a)$ and therefore $a^{-1}$ is not integral. Hence the
subcategory of integral elements is reduced. Let $x\in G$ and $e=s(x)$. Then
$a=x\wedge e\leq e$, hence $a$ is integral. Since $a\leq x$, also $x^{-1}a\leq
t(x)$ is integral. Set $b=x^{-1}a$. Then $x=ab^{-1}$ with $a,b\in G_{+}$.
Similarly one can find $c,d\in G_{+}$ with $x=d^{-1}c$.

  7. 7.

By P5 there exist integral $b^{\prime}\in G(f,e)$ and $c^{\prime}\in G(e,f)$.
Set $b=ab^{\prime}$ and $c=c^{\prime}a$. Then $b\leq a$, $c\leq a$ and $b\in
G(e)$, $c\in G(f)$. ∎

For $e,f\in G$ it is immediate from the definitions that
$G_{+}(e,\cdot)=G(e,\cdot)\cap G_{+}$, $G_{+}(\cdot,f)=G(\cdot,f)\cap G_{+}$
and $G_{+}(e,f)=G(e,f)\cap G_{+}$. Moreover, $G_{+}(e,\cdot)$ is a sublattice
of $G(e,\cdot)$, $G_{+}(\cdot,f)$ is a sublattice of $G(\cdot,f)$, and
$G_{+}(e,f)$ is a sublattice of $G(e,f)$.

If $a,b\in G_{+}(e,\cdot)$, then $a\leq b$ if and only if $b\mid_{G_{+}}^{l}a$
as $a=b(b^{-1}a)$, and $b^{-1}a$ is integral if and only if $a\leq b$.
Similarly, if $a,b\in G_{+}(\cdot,f)$, then $a\leq b$ if and only if
$b\mid_{G_{+}}^{r}a$. Correspondingly, for integral elements with the same
left (right) identity, we may view the join and meet operations as left
(right) gcd and lcm.

######  Definition & Lemma .

For $u\in G$ the following are equivalent:

  1. (a)

$u$ is maximal in $G_{+}(s(u),\cdot)\setminus\\{\,s(u)\,\\}$,

  2. (b)

$u$ is maximal in $G_{+}(\cdot,t(u))\setminus\\{\,t(u)\,\\}$,

  3. (c)

$u\in\mathcal{A}(G_{+})$.

An element $u\in G$ satisfying these equivalent conditions is called _maximal
integral_.

###### Proof.

(a) $\Rightarrow$ (b): By definition, $u$ is maximal in $G(s(u),\,\cdot)$ with
$u<s(u)$. If $u\leq y<t(u)$ with $y\in G(\cdot,\,t(u))$, then $uy^{-1}\leq
yy^{-1}=s(y)$, hence $uy^{-1}\in G(s(u),\,\cdot)$ is integral, and therefore
$u<uy^{-1}\leq s(u)$. By maximality of $u$ in the first set, therefore
$uy^{-1}=s(u)$, whence $y=u$ and $u$ is maximal in the second set.

(b) $\Rightarrow$ (c): Assume $u=vw$ with $v,w\in G_{+}\setminus G_{0}$. Then
$u<w<t(u)$, contradicting the maximality of $u$ in $G_{+}(\cdot,t(u))$.

(c) $\Rightarrow$ (a): Let $v\in G_{+}(s(u),\cdot)$ with $u\leq v<s(u)$. Then
$u=v(v^{-1}u)$ with $v$ and $v^{-1}u$ integral, and since $v\not\in G_{0}$
necessarily $v^{-1}u\in G_{0}$, i.e., $u=v$. ∎

######  Lemma .

Let $U$ be an l-group. For $p\in U$ the following are equivalent:

  1. (a)

$p$ is maximal integral,

  2. (b)

$p$ is a prime element in $U_{+}$,

  3. (c)

$p\in\mathcal{A}(U_{+})$.

###### Proof.

(a) $\Leftrightarrow$ (c) is shown as in Section 4. It suffices to show (a)
$\Rightarrow$ (b) and (b) $\Rightarrow$ (c). Let $e$ be the identity of $U$.

(a) $\Rightarrow$ (b): Let $p$ be maximal in $U_{+}$ with $p\neq e$. Assume
$p\mid ab$ for $a,b\leq e$. That means $ab\leq p$. Assume $a\not\leq p$. Then
$b\,=\,(a\vee p)b\,=\,ab\,\vee\,pb\,\leq\,p\,\vee\,pb\,=\,p$, i.e., $p\mid b$.

(b) $\Rightarrow$ (c): Let $p$ be a prime, $p=ab$ with $a,b\leq e$. Say $p\mid
a$, i.e., $a\leq p$. Then $a\leq p\leq a$ implies $p=a$ and therefore $b=e$. ∎

######  Proposition .

  1. 1.

If $G$ is a group (i.e., $\lvert G_{0}\rvert=1$), then $G$ is the free abelian
group with basis $\mathcal{A}(G_{+})$, and $G_{+}$ is the free abelian monoid
with basis $\mathcal{A}(G_{+})$. Moreover $\gcd(a,b)=a\vee b$ and
$\operatorname{lcm}(a,b)=a\wedge b$.

  2. 2.

Let $\mathcal{M}$ be a set, $F$ the free abelian group with basis
$\mathcal{M}$, and $H\subset F$ the free abelian monoid with the same basis. A
lattice order is defined on $F$ by $a\leq b$ if $a=cb$ with $c\in H$, and
$(F,\leq)$ is an arithmetical groupoid with $F_{+}=H$ and
$\mathcal{M}=\mathcal{A}(F_{+})$.

  3. 3.

For every $e\in G_{0}$, the group isomorphism $j_{e}\colon G(e)\to\mathbb{G}$
induces the structure of an arithmetical groupoid on $\mathbb{G}$, and the
induced structure on $\mathbb{G}$ is independent of the choice of $e$.

$\mathbb{G}$ ($G(e)$) is a free abelian group and $\mathbb{G}_{+}$
($G(e)_{+}$) is a free abelian monoid with basis $\mathcal{A}(\mathbb{G}_{+})$
($\mathcal{A}(G(e)_{+})$). Moreover, $j_{e}(G(e)_{+})=\mathbb{G}_{+}$ and
$j_{e}(\mathcal{A}(G(e)_{+}))=\mathcal{A}(\mathbb{G}_{+})$.

###### Proof.

  1. 1.

$G$ is an l-group, and by Lemma 4.5 it is conditionally complete. Therefore
$G$ is commutative ([51, Theorems 2.3.1(d) and 2.3.9]). Since it satisfies the
ACC on integral elements, $G_{+}$ is atomic (Section 3). By the previous
lemma, every atom of $G_{+}$ is a prime element, and therefore $G_{+}$ is
factorial. Because it is also reduced, $G_{+}$ is the free abelian monoid with
basis $\mathcal{A}(G_{+})$. Now $G=\mathbf{q}(G_{+})$ implies that $G$ is the
free abelian group with basis $\mathcal{A}(G_{+})$. Finally, $\gcd(a,b)=a\vee
b$ and $\gcd(a,b)=a\wedge b$ for $a,b\in G_{+}$ follow because $c\leq d$ if
and only if $d\mid c$ for all $c,d\in G_{+}$.

  2. 2.

Clearly $\leq$ defines a lattice order on $F$, and the properties of an
arithmetical groupoid are, except for P2, either trivial, or easily checked.
For P2 recall that every l-group is distributive, hence modular, as a lattice
([51, Theorem 2.1.3(a)]). Now $F_{+}=H$ and $\mathcal{A}(F_{+})=\mathcal{M}$
are immediate from the definitions.

  3. 3.

For every $e\in G_{0}$ the vertex group $G(e)$ is an arithmetical groupoid, as
is easily checked. Via the group isomorphism $j_{e}:G(e)\to\mathbb{G}$
therefore $\mathbb{G}$ gains the structure of an arithmetical groupoid. If
$f\in G_{0}$ and $c\in G(e,f)$, then for all $x,y\in G(e)$ we have $x\leq
y\Leftrightarrow c^{-1}xc\leq c^{-1}yc$, and since $j_{f}^{-1}\circ
j_{e}(x)=c^{-1}xc$, the induced order on $\mathbb{G}$ is independent of the
choice of $e$.

By 1., applied to $\mathbb{G}$, respectively $G(e)$, the remaining claims
follow (for $j_{e}(\mathcal{A}(\mathbb{G}_{+}))=\mathcal{A}(G(e)_{+})$ use the
characterization of atoms as maximal integral elements from Section 4). ∎

Let $[a,b]$, $[c,d]$ be intervals in a lattice. Recall that $[a,b]$ is _down-
perspective_ to $[c,d]$ if $c=a\wedge d$ and $b=a\vee d$. Moreover, $[a,b]$ is
_perspective_ to $[c,d]$ if either $[a,b]$ is down-perspective to $[c,d]$, or
$[c,d]$ is down-perspective to $[a,b]$. The intervals $[a,b]$, $[c,d]$ are
_projective_ if there exists a finite sequence of intervals
$[a,b]=[a_{0},b_{0}],[a_{1},b_{1}],\ldots,[a_{k},b_{k}]=[c,d]$ such that
$[a_{i-1},b_{i-1}]$ is perspective to $[a_{i},b_{i}]$ for all $i\in[1,k]$.
(See [30, Chapter I.3.5].)

######  Definition 4.3.

  1. 1.

An element $a\in G_{+}(e,f)$ is _transposable_ to an element $b\in
G_{+}(e^{\prime},f^{\prime})$ if there exists an element $c\in
G_{+}(e,e^{\prime})$ such that $[a,e]$ is down-perspective to $[cb,c]$.
Explicitly: $b=c^{-1}(c\wedge a)$ and $c\vee a=e$.

  2. 2.

An element $a\in G_{+}$ is _projective_ to an element $b\in G_{+}$ if there
exists a sequence of integral elements $a=c_{0},c_{1},\ldots,c_{n},c_{n+1}=b$,
such that for any pair of successive elements $(c_{i},c_{i+1})$ either $c_{i}$
is transposable to $c_{i+1}$ or $c_{i+1}$ is transposable to $c_{i}$.

It is easily checked that being transposable is a transitive and reflexive
relation (but not symmetric), and projectivity is an equivalence relation.
Note that in a modular lattice perspective intervals are isomorphic ([30,
p.308, Theorem 348]), and therefore in particular the lengths of $[a,s(a)]$
and $[b,s(b)]$ coincide if $a$ is projective to $b$.

######  Lemma .

If $a,b,c$ are as in the definition of transposability, then $cb=c\wedge a=ad$
for some $d\in G_{+}$ and $c$ is transposable to $d$.

###### Proof.

Since $c\wedge a\leq a$, there exists an integral $d$ with $ad=c\wedge a$, and
$cb=c\wedge a$ by definition of transposability. The claim follows from
$d=a^{-1}(c\wedge a)$ and $c\vee a=e$. ∎

######  Lemma .

If two lattice intervals $[a,b]$ and $[c,d]$ of $G(e,\cdot)$ are projective,
then the integral elements $b^{-1}a$ and $d^{-1}c$ are projective to each
other in the sense of the previous definition.

###### Proof.

It suffices to show that if the lattice interval $[a,b]$ is down-perspective
to $[c,d]$, then $b^{-1}a$ is transposable to $d^{-1}c$. Then $a=bx$, $d=by$
and $c=dz=byz$ with $x,y,z\in G_{+}$.

$\textstyle{b\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{d=by}$$\textstyle{bx=a}$$\textstyle{c=byz}$

Since $[a,b]$ is down-perspective to $[c,d]$, we get

$\displaystyle b$ $\displaystyle=a\vee d=bx\vee by=b(x\vee y)$ and therefore
$\displaystyle x\vee y$ $\displaystyle=s(x)=t(b),\text{ and }$ $\displaystyle
c$ $\displaystyle=a\wedge d=bx\wedge by=b(x\wedge y)$ and therefore
$\displaystyle x\wedge y$ $\displaystyle=b^{-1}c=yz.$

Thus $d^{-1}c=z=y^{-1}(x\wedge y)$ with $y\in G(t(b),t(d))$, and hence
$x=b^{-1}a$ is transposable to $d^{-1}c$. ∎

######  Definition & Lemma .

For every $a\in G$,

$\\{\,\mathcal{X}\in\mathbb{G}\mid\mathcal{X}_{s(a)}\leq
a\,\\}=\\{\,\mathcal{X}\in\mathbb{G}\mid\mathcal{X}_{t(a)}\leq a\,\\},$

and we write $\mathbb{G}_{\leq a}$ for this set. The _lower bound_ $\Phi\colon
G\to\mathbb{G}$ is defined by $\Phi(a)=\sup(\mathbb{G}_{\leq a})$.

###### Proof.

Let $\mathcal{X}\in\mathbb{G}$. Recall from Section 2.2 that
$\mathcal{X}_{s(a)}=j_{s(a)}^{-1}(\mathcal{X})$ denotes the unique
representative of $\mathcal{X}$ in $G(s(a))$, and that
$\mathcal{X}_{t(a)}=a^{-1}\mathcal{X}_{s(a)}a$. We have to show that
$\mathcal{X}_{s(a)}\leq a$ if and only if $\mathcal{X}_{t(a)}\leq a$, but this
follows from $\mathcal{X}_{s(a)}\leq
a\;\;\Leftrightarrow\;\;a^{-1}\mathcal{X}_{s(a)}a\leq a^{-1}aa=a$. ∎

With the definition of $\Phi$ and the notation of Section 2.2 we have: If
$a\in G(e,f)$, then $\Phi(a)_{e}=\sup\\{\,x\in G(e)\mid x\leq a\,\\}\in G(e)$,
$\Phi(a)_{f}=\sup\\{\,x\in G(f)\mid x\leq a\,\\}\in G(f)$,
$\Phi(a)_{f}=a^{-1}\Phi(a)_{e}a=b^{-1}\Phi(a)_{e}b$ for all $b\in G(e,f)$ and
$\Phi(a)=j_{e}(\Phi(a)_{e})=j_{f}(\Phi(a)_{f})$.

######  Lemma .

Let $e,f\in G_{0}$.

  1. 1.

If $a,b\in G(e,\cdot)$ or $a,b\in G(\cdot,f)$ with $a\leq b$, then
$\Phi(a)\leq\Phi(b)$. In particular, if $a\in G_{+}$, then
$\Phi(a)\in\mathbb{G}_{+}$.

  2. 2.

If $a\in G(e,f)$, $b\in G(f,\cdot)$ then $\Phi(a)\Phi(b)\leq\Phi(ab)$. If
moreover $a,b\in G_{+}$, then $\Phi(ab)\leq\Phi(a)\wedge\Phi(b)$, and if
furthermore $\Phi(a)$ and $\Phi(b)$ are coprime, then
$\Phi(ab)=\Phi(a)\Phi(b)$.

  3. 3.

If $u\in\mathcal{A}(G_{+})$, then $\Phi(u)\in\mathbb{G}_{+}$ is prime.

  4. 4.

Let $u,v\in G_{+}$ be projective elements. If $u\in\mathcal{A}(G_{+})$, then
$v\in\mathcal{A}(G_{+})$, and $\Phi(u)=\Phi(v)$.

###### Proof.

  1. 1.

Immediate from the definition of $\Phi$.

  2. 2.

Observe that $c=a^{-1}\Phi(a)_{e}\in G(f,e)$ is integral. Therefore

$\Phi(a)_{e}\Phi(b)_{e}=ac\Phi(b)_{e}=acc^{-1}\Phi(b)_{f}c=a\Phi(b)_{f}c\leq
a\Phi(b)_{f}\leq ab,$

and hence $\Phi(a)\Phi(b)\leq\Phi(ab)$.

Let now $a,b$ be integral. Then 1. implies $\Phi(ab)\leq\Phi(a)$ and
$\Phi(ab)\leq\Phi(b)$, so $\Phi(ab)\leq\Phi(a)\wedge\Phi(b)$. The last
statement follows because
$\Phi(a)\wedge\Phi(b)=\operatorname{lcm}(\Phi(a),\Phi(b))$ in
$\mathbb{G}_{+}$.

  3. 3.

By Section 4 it suffices to show $\Phi(u)\in\mathcal{A}(\mathbb{G}_{+})$. If
$e=s(u)$, then it suffices to prove $\Phi(u)_{e}\in\mathcal{A}(G(e)_{+})$ (by
Proposition 4.3). Assume that $\Phi(u)_{e}=ab$ with $a,b\in G(e)$ such that
$a<e$ and $b<e$. Then $b\vee u=e$, since $b>ab=\Phi(u)_{e}$, and therefore

$u\,\geq\,ab\vee au\,=\,a(b\vee u)\,=\,a,$

a contradiction to $a>ab=\Phi(u)_{e}$.

  4. 4.

We first show that $v$ is maximal integral, and may assume that either $u$ is
transposable to $v$ or $v$ is transposable to $u$. Let $e=s(u)$ and $f=s(v)$.
Assume first that $u$ is transposable to $v$ via $c\in G_{+}(e,f)$. Then
$[u,e]$ is down-perspective to $[cv,c]$, and since $G(e,\cdot)$ is modular,
the intervals are isomorphic, hence have the same length (namely $1$).
Multiplying from the left by $c^{-1}$ therefore also $[v,f]$ has length $1$,
and thus $v$ is maximal integral. If $v$ is transposable to $u$, one argues
along similar lines.

For the remainder of the claim we may now assume that $u$ is transposable to
$v$ (since we already know that $v$ is also maximal integral). Let again $c\in
G_{+}(e,f)$ be such that $cv=c\wedge u$ and $e=c\vee u$. If $p=\Phi(u)_{e}$,
then $c^{-1}pc=\Phi(u)_{f}$. Since $pc\leq c\wedge p\leq c\wedge u=cv$, we get
$c^{-1}pc\leq v$ and therefore
$\Phi(v)\geq\Phi(c^{-1}pc)=\Phi(\Phi(u)_{f})=\Phi(u)$. By 3., $\Phi(u)$ is
prime and thus maximal integral in $\mathbb{G}_{+}$, which implies
$\Phi(u)=\Phi(v)$. ∎

The converse of Lemma 4.3 is false in general: A non-maximal integral element
can have a prime lower bound.

######  Proposition .

  1. 1.

The category $G_{+}$ is half-factorial. Explicitly: Every $a\in G_{+}$
possesses a rigid factorization

$s(a)u_{1}*\ldots*u_{k}\in\mathsf{Z}^{*}(a)$

with $k\in\mathbb{N}_{0}$ and $u_{1},\ldots,u_{k}\in\mathcal{A}(G_{+})$ and
the number of factors, $k\in\mathbb{N}_{0}$, is uniquely determined by $a$.
Moreover, if $s(a)v_{1}*\ldots*v_{k}\in\mathsf{Z}^{*}(a)$ is another rigid
factorization with $v_{1},\ldots,v_{k}\in\mathcal{A}(G_{+})$, then there
exists a permutation $\tau\in\mathfrak{S}_{k}$ such that $u_{\tau(i)}$ is
projective to $v_{i}$ for all $i\in[1,k]$. In particular,
$\Phi(u_{\tau(i)})=\Phi(v_{i})$ for all $i\in[1,k]$.

  2. 2.

Any two rigid factorizations of $a\in G_{+}$ can be transformed into each
other by a number of steps, each of which only involves replacing two
successive elements by two new ones.

  3. 3.

(Transposition.) If $a=uv$ with $u,v\in\mathcal{A}(G_{+})$ and
$\Phi(u)=\mathcal{P}$, $\Phi(v)=\mathcal{Q}$, $\mathcal{Q}\neq\mathcal{P}$,
then there exist uniquely determined
$v^{\prime},u^{\prime}\in\mathcal{A}(G_{+})$ such that
$\Phi(v^{\prime})=\mathcal{Q}$, $\Phi(u^{\prime})=\mathcal{P}$ and
$uv=v^{\prime}u^{\prime}$.

Explicitly,

$\displaystyle u^{\prime}$ $\displaystyle=a\vee\mathcal{P}_{t(a)},$
$\displaystyle u^{\prime}\wedge v$ $\displaystyle=a,$ $\displaystyle
u^{\prime}\vee v$ $\displaystyle=t(a),$ $\displaystyle v^{\prime}$
$\displaystyle=a\vee\mathcal{Q}_{s(a)},$ $\displaystyle u\wedge v^{\prime}$
$\displaystyle=a,$ $\displaystyle u\vee v^{\prime}$ $\displaystyle=s(a).$

So $u$ is transposable to $u^{\prime}$ and $v^{\prime}$ is transposable to
$v$.

  4. 4.

Given any permutation $\tau^{\prime}\in\mathfrak{S}_{k}$, there exist
$w_{1},\ldots,w_{k}\in\mathcal{A}(G_{+})$, such that

$s(a)w_{1}*\ldots*w_{k}\in\mathsf{Z}^{*}(a)$

and $\Phi(w_{i})=\Phi(u_{\tau^{\prime}(i)})$ for all $i\in[1,k]$.

###### Proof.

  1. 1, 2.

We observe that rigid factorizations of $a$ correspond bijectively to maximal
chains of the sublattice $[a,s(a)]$ of $G(s(a),\cdot)$: If
$s(a)u_{1}*\ldots*u_{k}\in\mathsf{Z}^{*}(a)$, then
$s(a)>u_{1}>u_{1}u_{2}>\ldots>u_{1}\cdot\ldots\cdot u_{k}=a$ is a chain in
$[a,s(a)]$ and since $u_{1},\ldots,u_{k}$ are maximal integral, it is in fact
a maximal chain of $[a,s(a)]$. Conversely, if
$s(a)=x_{0}>x_{1}>x_{2}>\ldots>x_{k}=a$ is a maximal chain of $[a,s(a)]$ then
we set $u_{i}=x_{i-1}^{-1}x_{i}$ for all $i\in[1,k]$. These elements are
maximal integral, i.e., atoms of $G_{+}$, and
$a=x_{k}=s(a)x_{1}(x_{1}^{-1}x_{2})\cdot\ldots\cdot(x_{k-2}^{-1}x_{k-1})(x_{k-1}^{-1}x_{k})=s(a)u_{1}\cdot\ldots\cdot
u_{k}$.

By P6, $[a,s(a)]$ satisfies the ACC, but also the DCC because if
$s(a)=x_{0}\geq x_{1}\geq\ldots\geq a$ is a descending chain in $[a,s(a)]$,
then $x_{0}^{-1}a\leq x_{1}^{-1}a\leq\ldots\leq a^{-1}a=t(a)$ is an ascending
chain in $G_{+}(\cdot,t(a))$ and therefore becomes stationary again by P6.
Being a modular lattice, $[a,s(a)]$ is therefore of finite length.

The claims now follow from the Jordan-Hölder Theorem for modular lattices (see
e.g., [30, p.333, Theorem 377]). The existence of maximal chains implies that
$G_{+}$ is atomic (alternatively, use Section 3 together with the ACC on
integral elements). For half-factoriality, and projectivity of the factors,
assume that $s(a)=x_{0}>x_{1}>x_{2}>\ldots>x_{k}=a$ and
$s(a)=y_{0}>y_{1}>y_{2}>\ldots>y_{l}=a$ are two maximal chains from which
rigid factorizations with factors $u_{i}=x_{i-1}^{-1}x_{i}$ for $i\in[1,k]$
and $v_{i}=y_{i-1}^{-1}y_{i}$ for $i\in[1,l]$ are derived. Then the uniqueness
part of the Jordan-Hölder Theorem implies $k=l$ and that there exists a
permutation $\tau\in\mathfrak{S}_{k}$ such that $[x_{\tau(i)},x_{\tau(i)-1}]$
is projective to $[y_{i},y_{i-1}]$ for all $i\in[1,k]$. By Section 4, this
implies that $u_{\tau(i)}$ is projective to $v_{i}$ for all $i\in[1,k]$.

Finally, 2. follows in a similar manner by induction on the length of $a$. Fix
a composition series of $[u_{1}\wedge v_{1},a]$. This gives rise to
refinements of $s(a)>u_{1}>u_{1}\wedge v_{1}>a$ and $s(a)>v_{1}>u_{1}\wedge
v_{1}>a$ to composition series of $[a,s(a)]$. Applying the induction
hypothesis to $u_{2}*\ldots*u_{k}$ (respectively $v_{2}*\ldots*v_{k}$), and
the rigid factorization derived from the refined chain
$t(u_{1})>u_{1}^{-1}(u_{1}\wedge v_{1})>\ldots>u_{1}^{-1}a$ (respectively
$t(v_{1})>v_{1}^{-1}(u_{1}\wedge v_{1})>\ldots>v_{1}^{-1}a$) one proves the
claim.

  2. 3.

Let $e=s(u)$, $q=\mathcal{Q}_{e}$ and set $v^{\prime}=uv\vee q$. We first
show:

    1. Claim A.

$q\not\leq u$,

    2. Claim B.

$v^{\prime}\wedge u=uv$,

    3. Claim C.

$v^{\prime}$ is maximal integral.

_Proof of Claim A._ Suppose $q\leq u$. Then
$\mathcal{Q}\leq\Phi(u)=\mathcal{P}$, a contradiction to $\mathcal{P}$ and
$\mathcal{Q}$ being distinct prime elements of $\mathbb{G}_{+}$ (Lemma 4.3).

_Proof of Claim B._ Since $G(e,\cdot)$ is modular and $uv\leq u$,

$v^{\prime}\wedge u=(uv\vee q)\wedge u=uv\vee(q\wedge u).$

Because $q\not\leq u$ (Claim A), we have $q>q\wedge u\geq qu$ and thus, by
maximality of $u$, $qu=q\wedge u$. Therefore

$uv\vee(q\wedge u)=uv\vee qu=uv\vee
u\mathcal{Q}_{t(u)}=u(v\vee\mathcal{Q}_{t(u)})=uv.$

_Proof of Claim C._ Since $uv$ is a product of two atoms and $G_{+}$ is half-
factorial, it suffices to show $uv<v^{\prime}<e$. Suppose first
$uv=v^{\prime}$. Then $q\leq uv\leq u$, contradicting Claim A. Assume now
$v^{\prime}=e$. Then $u=e\wedge u=v^{\prime}\wedge u$, and by Claim B
therefore $u=uv$, contradicting $v<s(v)$.

_Existence._ We have $uv=v^{\prime}u^{\prime}$ with
$v^{\prime}\in\mathcal{A}(G_{+})$ (by Claim C) and
$u^{\prime}=v^{\prime-1}uv\in G_{+}$. Since $G_{+}$ is half-factorial, this
necessarily implies $u^{\prime}\in\mathcal{A}(G_{+})$. By definition of
$v^{\prime}$, $\mathcal{Q}\leq\Phi(v^{\prime})<1_{\mathbb{G}}$, where the
latter inequality is strict because $v^{\prime}<e$. Thus
$\Phi(v^{\prime})=\mathcal{Q}$ and 1. implies $\Phi(u^{\prime})=\mathcal{P}$.

_Uniqueness._ If $v^{\prime\prime}u^{\prime\prime}=uv$ with
$\Phi(v^{\prime\prime})=\mathcal{Q}$, then $v^{\prime\prime}<e$ and
$v^{\prime\prime}\geq uv\vee q=v^{\prime}$. By Claim C, $v^{\prime}$ is
maximal integral and thus $v^{\prime\prime}=v^{\prime}$, and then also
$u^{\prime\prime}=u^{\prime}$.

_Explicit formulas._ Since $e\geq u\vee
v^{\prime}\geq\Phi(u)_{e}\vee\Phi(v^{\prime})_{e}=\mathcal{P}_{e}\vee\mathcal{Q}_{e}=(\mathcal{P}\vee\mathcal{Q})_{e}=(1_{\mathbb{G}})_{e}=e$,
it follows that $u\vee v^{\prime}=e$. By Claim B, $u\wedge v^{\prime}=uv=a$.

The equalities $u^{\prime}=a\vee\mathcal{P}_{t(a)}$, $u^{\prime}\wedge v=a$
and $u^{\prime}\vee v=t(a)$ are shown similarly.

  3. 4.

Write $\tau^{\prime}$ as a product of transpositions and use 3. ∎

Proposition 4.3 gives an explicit and complete description of the possible
relations between two maximal integral elements with coprime lower bound. The
case where the lower bounds coincide is more complicated (there can be no
relations, or many), but in the case where we will need it, it is quite simple
(see Section 7.2).

######  Corollary .

If $H\subset G_{+}$ is a subcategory, then $\mathsf{L}_{H}(a)$ is finite and
non-empty for all $a\in H$. If for every prime $\mathcal{P}\in\mathbb{G}_{+}$
and all (equivalently, one) $e\in G_{0}$ the set
$\\{\,u\in\mathcal{A}(G_{+})\mid\Phi(u)=\mathcal{P}\text{ and }s(u)=e\,\\}$ is
finite, then $\mathsf{Z}^{*}_{H}(a)$ is finite for all $a\in H$.

###### Proof.

Using that $G_{+}$ is reduced, it follows from P6 that $H$ satisfies the ACC
on principal left and right ideals, and hence
$\mathsf{Z}_{H}^{*}(a)\neq\emptyset$. If
$s(a)u_{1}*\ldots*u_{k}\in\mathsf{Z}_{H}^{*}(a)$ with $k\in\mathbb{N}_{0}$ and
$u_{1},\ldots,u_{k}\in\mathcal{A}(H)$, then in particular $u_{i}<s(u_{i})$ for
all $i\in[1,k]$, and therefore $k$ is bounded by the length of the
factorization of $a$ in $G_{+}$. A similar argument shows the second claim. ∎

The properties that all sets of lengths, respectively that all sets of
factorizations, are finite have been studied a lot in the commutative setting.
Note, if $H$ is a commutative monoid and $a\in H$, then $\mathsf{Z}_{H}(a)$ is
finite if and only if $\mathsf{Z}^{*}_{H}(a)$ is finite.

######  Definition & Lemma .

There exists a unique groupoid epimorphism $\eta\colon G\to\mathbb{G}$ such
that $\eta(u)=\Phi(u)$ for all $u\in\mathcal{A}(G_{+})$. We call $\eta$ the
_abstract norm_ of $G$.

###### Proof.

We need to show existence and uniqueness of such a homomorphism. Let $a\in
G_{+}$, and let $s(a)u_{1}*\ldots*u_{k}\in\mathsf{Z}^{*}(a)$ with
$u_{1},\ldots,u_{k}\in\mathcal{A}(G_{+})$. Since the sequence of
$\Phi(u_{1}),\ldots,\Phi(u_{k})$ is, up to order, uniquely determined by $a$
(Section 4), it follows that we can define
$\eta(a)=\Phi(u_{1})\cdot\ldots\cdot\Phi(u_{k})$, and this is a homomorphism
$G_{+}\to\mathbb{G}_{+}$ with $\eta(u)=\Phi(u)$ for all
$u\in\mathcal{A}(G_{+})$. $G$ is the category of (left and right) fractions of
$G_{+}$, and hence $\eta$ extends to a unique groupoid homomorphism
$\eta\colon G\to\mathbb{G}$.

To verify that $\eta$ is surjective, let first $\mathcal{P}\in\mathbb{G}$ be a
prime element of $\mathbb{G}_{+}$, and let $e\in G_{0}$. Let $u\in
G_{+}(e,\cdot)$ be a maximal integral element with $\mathcal{P}_{u}\leq u$.
Then $\Phi(u)=\mathcal{P}$, and therefore $\eta(u)=\mathcal{P}$. The claim
follows since $\mathbb{G}$ is the free abelian group with basis
$\mathcal{A}(\mathbb{G}_{+})$. ∎

In general $\eta\neq\Phi$, since $\Phi$ need not be a homomorphism, but from
Lemma 4.2 it follows that for integral $a$ the prime factorizations of
$\Phi(a)$ and $\eta(a)$ have the same support and
$\mathsf{v}_{\mathcal{P}}(\eta(a))\geq\mathsf{v}_{\mathcal{P}}(\Phi(a))$ for
all primes $\mathcal{P}$ of $\mathbb{G}_{+}$.

######  Theorem .

Let $G$ be an arithmetical groupoid, $\eta\colon G\to\mathbb{G}$ the abstract
norm, $H$ a right-saturated subcategory of $G_{+}$, and
$C=\mathbb{G}/\mathbf{q}(\eta(H))$. For $\mathcal{G}\in\mathbb{G}$ set
$[\mathcal{G}]=\mathcal{G}\mathbf{q}(\eta(H))\in C$, and
$C_{M}=\\{\,[\eta(u)]\in C\mid\text{$u\in\mathcal{A}(G_{+})$}\,\\}$. Assume
that

  1. 1.

for $a\in G$ with $s(a)\in H_{0}$, $a\in HH^{-1}$ if and only if
$\eta(a)\in\mathbf{q}(\eta(H))$.

  2. 2.

for every $e\in G_{0}$ and $g\in C_{M}$, there exists an element
$u\in\mathcal{A}(G_{+})$ such that $s(u)=e$ and $[\eta(u)]=g$.

Then there exists a transfer homomorphism $\theta\colon
H\to\mathcal{B}(C_{M})$.

###### Proof.

Let $\theta\colon H\to\mathcal{B}(C_{M})$ be defined as follows: For $a\in H$
and $s(a)u_{1}*\ldots*u_{k}\in\mathsf{Z}^{*}_{G_{+}}(a)$ with
$u_{1},\ldots,u_{k}\in\mathcal{A}(G_{+})$, set
$\theta(a)=[\eta(u_{1})]\cdot\ldots\cdot[\eta(u_{k})]\in\mathcal{B}(C_{M})$
(in particular, identities are mapped to the empty sequence). We have to show
that this definition depends only on $a$, and not on the particular rigid
factorization into maximal integral elements chosen. Let
$s(a)v_{1}*\ldots*v_{k}\in\mathsf{Z}^{*}_{G_{+}}(a)$ be another such rigid
factorization. Then there exists a permutation $\tau\in\mathfrak{S}_{k}$ with
$\eta(u_{i})=\Phi(u_{i})=\Phi(v_{\tau(i)})=\eta(v_{\tau(i)})$ (due to
Proposition 4.1 and by definition of $\eta$). Therefore
$[\eta(u_{1})]\cdot\ldots\cdot[\eta(u_{k})]=[\eta(v_{1})]\cdot\ldots\cdot[\eta(v_{k})]$.

With this definition $\theta$ is a homomorphism: Obviously
$\theta(e)=\mathbf{1}_{\mathcal{B}(C_{M})}$ for all $e\in H_{0}$, and if $b\in
H$ with $t(a)=s(b)$ and $s(b)w_{1}*\ldots*w_{l}\in\mathsf{Z}^{*}(b)$ is a
rigid factorization of $b$ into maximal integral elements, then
$s(a)u_{1}*\ldots*u_{k}*w_{1}*\ldots*w_{l}$ is a rigid factorization of $ab$.
Thus

$\theta(ab)=[\eta(u_{1})]\cdot\ldots\cdot[\eta(u_{k})][\eta(w_{1})]\cdot\ldots\cdot[\eta(w_{l})]=\theta(a)\theta(b).$

We still have to check that $\theta$ has properties T1 and T2.

If $\theta(a)=\mathbf{1}_{\mathcal{B}(C_{M})}$, then $a$ possesses an empty
factorization into maximal elements, hence $a\in G_{0}\cap
H=H_{0}=H^{\times}$.

$\theta$ is surjective: $\theta(e)=\mathbf{1}_{\mathcal{B}(C_{M})}$ for any
$e\in H_{0}$. Let $k\in\mathbb{N}$ and $g_{1}\cdot\ldots\cdot
g_{k}\in\mathcal{B}(C_{M})$. By definition of $C_{M}$ and our second
assumption, there exists an element $u_{1}\in\mathcal{A}(G_{+})$ with
$[\eta(u_{1})]=g_{1}$ and $s(u_{1})\in H_{0}$. Again by our second assumption,
for all $i\in[2,k]$, there exist $u_{i}\in\mathcal{A}(G_{+})$ with
$s(u_{i})=t(u_{i-1})$ and $[\eta(u_{i})]=g_{i}$. With $a=u_{1}\cdot\ldots\cdot
u_{k}\in G$ we get
$[\eta(a)]=[\eta(u_{1})\cdot\ldots\cdot\eta(u_{k})]=[\eta(u_{1})]+\ldots+[\eta(u_{k})]=\mathbf{0}\in
C$ and $s(a)\in H_{0}$, and hence $\eta(a)\in\mathbf{q}(\eta(H))$. By our
first assumption, therefore $a\in HH^{-1}$, and since moreover $a$ is integral
in $G$ and $H$ is right-saturated in $G_{+}$, we get $a\in H$ and
$\theta(a)=g_{1}\cdot\ldots\cdot g_{k}$.

$\theta$ satisfies T2: Let $a\in H$, $\theta(a)=ST$ with
$S,T\in\mathcal{B}(C_{M})$ and $S=g_{1}\cdot\ldots\cdot g_{k}$,
$T=g_{k+1}\cdot\ldots\cdot g_{l}$, where $k\in\mathbb{N}_{0}$ and
$l\in\mathbb{N}_{\geq k}$. By Proposition 4.4, we can find a rigid
factorization $s(a)u_{1}*\ldots*u_{l}\in\mathsf{Z}^{*}(a)$ with
$u_{i}\in\mathcal{A}(G_{+})$ and $[\eta(u_{i})]=g_{i}$ for all $i\in[1,l]$.
Let $b=s(a)u_{1}\cdot\ldots\cdot u_{k}$ and $c=t(b)u_{k+1}\cdot\ldots\cdot
u_{l}$. Then $a=bc$. Since $s(b)\in H_{0}$ and
$[\eta(b)]=\sigma(S)=\mathbf{0}$, the first assumption implies $b\in
HH^{-1}\cap G_{+}=H$. Then $s(c)\in H_{0}$ and $c\in H$ follows similarly.
Finally, $\theta(b)=S$ and $\theta(c)=T$. ∎

The theorem remains true if $H$ is a left-saturated subcategory of $G_{+}$,
and in the first condition the set $HH^{-1}$ is replaced by $H^{-1}H$, and the
condition $s(a)\in H_{0}$ is replaced by $t(a)\in H_{0}$. Similarly, one can
replace the second condition by a symmetrical one, requiring $t(u)=e$ instead
of $s(u)=e$ (in the proof of the surjectivity of $\theta$ one then first
chooses $u_{k}$, followed by $u_{k-1}$ and so on).

######  Remark .

If $G$ is a group, then $G_{+}$ is the free abelian monoid with basis
$\mathcal{A}(G_{+})$ (Section 4). As a saturated submonoid of this free
abelian monoid, $H$ is therefore a reduced commutative Krull monoid [27,
Theorem 2.4.8]. Since $\eta=\operatorname{id}_{G}$ and
$HH^{-1}=\eta(H)\eta(H)^{-1}$ the first condition is trivially satisfied, and
because of $G_{0}=\\{\,1\,\\}$, the second condition is also trivially
satisfied.

Conversely, let $H$ be a normalizing Krull monoid. Then
$H_{\text{red}}=\\{\,aH^{\times}\mid a\in H\,\\}$ is a reduced commutative
Krull monoid, isomorphic to the monoid of its non-zero principal ideals ([25,
Corollary 4.14]). The latter is a submonoid of the divisorial fractional
ideals of $H$, which form the free abelian monoid of integral elements in the
free abelian group of divisorial ideals of $H$. In this way we recover the
well-known transfer homomorphism for Krull monoids as given for example in
[27, Proposition 3.4.8] for commutative Krull monoids, and in [25, Theorem
6.5] for normalizing Krull monoids.

We continue the discussion of normalizing Krull monoids in Remarks 5.2 and
5.1, where the divisorial two-sided ideal theory appears as a special case of
the divisorial one-sided ideal theory.

##  5\. Divisorial ideal theory in semigroups

In this section we develop a divisorial one-sided ideal theory in semigroups.
This follows again original ideas of Asano and Murata and generalizes the
corresponding theory in rings and the theory of divisorial two-sided ideals in
cancellative semigroups (see [2, 34, 3, 4, 6, 5, 17] for classical treatments,
and [42, 32, 33, 25, 36] for more modern treatments in this area). In
particular, the one-sided ideal theory of classical maximal orders over
Dedekind domains is a special case of the theory presented here.

The divisorial fractional one-sided ideals with left- and right-orders maximal
in a fixed equivalence class of orders form a groupoid as studied in the
previous section (this was in fact the motivation for Brandt to introduce the
notion of a groupoid, see [9, 10]). We connect the factorization theory of
elements of a maximal order $H$ with the one for the cancellative small
category of integral principal ideals with left- and right-order conjugate to
$H$, and apply results from the previous section to the factorization of
elements in $H^{\bullet}$. The main result in this section is Section 5. After
having derived it we discuss in detail the case of rings, and of classical
maximal orders (Section 5.1 and Section 5.2).

A semigroup $Q$ is called a _quotient semigroup_ if every cancellative element
is invertible in $Q$, in short $Q^{\bullet}=Q^{\times}$. A subsemigroup
$H\subset Q$ is a _right order_ in $Q$ if $H(H\cap Q^{\bullet})^{-1}=Q$, and
$H$ is a _left order_ in $Q$ if $(H\cap Q^{\bullet})^{-1}H=Q$. $H$ is an
_order_ in $Q$ if it is a left and a right order. We summarize the connection
between a subsemigroup $H\subset Q$ being an order, and $Q$ being a semigroup
of (left and right) fractions of $H$.

######  Lemma .

Let $Q$ be a quotient semigroup, and $H\subset Q$ a subsemigroup.

  1. 1.

If $H$ is an order in $Q$, then $H^{\bullet}=H\cap Q^{\bullet}$ and
$Q=\mathbf{q}(H)$.

  2. 2.

If $\mathbf{q}(H)=Q$, then $H^{\bullet}=H\cap Q^{\bullet}$ and $H$ is an order
in $Q$.

  3. 3.

If $H$ is an order in $Q$, $H^{\prime}$ is a subsemigroup of $Q$ and there
exist $a,b\in Q^{\bullet}$ with $aHb\subset H^{\prime}$, then $H^{\prime}$ is
an order in $Q$.

###### Proof.

  1. 1.

It suffices to show $H^{\bullet}=H\cap Q^{\bullet}$, and the inclusion $H\cap
Q^{\bullet}\subset H^{\bullet}$ is clear. Let $a\in H^{\bullet}$, and
$q,q^{\prime}\in Q$ with $aq=aq^{\prime}$. Since $H$ is a right order in $Q$,
there exist $c,d\in H$ and $s\in H\cap Q^{\bullet}$ with $q=cs^{-1}$ and
$q^{\prime}=ds^{-1}$, where we can choose a common denominator because $H\cap
Q^{\bullet}$ satisfies the right Ore condition in $H$. Then $ac=ad$, and,
because $a\in H^{\bullet}$, also $c=d$, showing $q=q^{\prime}$. Since $H$ is a
left order it follows in the same way that $a$ is right-cancellative in
$Q^{\bullet}$, and hence $a\in H\cap Q^{\bullet}$.

  2. 2.

It again suffices to show $H^{\bullet}=H\cap Q^{\bullet}$, and this follows in
the same way as in 1.

  3. 3.

It suffices to show that every $q\in Q$ has representations of the form
$q=cs^{-1}=t^{-1}d$ with $c,d\in H^{\prime}$ and $s,t\in H^{\prime}\cap
Q^{\bullet}$. Since $H$ is an order in $Q$, there exist
$c^{\prime},d^{\prime}\in H$ and $s^{\prime},t^{\prime}\in H\cap Q^{\bullet}$
with $a^{-1}qa=c^{\prime}s^{\prime-1}$ and $bqb^{-1}=t^{\prime-1}d^{\prime}$.
Setting $c=ac^{\prime}b$, $d=ad^{\prime}b$, $s=as^{\prime}b$ and
$t=at^{\prime}b$, the claim follows. ∎

_For the remainder of this section, let $Q$ be a quotient semigroup._

If $H$ and $H^{\prime}$ are orders in $Q$, then $H$ is (Asano-)equivalent to
$H^{\prime}$, written $H\sim H^{\prime}$, if there exist $a,b,c,d\in
Q^{\bullet}$ with $aHb\subset H^{\prime}$ and $cH^{\prime}d\subset H$. This is
an equivalence relation on the set of orders in $Q$. An order $H$ is _maximal_
if it is maximal within its equivalence class with respect to set inclusion.

A feature of the non-commutative theory is that often there is no unique
maximal order in a given equivalence class, and in fact in the most important
cases we study there are usually infinitely many, but only finitely many
conjugacy classes of them. In studying the divisorial one-sided ideal theory
of a maximal order $H$, one has to study the ideal theory of all maximal
orders in its equivalence class at the same time.

Let $H,H^{\prime}\subset Q$ be subsemigroups (not necessarily orders), and let
$X,Y\subset Q$. As in the previous sections $XY=\\{\,xy\mid x\in X,y\in
Y\,\\}$. $X$ is a _left $H$-module_ if $HX\subset X$, and a _right
$H^{\prime}$-module_ if $XH^{\prime}\subset X$. It is an _$(H,H^{\prime})$
-module_ if it is a left $H$\- and a right $H^{\prime}$-module, i.e.,
$HXH^{\prime}\subset X$. We define

${({Y}\\!\\!:_{r}\\!{X})}=\\{\,q\in Q\mid Xq\subset
Y\,\\}\quad\text{and}\quad{({Y}\\!\\!:_{l}\\!{X})}=\\{\,q\in Q\mid qX\subset
Y\,\\}.$

Every left $H$-module is an $(H,\\{1\\})$-module, and similarly every right
$H^{\prime}$-module is a $(\\{1\\},H^{\prime})$-module. We set
$\mathcal{O}_{l}(X)={({X}\\!\\!:_{l}\\!{X})}$ and
$\mathcal{O}_{r}(X)={({X}\\!\\!:_{r}\\!{X})}$.

######  Lemma .

Let $H,H^{\prime}$ be subsemigroups of $Q$ and let $X$ be an
$(H,H^{\prime})$-module.

  1. 1.

${({H}\\!\\!:_{r}\\!{X})}$ and ${({H^{\prime}}\\!\\!:_{l}\\!{X})}$ are
$(H^{\prime},H)$-modules.

  2. 2.

$X\subset{({H}\\!\\!:_{l}\\!{{({H}\\!\\!:_{r}\\!{X})}})}$ and
$X\subset{({H^{\prime}}\\!\\!:_{r}\\!{{({H^{\prime}}\\!\\!:_{l}\\!{X})}})}$.

  3. 3.

$\mathcal{O}_{l}(X)$ and $\mathcal{O}_{r}(X)$ are subsemigroups of $Q$.

  4. 4.

${({\mathcal{O}_{l}(X)}\\!\\!:_{r}\\!{X})}={({\mathcal{O}_{r}(X)}\\!\\!:_{l}\\!{X})}=\\{\,q\in
Q\mid XqX\subset X\,\\}$.

  5. 5.

$X\subset\mathcal{O}_{l}(X)$ if and only if $X\subset\mathcal{O}_{r}(X)$ if
and only if $X^{2}\subset X$.

###### Proof.

  1. 1.

$XH^{\prime}{({H}\\!\\!:_{r}\\!{X})}H\,\subset\,X{({H}\\!\\!:_{r}\\!{X})}H\,\subset
HH=H$ and similarly for ${({H^{\prime}}\\!\\!:_{l}\\!{X})}$.

  2. 2.

$X{({H}\\!\\!:_{r}\\!{X})}\subset H$ by definition of
${({H}\\!\\!:_{r}\\!{X})}$ and thus
$X\subset{({H}\\!\\!:_{l}\\!{{({H}\\!\\!:_{r}\\!{X})}})}$. The other identity
is proven analogously.

  3. 3.

Clearly $1\in\mathcal{O}_{l}(X)$ and
$\mathcal{O}_{l}(X)\mathcal{O}_{l}(X)X\subset\mathcal{O}_{l}(X)X\subset X$
implies $\mathcal{O}_{l}(X)\mathcal{O}_{l}(X)\subset\mathcal{O}_{l}(X)$. The
claim for $\mathcal{O}_{r}(X)$ is shown similarly.

  4. 4.

We show ${({\mathcal{O}_{l}(X)}\\!\\!:_{r}\\!{X})}=\\{\,q\in Q\mid XqX\subset
X\,\\}$. Let $q\in Q$. Then $XqX\subset X$ is equivalent to
$Xq\subset\mathcal{O}_{l}(X)$, which in turn is equivalent to
$q\in{({\mathcal{O}_{l}(X)}\\!\\!:_{r}\\!{X})}$.

  5. 5.

Immediate from the definitions of $\mathcal{O}_{l}(X)$ and
$\mathcal{O}_{r}(X)$. ∎

######  Definition 5.1.

For $X\subset Q$ as in Section 5, we define

$X^{-1}={({\mathcal{O}_{l}(X)}\\!\\!:_{r}\\!{X})}={({\mathcal{O}_{r}(X)}\\!\\!:_{l}\\!{X})}=\\{\,q\in
Q\mid XqX\subset X\,\\}\quad\text{and}\quad X_{v}=(X^{-1})^{-1}.$

######  Definition 5.2.

Let $H$ and $H^{\prime}$ be orders in $Q$.

  1. 1.

A _fractional left $H$-ideal_ is a left $H$-module $I$ such that $I\cap
Q^{\bullet}\neq\emptyset$ and ${({H}\\!\\!:_{r}\\!{I})}\cap
Q^{\bullet}\neq\emptyset$.

  2. 2.

A _fractional right $H^{\prime}$-ideal_ is a right $H^{\prime}$-module $I$
such that $I\cap Q^{\bullet}\neq\emptyset$ and
${({H^{\prime}}\\!\\!:_{l}\\!{I})}\cap Q^{\bullet}\neq\emptyset$.

  3. 3.

If $I$ is a fractional left $H$-ideal and a fractional right
$H^{\prime}$-ideal, then $I$ is a _fractional $(H,H^{\prime})$-ideal_.

  4. 4.

A _fractional $H$-ideal_ is a fractional $(H,H)$-ideal.

  5. 5.

$I$ is a _left $H$-ideal_ if it is a fractional left $H$-ideal and $I\subset
H$. A _right $H^{\prime}$-ideal_ is defined analogously.

  6. 6.

If $I$ is a left $H$-ideal and a right $H^{\prime}$-ideal, then $I$ is an
_$(H,H^{\prime})$ -ideal_.

  7. 7.

An _$H$ -ideal_ is an $(H,H)$-ideal.

  8. 8.

A fractional left $H$-ideal $I$ is _integral_ if $I\subset\mathcal{O}_{l}(I)$
(equivalently, $I\subset\mathcal{O}_{r}(I)$). The same definition is made for
fractional right $H^{\prime}$-ideals.

If $H$ is a maximal order, then the notions of a left $H$-ideal and that of an
integral fractional left $H$-ideal coincide (this will follow from Lemma 5.1
and 5.2). We will sometimes call a fractional left (right) $H$-ideal _one-
sided_ to emphasize that it need not be a fractional right (left) $H$-ideal,
or _two-sided_ to emphasize that it is indeed a fractional $H$-ideal.

We recall some properties of fractional left $H$-ideals and first observe the
following.

######  Lemma .

If $H$ is an order in $Q$ and $I$ is a fractional left $H$-ideal, then
$\mathcal{O}_{l}(I)$ and $\mathcal{O}_{r}(I)$ are orders. $I$ is a fractional
$(\mathcal{O}_{l}(I),\mathcal{O}_{r}(I))$-ideal.

###### Proof.

Let $a\in I\cap Q^{\bullet}$, and let $b\in{({H}\\!\\!:_{r}\\!{I})}\cap
Q^{\bullet}$. By definition, $H\subset\mathcal{O}_{l}(I)$ and since $H$ is an
order and $\mathcal{O}_{l}(I)$ a semigroup, it is also an order.
$\mathcal{O}_{l}(I)I\subset I$ and
$b\in{({\mathcal{O}_{l}(I)}\\!\\!:_{r}\\!{I})}$ imply that $I$ is a fractional
left $\mathcal{O}_{l}(I)$-ideal. Since
$b\in{({\mathcal{O}_{l}(I)}\\!\\!:_{r}\\!{I})}={({\mathcal{O}_{r}(I)}\\!\\!:_{l}\\!{I})}$,
it holds that $b\mathcal{O}_{l}(I)a\subset bI\subset\mathcal{O}_{r}(I)$, and
since $\mathcal{O}_{r}(I)$ is a semigroup and $\mathcal{O}_{l}(I)$ an order,
$\mathcal{O}_{r}(I)$ is also an order. Therefore $I$ is also a fractional
right $\mathcal{O}_{r}(I)$-ideal. ∎

The previous lemma implies that it is no restriction to require $I$ to be a
fractional $(H,H^{\prime})$-ideal over it, say, being a fractional left
$H$-ideal (set $H^{\prime}=\mathcal{O}_{r}(I)$).

######  Lemma .

Let $H$ and $H^{\prime}$ be orders in $Q$, and let $I$ be a fractional
$(H,H^{\prime})$-ideal.

  1. 1.

The orders $H$, $H^{\prime}$, $\mathcal{O}_{l}(I)$ and $\mathcal{O}_{r}(I)$
are all equivalent.

  2. 2.

If $H$ is maximal, then $\mathcal{O}_{l}(I)=H$, and similarly, if $H^{\prime}$
is maximal, then $\mathcal{O}_{r}(I)=H^{\prime}$.

  3. 3.

${({H}\\!\\!:_{r}\\!{I})}$ is a fractional right $H$-ideal, and
${({H^{\prime}}\\!\\!:_{l}\\!{I})}$ is a fractional left $H^{\prime}$-ideal.

  4. 4.

If $J$ is a fractional left $H$-ideal, then $I\cap J$ and $I\cup J$ are
fractional left $H$-ideals.

  5. 5.

If $(I_{m})_{m\in M}$ is a non-empty family of left $H$-ideals for some index
set $M$, then $\bigcup_{m\in M}I_{m}$ is a left $H$-ideal.

  6. 6.

If $H^{\prime\prime}$ is an order, and $K$ is a fractional
$(H^{\prime},H^{\prime\prime})$-ideal, then $IK$ is a fractional
$(H,H^{\prime\prime})$-ideal.

###### Proof.

  1. 1.

By definition of the right and left order, $H\subset\mathcal{O}_{l}(I)$ and
$H^{\prime}\subset\mathcal{O}_{r}(I)$. Let $a\in I\cap Q^{\bullet}$,
$b\in{({H}\\!\\!:_{r}\\!{I})}\cap Q^{\bullet}$ and
$c\in{({H^{\prime}}\\!\\!:_{l}\\!{I})}\cap Q^{\bullet}$. Then
$\mathcal{O}_{l}(I)ab\subset Ib\subset H$ and $ca\mathcal{O}_{r}(I)\subset
H^{\prime}$, so $H\sim\mathcal{O}_{l}(I)$ and
$H^{\prime}\sim\mathcal{O}_{r}(I)$. Finally, $cHa\subset cI\subset H^{\prime}$
and $aH^{\prime}b\subset H$ imply $H\sim H^{\prime}$.

  2. 2.

By 1., $\mathcal{O}_{l}(I)\sim H$ and by definition of the left order
$H\subset\mathcal{O}_{l}(I)$. Maximality of $H$ implies
$H=\mathcal{O}_{l}(I)$. Analogously, $H^{\prime}=\mathcal{O}_{r}(I)$ if
$H^{\prime}$ is maximal.

  3. 3.

${({H}\\!\\!:_{r}\\!{I})}$ is an $(H^{\prime},H)$-module, and
${({H}\\!\\!:_{r}\\!{I})}\cap Q^{\bullet}\neq\emptyset$ because $I$ is a
fractional left $H$-ideal. Since
$I\subset{({H}\\!\\!:_{l}\\!{{({H}\\!\\!:_{r}\\!{I})}})}$, also
${({H}\\!\\!:_{l}\\!{{({H}\\!\\!:_{r}\\!{I})}})}\cap
Q^{\bullet}\neq\emptyset$, and thus ${({H}\\!\\!:_{r}\\!{I})}$ is a fractional
right $H$-ideal. Similarly one shows that ${({H^{\prime}}\\!\\!:_{l}\\!{I})}$
is a fractional left $H^{\prime}$-ideal.

  4. 4.

Clearly $H(I\cap J)\subset I\cap J$ and if $c\in{({H}\\!\\!:_{r}\\!{I})}\cap
Q^{\bullet}$, then $(I\cap J)c\subset Ic\subset H$. It remains to show $I\cap
J\cap Q^{\bullet}\neq\emptyset$. Let $a\in I\cap Q^{\bullet}$ and $b\in J\cap
Q^{\bullet}$. Then $a=a^{\prime}s^{-1}$ and $b=b^{\prime}s^{-1}$ with
$a^{\prime},b^{\prime},s\in H\cap Q^{\bullet}$ (we can choose a common
denominator using the right Ore condition). By the left Ore condition there
exist $a^{\prime\prime}\in H\cap Q^{\bullet}$ and $b^{\prime\prime}\in H$ with
$a^{\prime\prime}a^{\prime}=b^{\prime\prime}b^{\prime}$. Then
$a^{\prime\prime}a^{\prime}s^{-1}=b^{\prime\prime}b^{\prime}s^{-1}\in I\cap
J\cap Q^{\bullet}$.

For the union, again $H(I\cup J)\subset I\cup J$, and there exists $a\in(I\cup
J)\cap Q^{\bullet}$. It remains to show ${({H}\\!\\!:_{r}\\!{(I\cup J)})}\cap
Q^{\bullet}\neq\emptyset$. But ${({H}\\!\\!:_{r}\\!{(I\cup
J)})}={({H}\\!\\!:_{r}\\!{I})}\cap{({H}\\!\\!:_{r}\\!{J})}$, and we are done
by applying our previous statement about the intersection to the fractional
right $H$-ideals ${({H}\\!\\!:_{r}\\!{I})}$ and ${({H}\\!\\!:_{r}\\!{J})}$.

  5. 5.

Set $I=\bigcup_{m\in M}I_{m}$. Then $HI\subset I$ and $I\cap
Q^{\bullet}\neq\emptyset$ are clear, and $I_{m}\subset H$ for all $m\in M$
implies $1\in{({H}\\!\\!:_{r}\\!{I})}$.

  6. 6.

Certainly $HIKH^{\prime\prime}\subset IK$. If $a\in I\cap Q^{\bullet}$ and
$b\in K\cap Q^{\bullet}$, then $ab\in IK\cap Q^{\bullet}$. Let
$c\in{({H}\\!\\!:_{r}\\!{I})}\cap Q^{\bullet}$ and
$d\in{({H^{\prime}}\\!\\!:_{r}\\!{K})}\cap Q^{\bullet}$. Then $IKdc\subset
IH^{\prime}c\subset Ic\subset H$, i.e., $dc\in{({H}\\!\\!:_{r}\\!{IK})}\cap
Q^{\bullet}$. If $c^{\prime}\in{({H^{\prime}}\\!\\!:_{l}\\!{I})}\cap
Q^{\bullet}$ and $d^{\prime}\in{({H^{\prime\prime}}\\!\\!:_{l}\\!{K})}\cap
Q^{\bullet}$, then $d^{\prime}c^{\prime}IK\subset d^{\prime}H^{\prime}K\subset
d^{\prime}K\subset H^{\prime\prime}$, i.e.,
$d^{\prime}c^{\prime}\in{({H^{\prime\prime}}\\!\\!:_{l}\\!{IK})}\cap
Q^{\bullet}$. ∎

######  Lemma .

Let $H$ and $H^{\prime}$ be orders in $Q$.

  1. 1.

If $H\sim H^{\prime}$, then there exist $a,b\in H^{\prime\bullet}$ with
$aH^{\prime}b\subset H$. If moreover $H\subset H^{\prime}$, then we can even
take $a,b\in H^{\bullet}$.

  2. 2.

If $H\sim H^{\prime}$ and $H\subset H^{\prime}$, then there exists an order
$H^{\prime\prime}$ and $a,b\in H^{\bullet}$ such that $H\subset
H^{\prime\prime}\subset H^{\prime}$ and $H^{\prime\prime}b\subset H$ and
$aH^{\prime}\subset H^{\prime\prime}$.

  3. 3.

The following statements are equivalent:

    1. (a)

$H\sim H^{\prime}$.

    2. (b)

There exists a fractional $(H,H^{\prime})$-ideal.

    3. (c)

There exists a fractional $(H^{\prime},H)$-ideal.

###### Proof.

  1. 1.

There exist $x,y\in Q^{\bullet}$ with $xH^{\prime}y\subset H$. Since
$H^{\prime}$ is an order in $Q$, $x=ac^{-1}$ and $y=d^{-1}b$ with $a,b\in
H^{\prime}$ and $c,d\in H^{\prime}\cap Q^{\bullet}$. If $H\subset H^{\prime}$
we even take $a,b,c,d\in H$. Then $aH^{\prime}b\subset
ac^{-1}H^{\prime}d^{-1}b\subset H$.

  2. 2.

Using 1. choose $a,b\in H^{\bullet}$ with $aH^{\prime}b\subset H$. Let
$H^{\prime\prime}=H\cup aH^{\prime}\cup HaH^{\prime}$. Then it is easily
checked that $H^{\prime\prime}H^{\prime\prime}\subset H^{\prime\prime}$ and
obviously $H\subset H^{\prime\prime}$, thus $H^{\prime\prime}$ is an order.
Moreover, $H^{\prime\prime}\subset H^{\prime}$, $H^{\prime\prime}b\subset H$
and $aH^{\prime}\subset H^{\prime\prime}$, as claimed.

  3. 3.

(a) $\Rightarrow$ (b): By 1. there exist $a,b\in H^{\bullet}$ with $aHb\subset
H^{\prime}$ and $c,d\in H^{\prime\bullet}$ with $cH^{\prime}d\subset H$.
Define $I=HbcH^{\prime}$. Clearly $I$ is an $(H,H^{\prime})$-module with
$bc\in I\cap Q^{\bullet}$. Since $aI=aHbcH^{\prime}\subset
H^{\prime}cH^{\prime}\subset H^{\prime}$ and $Id=HbcH^{\prime}d\subset
HbH\subset H$, $I$ is a fractional $(H,H^{\prime})$-ideal.

(b) $\Rightarrow$ (a): By Lemma 5.1.

(a) $\Leftrightarrow$ (c) follows by symmetry, swapping the roles of $H$ and
$H^{\prime}$. ∎

######  Lemma .

Let $H$ be an order in $Q$. The following statements are equivalent.

  1. (a)

$H$ is a maximal order.

  2. (b)

If $I$ is a fractional left $H$-ideal, then $\mathcal{O}_{l}(I)=H$ and if $J$
is a fractional right $H$-ideal, then $\mathcal{O}_{r}(I)=H$.

  3. (c)

If $I$ is a fractional $H$-ideal, then
$\mathcal{O}_{r}(I)=\mathcal{O}_{l}(I)=H$.

  4. (d)

If $I$ is an $H$-ideal, then $\mathcal{O}_{r}(I)=\mathcal{O}_{l}(I)=H$.

###### Proof.

(a) $\Rightarrow$ (b): By Lemma 5.2, $\mathcal{O}_{l}(I)=H$ and
$\mathcal{O}_{r}(J)=H$.

(b) $\Rightarrow$ (c) $\Rightarrow$ (d): Trivial.

(d) $\Rightarrow$ (a): Assume $H^{\prime}$ is an order equivalent to $H$ and
$H\subset H^{\prime}$. Applying Lemma 5.2 we find an equivalent order
$H^{\prime\prime}$ with $H\subset H^{\prime\prime}\subset H^{\prime}$ and
$a,b\in Q^{\bullet}$ with $aH^{\prime}\subset H^{\prime\prime}$ and
$H^{\prime\prime}b\subset H$. Let $I=\\{\,x\in Q\mid H^{\prime\prime}x\subset
H\,\\}$. Then $I$ is an $H$-ideal, and
$H^{\prime\prime}\subset\mathcal{O}_{l}(I)$, implying $H^{\prime\prime}=H$ by
(d). Set $J=\\{\,x\in Q\mid xH^{\prime}\subset H\,\\}$. Then $J$ is again an
$H$-ideal (we use $a\in J$ since $H^{\prime\prime}=H$), and
$H^{\prime}\subset\mathcal{O}_{r}(J)$ implies $H^{\prime}=H$. ∎

######  Lemma .

Let $H$ be a maximal order in $Q$, let $I$ and $J$ be fractional left
$H$-ideals, and let $K$ be a fractional left $\mathcal{O}_{r}(I)$-ideal.

  1. 1.

$\mathcal{O}_{l}(I)=H$, $I^{-1}$ is a fractional right $H$-ideal with
$\mathcal{O}_{r}(I^{-1})=H$ and $I_{v}$ is a fractional left $H$-ideal with
$\mathcal{O}_{l}(I_{v})=H$ and $I\subset I_{v}$. Moreover,
$\mathcal{O}_{l}(I)_{v}=H_{v}=H$.

  2. 2.

If $a\in Q^{\bullet}$, then $Ha$ is a fractional left $H$-ideal with
$\mathcal{O}_{r}(Ha)=a^{-1}Ha$, $(Ha)^{-1}=a^{-1}H$ and $(Ha)_{v}=Ha$. $Ha$ is
integral (equivalently, a left $H$-ideal), if and only if $a\in H^{\bullet}$.

  3. 3.

If $I\subset J$ then $J^{-1}\subset I^{-1}$ and $I_{v}\subset J_{v}$.

  4. 4.

$I\subset I_{v}=(I_{v})_{v}$, $I_{v}^{-1}=(I^{-1})_{v}=I^{-1}$ and
$\mathcal{O}_{l}(I_{v})=\mathcal{O}_{l}(I)=\mathcal{O}_{l}(I)_{v}=H$.

  5. 5.

$(I_{v}\cap J_{v})_{v}=I_{v}\cap J_{v}$.

  6. 6.

$(I\cup J)_{v}=(I_{v}\cup J)_{v}=(I\cup J_{v})_{v}=(I_{v}\cup J_{v})_{v}$.

  7. 7.

If $\mathcal{O}_{r}(I)$ and $\mathcal{O}_{r}(K)$ are also maximal, then
$(IK)_{v}=(I_{v}K)_{v}=(IK_{v})_{v}=(I_{v}K_{v})_{v}$.

###### Proof.

  1. 1.

By Lemma 5.2, $\mathcal{O}_{l}(I)=H$, and thus Lemma 5.3 implies that
$I^{-1}={({H}\\!\\!:_{r}\\!{I})}$ is a right $H$-ideal. By the symmetric
statement of what we just showed for fractional right $H$-ideals, therefore
$\mathcal{O}_{r}(I^{-1})=H$ and $I_{v}$ is a fractional left $H$-ideal.
Applying the first part of the statement to $I_{v}$ yields
$\mathcal{O}_{l}(I_{v})=H$. Now $I\subset I_{v}$ follows from Lemma 5.2, and
$H_{v}=H$ from $H^{-1}={({H}\\!\\!:_{r}\\!{H})}=H$.

  2. 2.

Since $a\in Ha\cap Q^{\bullet}$ and $a^{-1}\in{({H}\\!\\!:_{r}\\!{Ha})}$, $Ha$
is a fractional left $H$-ideal. Certainly
$a^{-1}Ha\subset\mathcal{O}_{r}(Ha)$. Conversely, if
$x\in\mathcal{O}_{r}(Ha)$, then $ax\in Ha$ and thus $x\in a^{-1}Ha$, so that
altogether $\mathcal{O}_{r}(Ha)=a^{-1}Ha$. Moreover, $(Ha)a^{-1}H\subset H$
and if $Hax\subset H$ for $x\in Q^{\bullet}$, then $ax\in H$ and hence $x\in
a^{-1}H$, implying $(Ha)^{-1}=a^{-1}H$. Finally, $(Ha)_{v}=Ha$ because
$(Ha)_{v}=((Ha)^{-1})^{-1}=Ha$. $Ha$ is a left $H$-ideal if and only if it is
integral due to maximality of $H$, and $Ha\subset H$ if and only if $a\in
H\cap Q^{\bullet}=H^{\bullet}$.

  3. 3.

If $x\in Q$ with $Jx\subset H$, then $Ix\subset Jx\subset H$, and hence
$J^{-1}\subset I^{-1}$. By 1., $J^{-1}$ and $I^{-1}$ are fractional right
$H$-ideals with $\mathcal{O}_{r}(I^{-1})=\mathcal{O}_{r}(J^{-1})=H$, and we
apply the symmetric statement for right fractional $H$-ideals to obtain
$I_{v}\subset J_{v}$.

  4. 4.

By 1., $I^{-1}$ is a fractional right $H$-ideal with
$\mathcal{O}_{r}(I^{-1})=H$, and $I_{v}$ is a fractional left $H$-ideal with
$\mathcal{O}_{l}(I_{v})=H$. Moreover, also by 1, $I\subset I_{v}$ and
$I^{-1}\subset(I^{-1})_{v}=[(I^{-1})^{-1}]^{-1}=I_{v}^{-1}$. It follows from
3., that $I_{v}\subset(I_{v})_{v}$ and $I^{-1}=I_{v}^{-1}$. Therefore
$(I_{v})_{v}=(I_{v}^{-1})^{-1}\subset(I^{-1})^{-1}=I_{v}$, whence
$I_{v}=(I_{v})_{v}$.

  5. 5.

$I_{v}\cap J_{v}\subset(I_{v}\cap
J_{v})_{v}\subset(I_{v})_{v}\cap(J_{v})_{v}=I_{v}\cap J_{v}$.

  6. 6.

$I\cup J\subset I_{v}\cup J\subset I_{v}\cup J_{v}\subset(I\cup J)_{v}$ and by
taking divisorial closures, and 4., the claim follows.

  7. 7.

We use $\mathcal{O}_{r}(I_{v})=\mathcal{O}_{r}(I)$ and
$\mathcal{O}_{l}(K)=\mathcal{O}_{l}(K_{v})$ (from 1.). We have $IK\subset
I_{v}K\subset I_{v}K_{v}$, and similarly $IK\subset IK_{v}\subset I_{v}K_{v}$
(by 1.). By 3., this implies
$(IK)_{v}\subset(IK_{v})_{v}\subset(I_{v}K_{v})_{v}$ and
$(IK)_{v}\subset(I_{v}K)_{v}\subset(I_{v}K_{v})_{v}$.

To prove the claim it suffices to show $(I_{v}K_{v})_{v}\subset(IK)_{v}$,
which will follow from 3. and 4. if we show $I_{v}K_{v}\subset(IK)_{v}$. Since
$IK(IK)^{-1}\subset\mathcal{O}_{l}(I)$, we have
$K(IK)^{-1}\subset{({\mathcal{O}_{l}(I)}\\!\\!:_{r}\\!{I})}=I^{-1}=I_{v}^{-1}$,
where the last equality is due to 4. Multiplying by $I_{v}$ from the right
gives $K(IK)^{-1}I_{v}\subset I_{v}^{-1}I_{v}$. By definition,
$I_{v}^{-1}I_{v}\subset\mathcal{O}_{r}(I_{v})$. Since
$\mathcal{O}_{r}(I_{v})=\mathcal{O}_{r}(I)=\mathcal{O}_{l}(K)$, therefore
$K(IK)^{-1}I_{v}\subset\mathcal{O}_{l}(K)$. Now $(IK)^{-1}I_{v}\subset
K^{-1}=K_{v}^{-1}$ (using 4. again). Multiplying by $K_{v}$ from the right and
using $\mathcal{O}_{r}(K_{v})=\mathcal{O}_{r}(K)$, we obtain
$(IK)^{-1}I_{v}K_{v}\subset\mathcal{O}_{r}(K)$. Since
$\mathcal{O}_{l}((IK)^{-1})=\mathcal{O}_{r}(K)$, this implies
$I_{v}K_{v}\subset((IK)^{-1})^{-1}=(IK)_{v}$. ∎

######  Definition 5.3.

Let $H$ be an order in $Q$. A fractional left or right $H$-ideal $I$ is called
_divisorial_ if $I=I_{v}$.

If $I$ is a fractional left $H$-ideal for a maximal order $H$, then it is not
necessarily true that $\mathcal{O}_{r}(I)$ is again a maximal order. The next
proposition shows that for divisorial fractional left or right $H$-ideals with
$H$ maximal, already both, $\mathcal{O}_{l}(I)$ and $\mathcal{O}_{r}(I)$, are
maximal. We can define an associative partial operation, the _$v$ -product_,
by $I\cdot_{v}J=(IJ)_{v}$ when $J$ is a divisorial fractional left
$\mathcal{O}_{r}(I)$-ideal. Moreover it shows that every divisorial fractional
left or right ideal is _$v$ -invertible_, i.e., invertible with respect to
this operation.

######  Proposition .

Let $H$ be a maximal order in $Q$. Let $I$ be a fractional left $H$-ideal.
Then

  1. 1.

$\mathcal{O}_{l}(I^{-1})$ is a maximal order. In particular,
$\mathcal{O}_{r}(I_{v})$ is a maximal order.

  2. 2.

$(II^{-1})_{v}=\mathcal{O}_{l}(I)$ and if $\mathcal{O}_{r}(I)$ is also
maximal, then $(I^{-1}I)_{v}=\mathcal{O}_{r}(I)$.

  3. 3.

If $I$ is a divisorial fractional left $H$-ideal, $J$ a divisorial fractional
left $\mathcal{O}_{r}(I)$-ideal and $K$ a divisorial fractional left
$\mathcal{O}_{r}(J)$-ideal, then

$(I\cdot_{v}J)\cdot_{v}K=I\cdot_{v}(J\cdot_{v}K).$

###### Proof.

  1. 1.

Because $H$ is maximal, $\mathcal{O}_{l}(I)=H$. Trivially,
$\mathcal{O}_{r}(I)\subset\mathcal{O}_{l}(I^{-1})$. Let
$H^{\prime}\supset\mathcal{O}_{l}(I^{-1})$ be an order with
$H^{\prime}\sim\mathcal{O}_{l}(I^{-1})$. Then $J=IH^{\prime}I^{-1}$ is an
$(H,H)$-module and if $a\in I\cap Q^{\bullet}$ and $b\in I^{-1}\cap
Q^{\bullet}$, then $ab\in J$ and moreover

$J^{2}=IH^{\prime}I^{-1}IH^{\prime}I^{-1}\subset
IH^{\prime}\mathcal{O}_{l}(I^{-1})H^{\prime}I^{-1}=IH^{\prime}I^{-1}=J,$

showing that $J$ is an integral left $\mathcal{O}_{l}(J)$-ideal.

We claim $H=\mathcal{O}_{l}(J)$. Since
$H=\mathcal{O}_{l}(I)\subset\mathcal{O}_{l}(J)$ and $H$ is maximal, it
suffices to show $\mathcal{O}_{l}(J)\sim H$. To this end we first show
$bJa\subset H^{\prime}$:

$bJa=bIH^{\prime}I^{-1}a\subset
I^{-1}IH^{\prime}I^{-1}I\subset\mathcal{O}_{r}(I)H^{\prime}\mathcal{O}_{r}(I)=H^{\prime}.$

Since $H^{\prime}\sim H$, there exist $c,d\in Q^{\bullet}$ with
$cH^{\prime}d\subset H$. Since $ab\in J\cap Q^{\bullet}$ therefore
$cb(\mathcal{O}_{l}(J)ab)ad\subset cbJad\subset H$, proving the claim.

Therefore, from the definition of $J$,
$H^{\prime}I^{-1}\subset{({J}\\!\\!:_{r}\\!{I})}\subset{({\mathcal{O}_{l}(J)}\\!\\!:_{r}\\!{I})}={({H}\\!\\!:_{r}\\!{I})}=I^{-1}$
and thus $H^{\prime}\subset\mathcal{O}_{l}(I^{-1})$, and, because we started
out with the converse inclusion, also $H^{\prime}=\mathcal{O}_{l}(I^{-1})$.

Now $\mathcal{O}_{r}(I_{v})=\mathcal{O}_{l}(I^{-1})$ implies the “in
particular” statement.

  2. 2.

We have to show $(II^{-1})_{v}=\mathcal{O}_{l}(I)$ and
$(I^{-1}I)_{v}=\mathcal{O}_{r}(I)$, and we check the first equality as the
second one then follows analogously. The inclusion
$II^{-1}\subset\mathcal{O}_{l}(I)$ implies
$(II^{-1})_{v}\subset\mathcal{O}_{l}(I)_{v}=\mathcal{O}_{l}(I)$. It remains to
prove $\mathcal{O}_{l}(I)\subset(II^{-1})_{v}$. Due to maximality of
$\mathcal{O}_{l}(I)$, it holds that
$\mathcal{O}_{l}(II^{-1})=\mathcal{O}_{l}(I)$, and therefore
$II^{-1}(II^{-1})^{-1}\subset\mathcal{O}_{l}(I)$. Thus
$I^{-1}(II^{-1})^{-1}\subset I^{-1}$, and
$(II^{-1})^{-1}\subset\mathcal{O}_{r}(I^{-1})=\mathcal{O}_{l}(I)$. By Lemma
5.3, therefore $\mathcal{O}_{l}(I)\subset(II^{-1})_{v}$.

  3. 3.

Using Lemma 5.7, which can be applied due to 1.,
$((IJ)_{v}K)_{v}=(IJK)_{v}=(I(JK)_{v})_{v}$. ∎

######  Corollary .

If $H$ is a maximal order, then every order $H^{\prime}$ with $H^{\prime}\sim
H$ is contained in a maximal order equivalent to $H$.

###### Proof.

By Lemma 5.3, there exists a fractional $(H,H^{\prime})$-ideal $I$. Then
$I_{v}$ is divisorial, $\mathcal{O}_{r}(I_{v})\sim\mathcal{O}_{l}(I_{v})=H$,
$H^{\prime}\subset\mathcal{O}_{r}(I_{v})$ and by Proposition 5.1
$\mathcal{O}_{r}(I_{v})$ is maximal. ∎

######  Corollary .

Let $\alpha$ denote an equivalence class of maximal orders of $Q$. Let

$\mathcal{F}_{v}(\alpha)=\\{\,I\mid\text{$I$ is a divisorial fractional left
(or right) $H$-ideal with $H\in\alpha$}\,\\}$

and

$\mathcal{I}_{v}(\alpha)=\\{\,I\mid\text{$I$ is a divisorial left (or right)
$H$-ideal with $H\in\alpha$}\,\\}.$

Then $(\mathcal{F}_{v}(\alpha),\cdot_{v},\subset)$ is a lattice-ordered
groupoid, with identity elements the maximal orders in $\alpha$. If $I,J$ are
in $\mathcal{F}_{v}(\alpha)$ with $\mathcal{O}_{l}(I)=\mathcal{O}_{l}(J)$ or
$\mathcal{O}_{r}(I)=\mathcal{O}_{r}(J)$, then $I\wedge J=I\cap J$ and $I\vee
J=(I\cup J)_{v}$. Moreover, $\mathcal{I}_{v}(\alpha)$ is the subcategory of
integral elements.

###### Proof.

For $I\in\mathcal{F}_{v}(\alpha)$ we have
$\mathcal{O}_{l}(I)\cdot_{v}I=I=I\cdot_{v}\mathcal{O}_{r}(I)$. If
$J\in\mathcal{F}_{v}(\alpha)$, then the $v$-product $I\cdot_{v}J$ is defined
whenever $\mathcal{O}_{r}(I)=\mathcal{O}_{l}(J)$, and then $I\cdot_{v}J$ is a
divisorial fractional $(\mathcal{O}_{l}(I),\mathcal{O}_{r}(J))$-ideal. The
$v$-product is associative when it is defined (Proposition 5.3). Therefore
$\mathcal{F}_{v}(\alpha)$ with $\cdot_{v}$ as composition is a category where
the set of identities is the set of maximal orders, $\alpha$, and for
$I\in\mathcal{F}_{v}(\alpha)$ we have $s(I)=\mathcal{O}_{l}(I)$ and
$t(I)=\mathcal{O}_{r}(I)$. This category is a groupoid due to Proposition 5.2.

On $\mathcal{F}_{v}(\alpha)$ set inclusion defines a partial order, and
obviously also the restrictions to
$\\{\,I\in\mathcal{F}_{v}(\alpha)\mid\mathcal{O}_{l}(I)=H\,\\}$ and
$\\{\,I\in\mathcal{F}_{v}(\alpha)\mid\mathcal{O}_{r}(I)=H\,\\}$ for
$H\in\alpha$, given by set inclusion in these subsets, are partial orders. Let
$I,J\in\mathcal{F}_{v}(\alpha)$ with $\mathcal{O}_{l}(I)=\mathcal{O}_{l}(J)$.
Then $I\cap J\in\mathcal{F}_{v}(\alpha)$ and $(I\cup
J)_{v}\in\mathcal{F}_{v}(\alpha)$ (by Lemmas 5 and 5), and clearly they are
the infimum respectively supremum of $\\{\,I,J\,\\}$ in
$\\{\,I\in\mathcal{F}_{v}(\alpha)\mid\mathcal{O}_{l}(I)=H\,\\}$, making this
set lattice-ordered. Symmetric statements hold if
$\mathcal{O}_{r}(I)=\mathcal{O}_{r}(J)$. If
$\mathcal{O}_{l}(I)=\mathcal{O}_{l}(J)$ and
$\mathcal{O}_{r}(I)=\mathcal{O}_{r}(J)$ both hold, then also
$\mathcal{O}_{l}(I\cap J)=\mathcal{O}_{l}((I\cup J)_{v})=\mathcal{O}_{l}(I)$
and $\mathcal{O}_{r}(I\cap J)=\mathcal{O}_{r}((I\cup
J)_{v})=\mathcal{O}_{r}(I)$ both hold. Therefore
$(\mathcal{F}_{v}(\alpha),\subset)$ is a lattice-ordered groupoid with the
claimed meet and join. It is immediate from the definitions that
$\mathcal{I}_{v}(\alpha)$ is the subcategory of integral elements of this
lattice-ordered groupoid. ∎

######  Definition & Lemma .

An order $H$ is _bounded_ if it satisfies the following equivalent conditions:

  1. (a)

Every fractional left $H$-ideal and every fractional right $H$-ideal contains
a fractional (two-sided) $H$-ideal.

  2. (b)

Every left $H$-ideal and every right $H$-ideal contains a (two-sided)
$H$-ideal.

  3. (c)

For all $a\in Q^{\bullet}$ there exist $b,c\in Q^{\bullet}$ such that
$bH\subset Ha$ and $Hc\subset aH$.

  4. (d)

For all $a\in Q^{\bullet}$ there exist $b,c\in Q^{\bullet}$ such that
$Ha\subset bH$ and $aH\subset Hc$.

  5. (e)

For all $a\in Q^{\bullet}$, $HaH$ is a fractional (two-sided) $H$-ideal.

  6. (f)

For all $a\in Q^{\bullet}$ there exists a fractional (two-sided) $H$-ideal $I$
with $a\in I$.

  7. (g)

If $M\subset Q$ and $a,b\in Q^{\bullet}$ with $aMb\subset H$, then there exist
$c,d\in Q^{\bullet}$ with $cM\subset H$ and $Md\subset H$.

###### Proof.

(a) $\Rightarrow$ (b): Trivial.

(b) $\Rightarrow$ (c): Let $a\in Q^{\bullet}$. Then $a=d^{-1}c$ with $c,d\in
H\cap Q^{\bullet}$, and $Ha=Hd^{-1}c\supset Hc$. By (b), $Hc$ contains an
$H$-ideal $J$. If $b\in J\cap Q^{\bullet}$, then $bH\subset J\subset Ha$. The
symmetric claim follows similarly.

(c) $\Rightarrow$ (d): By (c) applied to $a^{-1}$, there exist $b,c\in
Q^{\bullet}$ with $b^{-1}H\subset Ha^{-1}$ and $Hc^{-1}\subset a^{-1}H$. Then
$Ha\subset bH$ and $aH\subset Hc$.

(d) $\Rightarrow$ (e): $HaH$ is an $(H,H)$-module and contains the element
$a\in Q^{\bullet}$. Let $b,c\in Q^{\bullet}$ with $Ha\subset bH$ and
$aH\subset Hc$. Then $HaH\subset bH$ and $HaH\subset Hc$, hence
$b^{-1}\in{({H}\\!\\!:_{l}\\!{HaH})}$ and
$c^{-1}\in{({H}\\!\\!:_{r}\\!{HaH})}$.

(e) $\Rightarrow$ (f): Trivial.

(f) $\Rightarrow$ (g): $aM\subset Hb^{-1}\subset Hb^{-1}H$, and the latter
being contained in a fractional $H$-ideal, there exists $a^{\prime}\in
Q^{\bullet}\cap{({H}\\!\\!:_{l}\\!{Hb^{-1}H})}$ and thus $a^{\prime}aM\subset
H$. Similarly, $Mb\subset a^{-1}H\subset Ha^{-1}H$, and there exists a
$b^{\prime}\in Q^{\bullet}\cap{({H}\\!\\!:_{r}\\!{Ha^{-1}H})}$. Thus
$Mbb^{\prime}\subset H$.

(g) $\Rightarrow$ (a): Let $I$ be a fractional left $H$-ideal and $a\in I\cap
Q^{\bullet}$. Then $(Ha^{-1})a\subset H$ and so there exists a $b\in
Q^{\bullet}$ such that $b(Ha^{-1})\subset H$, and thus $bH\subset Ha$.
Therefore $HbH\subset Ha\subset I$ and $HbH$ is a fractional $H$-ideal
contained in $I$ (as $b\in HbH$, $a^{-1}\in{({H}\\!\\!:_{r}\\!{HbH})}$ and, by
(g) again, there exists a $c\in Q^{\bullet}$ with $c(Hb)\subset H$, whence
$c\in{({H}\\!\\!:_{l}\\!{HbH})}$). The case where $I$ is a fractional right
$H$-ideal is similar. ∎

######  Lemma .

  1. 1.

Let $H$ and $H^{\prime}$ be orders in $Q$. If $H$ is bounded and $H\sim
H^{\prime}$, then $H^{\prime}$ is also bounded.

  2. 2.

Let $H$ and $H^{\prime}$ be bounded equivalent maximal orders of $Q$. Then
there exists an $(H,H^{\prime})$-ideal $I$.

###### Proof.

  1. 1.

Because $H\sim H^{\prime}$ and $H$ is bounded, there exist $c,d\in
Q^{\bullet}$ with $cH\subset H^{\prime}\subset dH$ (using (c) and (d) of the
equivalent characterizations of boundedness). We verify condition (c) for
$H^{\prime}$. Let $a\in Q^{\bullet}$. Then $cHa\subset H^{\prime}a$, and there
exists an $x\in Q^{\bullet}$ with $xH\subset Ha$. Then $cxH\subset
H^{\prime}a$ and finally $cxd^{-1}H^{\prime}\subset cxH\subset H^{\prime}a$.
Similarly, one finds $z\in Q^{\bullet}$ with $H^{\prime}z\subset aH^{\prime}$.

  2. 2.

We show: If $H$ and $H^{\prime}$ are bounded equivalent maximal orders, then
$H^{\prime}H$ is a fractional $(H^{\prime},H)$-ideal. Then
$(H^{\prime}H)^{-1}$ is an $(H,H^{\prime})$-ideal (since
$\mathcal{O}_{r}(H^{\prime}H)=H$ by maximality of $H$ and $H\subset
H^{\prime}H$, Lemma 5.3 implies $(H^{\prime}H)^{-1}\subset H$; similarly, one
shows $(H^{\prime}H)^{-1}\subset H^{\prime}$).

Clearly $H^{\prime}H$ is an $(H^{\prime},H)$-module and $1\in H^{\prime}H$. We
need to show that there exist $a,b\in Q^{\bullet}$ with $H^{\prime}Ha\subset
H^{\prime}$ and $bH^{\prime}H\subset H$. Since $H$ and $H^{\prime}$ are
bounded and equivalent there exist $a,b\in Q^{\bullet}$ with $Ha\subset
H^{\prime}$ and $bH^{\prime}\subset H$, and the claim follows. ∎

######  Proposition .

Let $\alpha$ be an equivalence class of maximal orders in $Q$.
$(\mathcal{F}_{v}(\alpha),\cdot_{v},\subset)$ is an arithmetical groupoid if
and only if all $H\in\alpha$ (equivalently, one $H\in\alpha$) satisfy the
following three conditions:

  1. (A1)

$H$ satisfies the ACC on divisorial left $H$-ideals and the ACC on divisorial
right $H$-ideals,

  2. (A2)

$H$ is bounded,

  3. (A3)

the lattice of divisorial fractional left $H$-ideals is modular, and the
lattice of divisorial right $H$-ideals is modular.

###### Proof.

From Section 5 we already know that $\mathcal{F}_{v}(\alpha)$ is a lattice-
ordered groupoid. As in the discussion after Definition 4.2 and from Section
5, we see that if one representative $H\in\alpha$ satisfies A1–A3, then the
same is true for all $H^{\prime}\in\alpha$.

Assume first that A1–A3 hold. Then P1 holds due to Lemma 5.5, property P2 is
just A3 in the present setting. P3 follows easily: If $I\subset J$ are
divisorial fractional left $H$-ideals, and $K$ is a divisorial fractional
right $H$-ideal, then $KI\subset KJ$ and therefore
$K\cdot_{v}I=(KI)_{v}\subset(KJ)_{v}=K\cdot_{v}J$, and similarly for the
symmetric statement. P4 also holds: Let $(I_{m})_{m\in M}$ be a non-empty
family of divisorial left $H$-ideals. Then $(\bigcup_{m\in M}I_{m})_{v}\subset
H$ is also a divisorial left $H$-ideal, and if $(I_{m})_{m\in M}$ is a family
of $(H,H^{\prime})$-ideals with $H^{\prime}\in\alpha$, then the divisorial
closure of the union is again an $(H,H^{\prime})$-ideal. P6 is just A1. A2
implies P5: If $H,H^{\prime}\in\alpha$, then there exists an
$(H,H^{\prime})$-ideal $I$ by Lemma 5.2. Then $I_{v}$ is as required in P5.

Assume now that $(\mathcal{F}_{v}(\alpha),\cdot_{v},\subset)$ is an
arithmetical groupoid, and $H\in\alpha$. Then P2 implies A3, and P6 implies
A3. From P5 we can derive A2: Let $I$ be a fractional left $H$-ideal, and
$x\in I\cap Q^{\bullet}$. Then $Hx\in\mathcal{F}_{v}(\alpha)$. By P5, there
exists $J\in\mathcal{I}_{v}(\alpha)$ with
$\mathcal{O}_{l}(J)=\mathcal{O}_{r}(Hx)$ and $\mathcal{O}_{r}(J)=H$. Thus
$HxJ$ is a fractional $H$-ideal, and $HxJ\subset I$. We proceed similarly if
$I$ is a fractional right $H$-ideal. ∎

######  Remark .

  1. 1.

From the discussion after Definition 4.2, we also see that we can equivalently
formulate A3 as “the lattice of divisorial fractional left (right) $H$-ideals
is modular”, as the property for the other side then holds automatically.

  2. 2.

Let $H$ be a normalizing monoid. By definition of a monoid, $H$ satisfies the
left and right Ore condition, hence it is an order in its quotient group.
Lemma 5.2 shows that every fractional left or right $H$-ideal is in fact
already a two-sided $H$-ideal, and thus $H$ is bounded.

Assume that $H$ is a normalizing Krull monoid. Then $\alpha=\\{\,H\,\\}$, and
the lattice-ordered groupoid $\mathcal{F}_{v}(\alpha)$ is in fact a group. The
lattice of divisorial fractional $H$-ideals is then modular, even distributive
[51, Theorem 2.1.3(a)], and hence by the previous theorem an arithmetical
groupoid.

######  Definition 5.4.

We call a maximal order $H$ satisfying A1–A3 an _arithmetical maximal order_.
If $\alpha$ is its equivalence class of arithmetical maximal orders, then we
denote by $\mathcal{M}_{v}(\alpha)\subset\mathcal{I}_{v}(\alpha)$ the (quiver
of) maximal integral elements.

_Let from here on $H$ be an arithmetical maximal order in $Q$, and let
$\alpha$ be its equivalence class of arithmetical maximal orders._

By Lemma 5.2, every principal left ideal $Ha$ with $a\in H^{\bullet}$ is a
divisorial left $H$-ideal with inverse $a^{-1}H\in\mathcal{F}_{v}(\alpha)$.
Let

$\mathcal{H}(\alpha)=\\{\,H^{\prime}a\in\mathcal{I}_{v}(\alpha)\mid
H^{\prime}\in\alpha,\,a\in H^{\prime\bullet}\,\\}.$

The $v$-product coincides with the usual proper product on
$\mathcal{H}(\alpha)$. Thus
$(\mathcal{H}(\alpha),\cdot)\subset(\mathcal{I}_{v}(\alpha),\cdot_{v})$ is a
wide subcategory, with the product $IJ=I\cdot J=I\cdot_{v}J$ for
$I,J\in\mathcal{H}(\alpha)$ defined whenever
$\mathcal{O}_{r}(I)=\mathcal{O}_{l}(J)$, and then
$\mathcal{O}_{l}(IJ)=\mathcal{O}_{l}(I)$ and
$\mathcal{O}_{r}(IJ)=\mathcal{O}_{r}(J)$. The inclusion
$(\mathcal{H}(\alpha),\cdot)\subset(\mathcal{I}_{v}(\alpha),\cdot_{v})$ is
left- and right-saturated. By $\mathcal{H}_{H}(\alpha)$ (or shorter,
$\mathcal{H}_{H}$, since $H$ determines $\alpha$) we denote the subcategory of
$\mathcal{H}(\alpha)$ where the left and right orders of every element are not
only equivalent but in fact conjugate to $H$. Explicitly,

$\mathcal{H}_{H}=\mathcal{H}_{H}(\alpha)=\\{\,d(Ha)d^{-1}\in\mathcal{I}_{v}(\alpha)\mid
a\in H^{\bullet},\,d\in Q^{\bullet}\,\\}.$

If $H^{\prime}\in\alpha$, then $\mathcal{H}_{H}=\mathcal{H}_{H^{\prime}}$ if
and only if $H^{\prime}$ and $H$ are conjugate. Again the inclusion
$(\mathcal{H}_{H},\cdot)\subset(\mathcal{H}(\alpha),\cdot)$ is left- and
right-saturated, and thus so is the inclusion
$(\mathcal{H}_{H},\cdot)\subset(\mathcal{I}_{v}(\alpha),\cdot_{v})$.

The following simple lemma gives a correspondence between $H$ and
$\mathcal{H}_{H}$.

######  Lemma .

Let $d\in Q^{\bullet}$.

  1. 1.

If $a,a_{1},a_{2}\in H^{\bullet}$ with $a=a_{1}a_{2}$, then
$d^{-1}(Ha)d=d^{-1}(Ha_{2})d\cdot
d^{-1}a_{2}^{-1}(Ha_{1})a_{2}d\in\mathcal{H}_{H}$ with
$d^{-1}(Ha_{2})d,\,d^{-1}a_{2}^{-1}(Ha_{1})a_{2}d\in\mathcal{H}_{H}$.

  2. 2.

If $a\in H^{\bullet}$ and $d^{-1}(Ha)d=I_{2}\cdot I_{1}$ with
$I_{1},I_{2}\in\mathcal{H}_{H}$, then there exist $a_{1},a_{2}\in H^{\bullet}$
with $I_{2}=d^{-1}(Ha_{2})d$, $I_{1}=d^{-1}a_{2}^{-1}(Ha_{1})a_{2}d$ and
$a=a_{1}a_{2}$.

  3. 3.

If $a_{1},a_{2},b_{1},b_{2}\in H^{\bullet}$ with $Ha_{2}=Hb_{2}$ and
$a_{2}^{-1}(Ha_{1})a_{2}=b_{2}^{-1}(Hb_{1})b_{2}$, then there exist
$\varepsilon_{1},\varepsilon_{2}\in H^{\times}$ with
$b_{1}=\varepsilon_{1}a_{1}\varepsilon_{2}^{-1}$ and
$b_{2}=\varepsilon_{2}a_{2}$.

In particular, for $a\in H^{\bullet}$ we have $a\in\mathcal{A}(H^{\bullet})$
if and only if $d^{-1}(Ha)d\in\mathcal{A}(\mathcal{H}_{H})$.

###### Proof.

  1. 1.

The multiplication is defined because
$\mathcal{O}_{r}(d^{-1}(Ha_{2})d)=d^{-1}a_{2}^{-1}Ha_{2}d=\mathcal{O}_{l}(d^{-1}a_{2}^{-1}(Ha_{1})a_{2}d)$.
The remaining statements are then clear.

  2. 2.

Since $\mathcal{O}_{l}(I_{2})=d^{-1}Hd$ we have $I_{2}=d^{-1}Hda_{2}^{\prime}$
with $a_{2}^{\prime}\in(d^{-1}Hd)^{\bullet}$, and hence, with
$a_{2}=da_{2}^{\prime}d^{-1}\in H^{\bullet}$, $I_{2}=d^{-1}(Ha_{2})d$. Then
$\mathcal{O}_{l}(I_{1})=\mathcal{O}_{r}(I_{2})=d^{-1}a_{2}^{-1}Ha_{2}d$, and
therefore similarly $I_{1}=d^{-1}a_{2}^{-1}(Ha_{1}^{\prime})a_{2}d$ with
$a_{1}^{\prime}\in H^{\bullet}$. Hence
$d^{-1}(Ha)d=d^{-1}(Ha_{1}^{\prime}a_{2})d$, and thus $a=\varepsilon
a_{1}^{\prime}a_{2}$ with $\varepsilon\in H^{\times}$. Taking
$a_{1}=\varepsilon a_{1}^{\prime}$ the claim follows.

  3. 3.

Since $Ha_{2}=Hb_{2}$, there exists an $\varepsilon_{2}\in H^{\times}$ with
$b_{2}=\varepsilon_{2}a_{2}$. Then

$a_{2}^{-1}(Ha_{1})a_{2}=b_{2}^{-1}(Hb_{1})b_{2}=a_{2}^{-1}\varepsilon_{2}^{-1}(Hb_{1})\varepsilon_{2}a_{2}=a_{2}^{-1}(Hb_{1}\varepsilon_{2})a_{2},$

and thus there exists $\varepsilon_{1}\in H^{\times}$ with
$\varepsilon_{1}a_{1}=b_{1}\varepsilon_{2}$, i.e.,
$b_{1}=\varepsilon_{1}a_{1}\varepsilon_{2}^{-1}$. ∎

Observe that we may view a rigid factorization
$Ha_{2}*a_{2}^{-1}(Ha_{1})a_{2}\in\mathsf{Z}^{*}(\mathcal{H}_{H})$ as a
multiplicative way of writing the chain $H\supset Ha_{2}\supset Ha_{1}a_{2}$.

######  Proposition .

Let $a\in H^{\bullet}$. For every $d\in Q^{\bullet}$ there is a bijection
$\mathsf{Z}_{H}^{*}(a)\to\mathsf{Z}_{\mathcal{H}_{H}}^{*}(d^{-1}(Ha)d)$, given
by

$u_{1}*u_{2}*\ldots*u_{k}\,\mapsto\,d^{-1}(Hu_{k})d*(d^{-1}u_{k}^{-1}(Hu_{k-1})u_{k}d)*\ldots*(d^{-1}u_{k}^{-1}\cdot\ldots\cdot
u_{2}^{-1}(Hu_{1})u_{2}\cdot\ldots\cdot u_{k}d).$

If $\overline{\theta}\colon\mathcal{H}_{H}\to B$ is a transfer homomorphism to
a reduced cancellative small category $B$ and having the additional property
that $\overline{\theta}(d^{-1}(Ha)d)=\overline{\theta}(Ha)$ for all $a\in
H^{\bullet}$ and $d\in Q^{\bullet}$, then it induces a transfer homomorphism
$\theta\colon H^{\bullet}\to B^{\text{op}}$ given by
$\theta(a)=\overline{\theta}(Ha)$.

###### Proof.

The claimed bijection follows by iterating the previous lemma.

We need to verify that $\theta$ is a transfer homomorphism and first check
that $\theta$ is a homomorphism: For $a,b\in H^{\bullet}$

$\theta(ab)=\overline{\theta}(Hab)=\overline{\theta}(Hb\cdot
b^{-1}(Ha)b)=\overline{\theta}(Hb)\cdot\overline{\theta}(b^{-1}(Ha)b)=\overline{\theta}(Hb)\cdot\overline{\theta}(Ha)=\theta(a)\cdot^{\text{op}}\theta(b),$

and if $a\in H^{\times}$ then $Ha=H$, whence
$\theta(a)=\overline{\theta}(H)\in B_{0}$. We verify T1: Let $b\in B$. Then
there exist $d\in Q^{\bullet}$ and $a\in H^{\bullet}$ with
$\overline{\theta}(d^{-1}(Ha)d)=b$, hence $\theta(a)=b$. If $a\in H^{\bullet}$
with $\theta(a)\in B_{0}$, then $\overline{\theta}(Ha)\in B_{0}$, hence
$Ha\in(\mathcal{H}_{H})_{0}$, i.e., $Ha=H$ and $a\in H^{\times}$. It remains
to check T2: Let $a\in H^{\bullet}$ and $b_{1},b_{2}\in B$ with
$\theta(a)=b_{1}\cdot^{\text{op}}b_{2}$. Then
$\overline{\theta}(Ha)=b_{2}b_{1}$, hence there exist $a_{1},a_{2}\in
H^{\bullet}$ with $Ha=Ha_{2}\cdot a_{2}^{-1}(Ha_{1})a_{2}$ and
$\overline{\theta}(Ha_{2})=b_{2}$,
$\overline{\theta}(a_{2}^{-1}(Ha_{1})a_{2})=b_{1}$. This implies
$a=\varepsilon a_{1}a_{2}$ with $\varepsilon\in H^{\times}$, and
$\theta(\varepsilon a_{1})=b_{1}$, $\theta(a_{2})=b_{2}$. ∎

######  Remark .

The condition $\overline{\theta}(d^{-1}(Ha)d)=\overline{\theta}(Ha)$ implies
in particular $\lvert\overline{\theta}(\mathcal{H}_{H})_{0}\rvert=1$. Thus in
fact $B$ is necessarily a semigroup.

Let $\mathbb{G}$ be the universal vertex group of $\mathcal{F}_{v}(\alpha)$,
and let $\eta\colon\mathcal{F}_{v}(\alpha)\to\mathbb{G}$ be the abstract norm,
as defined in the previous section.

######  Lemma .

  1. 1.

If $I\in\mathcal{F}_{v}(\alpha)$ and $d\in Q^{\bullet}$, then
$\eta(d^{-1}Id)=\eta(I)$.

  2. 2.

$\mathbf{q}(\eta(\mathcal{H}_{H}))=\mathbf{q}(\\{\,\eta(Ha)\mid a\in
H^{\bullet}\,\\})=\\{\,\eta(Hq)\mid q\in Q^{\bullet}\,\\}$.

###### Proof.

  1. 1.

It suffices to verify the claim for maximal integral
$I\in\mathcal{I}_{v}(\alpha)$. If $P\in\mathcal{I}_{v}(\alpha)$ is the maximal
divisorial two-sided $\mathcal{O}_{l}(I)$-ideal contained in $I$, then
$d^{-1}Pd$ is the maximal divisorial two-sided ideal contained in $d^{-1}Id$,
and since
$d^{-1}Pd=(\mathcal{O}_{l}(d^{-1}Id)d^{-1}\mathcal{O}_{l}(I))\cdot_{v}P\cdot_{v}(\mathcal{O}_{l}(I)d\mathcal{O}_{l}(d^{-1}Id))$
we have $\eta(I)=(P)=(d^{-1}Pd)=\eta(d^{-1}Id)\in\mathbb{G}$.

  2. 2.

The first equality is immediate from 1. For the second equality, note that if
$q=ab^{-1}$ with $a,b\in H^{\bullet}$, then (using 1. multiple times and the
fact that $\eta$ is a homomorphism)

$\begin{split}\eta(Hq)&=\eta(Hab^{-1})=\eta(Hb^{-1}\cdot
b(Ha)b^{-1})=\eta(Hb^{-1})\eta(b(Ha)b^{-1})=\eta(bH)^{-1}\eta(Ha)\\\
&=\eta(b^{-1}(bH)b)^{-1}\eta(Ha)=\eta(Ha)\eta(Hb)^{-1}.\qed\end{split}$

Applying Section 4 to the present situation, we obtain a transfer homomorphism
$\mathcal{H}_{H}\to\mathcal{B}(C_{M})$ if we impose some additional crucial
conditions on $H$.

######  Theorem .

Let $Q$ be a quotient semigroup, $H$ an arithmetical maximal order in $Q$, and
$\alpha$ its equivalence class of arithmetical maximal orders.

  1. 1.

For all $a\in H^{\bullet}$, $\mathsf{L}_{H^{\bullet}}(a)$ is finite and non-
empty. If, for every maximal divisorial $H$-ideal $P$, the number of maximal
divisorial left $H$-ideals $I$ with $P\subset I$ is finite, then
$\mathsf{Z}^{*}_{H^{\bullet}}(a)$ is finite for all $a\in H^{\bullet}$.

  2. 2.

Let $P_{H^{\bullet}}=\\{\,\eta(Hq)\mid q\in
Q^{\bullet}\,\\}\subset\mathbb{G}$, $C=\mathbb{G}/P_{H^{\bullet}}$, and
$C_{M}=\\{\,[\eta(I)]\in C\mid I\in\mathcal{I}_{v}(\alpha)\text{ maximal
integral}\,\\}$. Assume:

    1. (i)

A divisorial fractional left $H$-ideal $I$ is principal if and only if
$\eta(I)\in P_{H^{\bullet}}$.

    2. (ii)

For all $H^{\prime}\in\alpha$ and all $g\in C_{M}$ there exists a maximal
divisorial left $H^{\prime}$-ideal with $[\eta(I)]=g$.

Then there exists a transfer homomorphism $\theta\colon
H^{\bullet}\to\mathcal{B}(C_{M})$.

###### Proof.

By Section 5, $(\mathcal{F}_{v}(\alpha),\cdot_{v},\subset)$ is an arithmetical
groupoid, and $\mathcal{I}_{v}(\alpha)$ is its subcategory of integral
elements. $(\mathcal{H}_{H},\cdot)$ is a left- and right-saturated subcategory
of $(\mathcal{I}_{v},\cdot_{v})$.

  1. 1.

This follows immediately from Section 4 and Section 5.

  2. 2.

Let $I$ be a fractional left $H^{\prime}$-ideal with $H^{\prime}=dHd^{-1}$.
Then $I$ is principal if and only if the fractional left $H$-ideal $d^{-1}Id$
is, and this is the case if and only if $\eta(I)=\eta(d^{-1}Id)\in
P_{H^{\bullet}}=\mathbf{q}(\eta(\mathcal{H}_{H}))$ (where the last equality is
due to the previous lemma). Therefore the first condition of Section 4 is
satisfied. Condition (ii) of the present theorem is equivalent to the second
condition of Section 4. Thus there exists a transfer homomorphism
$\overline{\theta}\colon\mathcal{H}_{H}\to\mathcal{B}(C_{M})$ as in Section 4.
By Section 5, there exists a transfer homomorphism $\theta\colon
H^{\bullet}\to\mathcal{B}(C_{M})$. ∎

######  Remark .

  1. 1.

We continue our discussion from Section 5. Let $H$ be a normalizing Krull
monoid. Then $\alpha=\\{\,H\,\\}$, $Ha=HaH=aH$ for all $a\in Q^{\bullet}$ and
associativity is a congruence relation [25, Lemma 4.4.1], thus
$H_{\text{red}}=\\{\,H^{\times}a\mid a\in H\,\\}$ with the induced operation
is also a monoid. Therefore $\mathcal{H}=\mathcal{H}_{H}=\\{\,HaH\mid a\in
H\,\\}\cong H_{\text{red}}$ and $G=\mathcal{F}_{v}(\alpha)$ is the free
abelian group on the maximal divisorial (two-sided) $H$-ideals, while
$\mathcal{I}_{v}(\alpha)$ is the free abelian monoid on the same basis.

In the previous theorem we therefore have $\mathbb{G}=G$,
$\eta=\operatorname{id}$, $P_{H^{\bullet}}=\\{\,Hq\mid q\in Q^{\bullet}\,\\}$,
and hence $C$ is the divisorial class group of $H$, and $C_{M}$ is the set of
divisorial ideal classes that contain a maximal divisorial $H$-ideal. The
second condition of the theorem is trivially true by virtue of $\lvert
G_{0}\rvert=1$ and the definition of $C_{M}$, and the first condition is
trivially true because $\eta=\operatorname{id}$. We thus get a transfer
homomorphism $H\to\mathcal{B}(C_{M})$ (induced from the transfer homomorphism
$H_{\text{red}}\cong\mathcal{H}_{H}\to\mathcal{B}(C_{M})$), which is the same
one as in [25, Theorem 6.5].

  2. 2.

If $H$ is a maximal order satisfying only A1 and A3, then
$\mathsf{L}_{H^{\bullet}}(a)$ is finite and non-empty for all $a\in
H^{\bullet}$. In Section 4 one may drop P5 and P6, and still obtain
Proposition 4.1 in the weaker form that, for each $a\in G_{+}$, either
$\mathsf{Z}^{*}_{G_{+}}(a)=\emptyset$ or $\lvert\mathsf{L}_{G_{+}}(a)\rvert=1$
(and of course without any statement about $\Phi$, which can only be defined
in the presence of P5). This is possible because P6 is only used to show
existence of a rigid factorization of $a$. A sufficient condition for
$\mathsf{Z}^{*}_{G_{+}}(a)\neq\emptyset$ is that $G_{+}(s(a),\cdot)$ and
$G_{+}(\cdot,t(a))$ satisfy the ACC. If $H$ satisfies A1, then
$G_{+}(e,\cdot)$ and $G_{+}(\cdot,e)$ with $e\in(\mathcal{H}_{H})_{0}$
(corresponding to conjugate orders of $H$) satisfy the ACC, and as in Section
4 one shows that $\mathsf{L}_{\mathcal{H}_{H}}(a)$ is finite and non-empty for
all $a\in\mathcal{H}_{H}$. Hence the same is true for $H^{\bullet}$.

###  5.1. Rings

Suppose that $Q=(Q,+,\cdot)$ is a quotient ring in the sense of [42, Chapter
3] (but recall that we in addition require it to be unital, as we do for all
rings). Then $(Q,\cdot)$ is a quotient semigroup. In the remainder of this
section we show that the ring-theoretic divisorial one-sided ideal theory for
maximal orders in $(Q,+,\cdot)$ coincides with the semigroup-theoretic one.
222In [42, Chapter 5] the terminology “reflexive” is used in place of
“divisorial”. If $R$ is a ring-theoretic order in $Q$, then a fractional left
$R$-ideal $I$ in the semigroup-theoretic sense is a fractional left $R$-ideal
in the ring-theoretic sense if and only if $I-I\subset I$ (see [42, §3.1.11]
for the usual definition).

_Let for the remainder of this subsection $Q=(Q,+,\cdot)$ be a quotient ring._

######  Lemma .

Let $H$ be an order in the multiplicative semigroup $(Q,\cdot)$ and $I$ a
fractional left $H$-ideal (in the semigroup-theoretic sense). Consider the
following statements:

  1. (a)

$I-I\subset I$.

  2. (b)

$\mathcal{O}_{l}(I)$ is a subring of $Q$.

  3. (c)

$\mathcal{O}_{r}(I)$ is a subring of $Q$.

Then (a) $\Rightarrow$ (b) and (a) $\Rightarrow$ (c). If $H$ is a maximal
order and $I$ is divisorial, then (a) $\Leftrightarrow$ (b) $\Leftrightarrow$
(c).

###### Proof.

Assume that (a) holds. We show (b): Let $a,b\in\mathcal{O}_{l}(I)$. Then
$aI\subset I$ and $bI\subset I$ and hence $(a-b)I\subset aI-bI\subset
I-I\subset I$, thus $a-b\in\mathcal{O}_{l}(I)$.

Assume now that $H$ is maximal, $I=I_{v}$ and (b) holds. We show (a). Let
$a,b\in I=I_{v}=(I^{-1})^{-1}$. Then $aI^{-1}\subset\mathcal{O}_{l}(I)$, and
$bI^{-1}\subset\mathcal{O}_{l}(I)$, whence $(a-b)I^{-1}\subset
aI^{-1}-bI^{-1}\subset\mathcal{O}_{l}(I)-\mathcal{O}_{l}(I)=\mathcal{O}_{l}(I)$
and thus $a-b\in(I^{-1})^{-1}=I$. ∎

######  Lemma .

A ring-theoretic order $R$ in $Q$ is maximal in the ring-theoretic sense if
and only if it is maximal in the semigroup-theoretic sense.

###### Proof.

We show that if $R$ is maximal in the ring-theoretic sense, then it is maximal
in the semigroup-theoretic sense, as the other direction is trivial. Let $I$
be a fractional left $R$-ideal in the semigroup-theoretic sense. Then
${}_{R}\langle I\rangle$ is a fractional left $R$-ideal in the ring-theoretic
sense, and using $R\subset\mathcal{O}_{l}(I)$, it follows that
$\mathcal{O}_{l}(I)\subset\mathcal{O}_{l}({}_{R}\langle I\rangle)$. Maximality
of $R$ in the ring-theoretic sense implies $R=\mathcal{O}_{l}({}_{R}\langle
I\rangle)$, hence also $R=\mathcal{O}_{l}(I)$. Similarly, if $J$ is a
fractional right $R$-ideal in the ring-theoretic sense then
$\mathcal{O}_{r}(J)=R$. Therefore Section 5 implies that $R$ is maximal in the
semigroup-theoretic sense. ∎

As before let $\alpha$ be an equivalence class of maximal orders of
$(Q,\cdot)$ in the semigroup-theoretic sense.

######  Lemma .

Let $H\in\alpha$ and assume that $H$ is a subring of $Q$ (i.e., an order in
$Q$ in the ring-theoretic sense).

  1. 1.

Every $H^{\prime}\in\alpha$ is a subring of $Q$ (and therefore an order in $Q$
in the ring-theoretic sense).

  2. 2.

If $I$ is a divisorial fractional left $H$-ideal and $J$ is a divisorial
fractional left $\mathcal{O}_{r}(I)$-ideal, then

$\\{\,ab\mid a\in I,b\in J\,\\}_{v}=\big{(}{}_{H}\langle\\{\,ab\mid a\in
I,b\in J\,\\}\rangle\big{)}_{v},$

i.e., the semigroup-theoretic $v$-product coincides with the ring-theoretic
one.

  3. 3.

If $I$ and $J$ are divisorial fractional left $H$-ideals, then $(I\cup
J)_{v}=(I+J)_{v}$.

###### Proof.

  1. 1.

By Lemma 5.3 there exists a fractional $(H,H^{\prime})$-ideal $I$. By
maximality of $H$ and $H^{\prime}$, also $\mathcal{O}_{l}(I_{v})=H$ and
$\mathcal{O}_{r}(I_{v})=H^{\prime}$ and the claim follows from Section 5.1
applied to $I_{v}$.

  2. 2.

Write $I\cdot_{S}J=\\{\,ab\mid a\in I,b\in J\,\\}$ for the semigroup-theoretic
ideal product and $I\cdot_{R}J={}_{H}\langle\\{\,ab\mid a\in I,b\in
J\,\\}\rangle$ for the ring-theoretic one. Then $I\cdot_{S}J\subset
I\cdot_{R}J$, and both of these sets are fractional left $H$-ideals (in the
semigroup-theoretic sense). Therefore
$(I\cdot_{S}J)_{v}\subset(I\cdot_{R}J)_{v}$. For the converse inclusion, it
suffices to show $I\cdot_{R}J\subset(I\cdot_{S}J)_{v}$, but this is true
because by Section 5.1 $(I\cdot_{S}J)_{v}$ is additively closed.

  3. 3.

Clearly $I\cup J\subset I+J$ and both sets are fractional left $H$-ideals (for
$I+J$ proceed as in the proof of Lemma 5.4; in particular observe
${({H}\\!\\!:_{r}\\!{I\cup J})}\subset{({H}\\!\\!:_{r}\\!{I+J})}$). As before
it therefore suffices to show $I+J\subset(I\cup J)_{v}$. This again holds due
to Section 5.1. ∎

Altogether, if $R$ is a maximal order in $Q$ in the ring-theoretic sense, then
it does not matter whether we form $\mathcal{F}_{v}(\alpha)$ by using the
ring-theoretic or the semigroup-theoretic notions. We use the same notion of
boundedness for ring-theoretic orders as in Section 5; for semiprime Goldie
rings this coincides with the notion in [42].

######  Theorem .

Let $R$ be a maximal order in a quotient ring $Q$, $\alpha$ its equivalence
class of maximal orders in the semigroup-theoretic sense, and $\beta$ its
equivalence class of maximal orders in the ring-theoretic sense. Then
$\alpha=\beta$ and $\mathcal{F}_{v}(\alpha)=\mathcal{F}_{v}(\beta)$, where the
latter is the ring-theoretic analogue of $\mathcal{F}_{v}(\alpha)$.

If $R$ is bounded, satisfies the ACC on divisorial left $R$-ideals and on
divisorial right $R$-ideals, and the lattice of divisorial fractional left
(right) $R$-modules is modular, then $(R,\cdot)$ is an arithmetical maximal
order in $(Q,\cdot)$ in the semigroup-theoretic sense. In particular, the
conclusions of Section 5 hold for $R$.

###### Proof.

By Lemma 5.1.1, $\alpha=\beta$, and by Section 5.1,
$\mathcal{F}_{v}(\alpha)=\mathcal{F}_{v}(\beta)$ as sets. By Section 5.1, the
$v$-product, meet and join coincide, and hence
$\mathcal{F}_{v}(\alpha)=\mathcal{F}_{v}(\beta)$ as lattice-ordered groupoids.
The remaining claims follow from this. ∎

In [46, §5(d)], Rehm gives examples for bounded maximal orders $E$, that are
prime and satisfy the ACC on divisorial two-sided $E$-ideals, but do not
satisfy the ACC on divisorial left $E$-ideals or the ACC on divisorial right
$E$-ideals. In fact (unless one takes the special case where $E$ itself is a