# Equivalent generating vectors of finitely generated modules over commutative
rings

Luc Guyot

###### Abstract.

Let $R$ be a commutative ring with identity and let $M$ be an $R$-module which
is generated by $\mu$ elements but not fewer. We denote by
$\textnormal{SL}_{n}(R)$ the group of the $n\times n$ matrices over $R$ with
determinant $1$. We denote by $\textnormal{E}_{n}(R)$ the subgroup of
$\textnormal{SL}_{n}(R)$ generated by the the matrices which differ from the
identity by a single off-diagonal coefficient. Given $n\geq\mu$ and
$G\in\left\\{\textnormal{SL}_{n}(R),\textnormal{E}_{n}(R)\right\\}$, we study
the action of $G$ by matrix right-multiplication on $\operatorname{V}_{n}(M)$,
the set of elements of $M^{n}$ whose components generate $M$. Assuming that
$M$ is finitely presented and that $R$ is an elementary divisor ring or an
almost local-global coherent Prüfer ring, we obtain a description of
$\operatorname{V}_{n}(M)/G$ which extends the author’s earlier result on
finitely generated modules over quasi-Euclidean rings [Guy17].

###### Key words and phrases:

elementary divisor ring; semihereditary ring; coherent ring; Prüfer domain;
local-global ring; generalized Euclidean ring; special Whitehead group;
Nielsen equivalence;

###### 2020 Mathematics Subject Classification:

Primary 13E15, Secondary 13F05, 13F05, 19B10

##  1\. Introduction

Rings are supposed unital and commutative. The unit group of a ring $R$ is
denoted by $R^{\times}$. Let $M$ be a finitely generated $R$-module. We denote
by $\mu(M)$ the minimal number of generators of $M$. For $n\geq\mu(M)$, we
denote by $\operatorname{V}_{n}(M)$ the set of _generating vectors of $M$ of
length $n$_, i.e., the set of $n$-tuples in $M^{n}$ whose components generate
$M$. We consider the action of $\textnormal{GL}_{n}(R)$ on
$\operatorname{V}_{n}(M)$ by matrix right-multiplication. Let
$\textnormal{SL}_{n}(R)$ be the subgroup of $\textnormal{GL}_{n}(R)$ of the
matrices with determinant $1$. Let $\textnormal{E}_{n}(R)$ be the subgroup of
$\textnormal{SL}_{n}(R)$ generated by the _elementary matrices_ , i.e., the
matrices which differ from the identity by a single off-diagonal element. Let
$G$ be a subgroup of $\textnormal{GL}_{n}(R)$. Two generating vectors
$\mathbf{m}$, $\mathbf{m}^{\prime}\in\operatorname{V}_{n}(M)$ are said to be
_$G$ -equivalent_, which we also denote by
$\mathbf{m}^{\prime}\sim_{G}\mathbf{m}$, if there exists $g\in G$ such that
$\mathbf{m}^{\prime}=\mathbf{m}g$. Our chief concern is the description of the
quotient $\operatorname{V}_{n}(M)/G$ with
$G\in\left\\{\textnormal{SL}_{n}(R),\textnormal{E}_{n}(R),\textnormal{GL}_{n}(R)\right\\}$
for $R$ in a class of rings comprising the Dedekind rings.

Let us highlight earlier results pertaining to this topic. When $M=R$, the
elements of $\operatorname{V}_{n}(M)$ are usually called the unimodular rows
of size $n$ and the orbit set
$\textnormal{W}_{n}(R)\Doteq\operatorname{V}_{n}(R)/\textnormal{E}_{n}(R)$ has
been extensively studied [Lam06, Section VIII.5]. Suslin and Vaserstein
[Bas75, Théorème 4] [VS76, Corollary 7.4] discovered that
$\textnormal{W}_{3}(R)$ can be identified with the elementary symplectic Witt
group of $R$ if $\textnormal{E}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(R)$ for every $n>3$. In [vdK83, Theorem 3.6], van der
Kallen used the former structure inductively to define a structure of Abelian
group on $\textnormal{W}_{n}(R)$ when $R$ has finite stable rank and $n$ is
large enough. The interested reader is referred to [DTZ18] for the latest
developments on the van der Kallen group structure of $\textnormal{W}_{n}(R)$
for an affine domain $R$. In the latter article, Das, Tikader and Zinna
describe moreover the Abelian group structure of
$\operatorname{V}_{n+1}(R)/\textnormal{SL}_{n+1}(R)$ induced by
$\textnormal{W}_{n+1}(R)$ for $R$ a smooth affine real algebra of dimension
$n\geq 2$ [DTZ18, Theorem 1.2]; see also [Fas11] for seminal results of this
flavor.

For $R=S[x_{1},\dots,x_{m}]$, the ring of polynomials in $m$ indeterminates
over a ring $S$, Murthy determined conditions under which
$\textnormal{GL}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(\sum_{k=1}^{r}Rx_{i})$ for $r\leq n\leq m$ [Mur03].

To introduce further known results, let us define the _determinant_ $\det(M)$
of a finitely generated $R$-module $M$ as the exterior product
$\bigwedge^{\mu}M$ where $\mu=\mu(M)$. The module $\det(M)$ is a cyclic
$R$-module whose annihilator is $\operatorname{Fitt}_{\mu-1}(M)$, the $\mu$-th
Fitting ideal of $M$ (see Section 2.1 for definition). The _determinant map_
$\det:M^{n}\rightarrow\bigwedge^{n}M$ is defined by
$\det(m_{1},\dots,m_{\mu})=m_{1}\wedge\cdots\wedge m_{n}$. It is immediate to
check that
$\det(\operatorname{V}_{\mu}(M))\subseteq\operatorname{V}_{1}(\det(M))$. For
$I$ an ideal of a ring $R$, we showed that
$\operatorname{V}_{2}(I)/\textnormal{SL}_{2}(R)$ is equipotent with a subset
of $\operatorname{V}_{1}(\det(I))$ [Guy20, Theorem A] if $I$ can be generated
by two elements which are not zero-divisors. For $R$ a quasi-Euclidean ring
and $M$ an arbitrary finitely generated $R$-module, we proved that the
determinant map induces a bijection from
$\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ onto
$\operatorname{V}_{1}(\det(M))$ and that $\textnormal{E}_{n}(R)$ acts
transitively on $\operatorname{V}_{n}(M)$ for every $n>\mu$ [Guy17, Theorem A
and Corollary B]. In this article we generalize [Guy17, Theorem A and
Corollary B] to two classes of rings: the elementary divisor rings and the
rings considered by Couchot in [Cou07, Theorems 1 and 2]. For the time being,
we denote by $\mathcal{D}$ the union of these two classes. Their definitions
are postponed to Section 2.3. The reader should only bear in mind that
$\mathcal{D}$ contains the Dedekind rings and some natural generalizations.

The rings in $\mathcal{D}$ share three features which make them particularly
amenable to the study of the generating vectors of their finitely presented
modules. For any such ring $R$, the following hold (see Proposition 2.15 and
Section 2.3’s theorems):

  * (1)

Every finitely generated ideal of $R$ can be generated by two elements.

  * (2)

The stable rank of $R$ is at most $2$.

  * (3)

Finitely presented $R$-modules have an _invariant decomposition_.

Our first result relates $\textnormal{SL}_{n}(R)$-orbits of generating vectors
to the image of the determinant map.

######  Theorem A.

Let $R$ be a ring in $\mathcal{D}$. Let $M$ be a finitely presented $R$-module
and set $\mu=\mu(M)$. Then the determinant map induces a bijection from
$\operatorname{V}_{\mu}(M)/\textnormal{SL}_{\mu}(R)$ onto
$\operatorname{V}_{1}(\det(M))$. In addition, the group
$\textnormal{SL}_{n}(R)$ acts transitively on $\operatorname{V}_{n}(M)$ for
every $n>\mu$.

In order to generalize Theorem A to direct products with factors in
$\mathcal{D}$, we introduce the following definition.

######  Definition B.

A finitely generated $R$-module $M$ is said to satisfy the property
$\Delta_{\textnormal{SL}}$ if the determinant map induces a bijection from
$\operatorname{V}_{n}(M)/\textnormal{SL}_{n}(R)$ to
$\operatorname{V}_{1}(\bigwedge^{n}M)$ for every $n\geq\mu(M)$. The ring $R$
is said to satisfy the property $\Delta_{\textnormal{SL}}$ if every finitely
presented $R$-module satisfies $\Delta_{\textnormal{SL}}$.

We define likewise the property $\Delta_{\textnormal{E}}$ for which an analog
of Proposition C below holds (see Propositions 5.3 and 5.4).

######  Proposition C (Propositions 5.3 and 5.4).

The two following hold.

  * $(i)$

Let $\\{R_{x}\\}_{x\in X}$ be a family of rings satisfying
$\Delta_{\textnormal{SL}}$ and let $R=\prod_{x}R_{x}$. Then $R$ satisfies
$\Delta_{\textnormal{SL}}$.

  * $(ii)$

Let $R$ be a ring which satisfies $\Delta_{\textnormal{SL}}$. Let
$\overline{R}$ be a quotient of $R$ such that the natural map
$\textnormal{SL}_{n}(R)\rightarrow\textnormal{SL}_{n}(\overline{R})$ is
surjective for every $n$. Then $\overline{R}$ satisfies
$\Delta_{\textnormal{SL}}$.

Proposition C implies that a direct product of rings, each factor of which is
in $\mathcal{D}$ or is the quotient of a ring in $\mathcal{D}$, has property
$\Delta_{\textnormal{SL}}$.

Let us now outline a class of rings $R$ for which the action of
$\textnormal{SL}_{n}(R)$ on $\operatorname{V}_{n}(M)$ is always transitive. We
say that a ring $R$ has property $T_{\textnormal{SL}}$ if for every finitely
presented $R$-module $M$, the group $\textnormal{SL}_{n}(R)$ acts transitively
on $\operatorname{V}_{n}(M)$ for every $n\geq\mu(M)$. We define likewise the
property $T_{G}$ for $G\in\\{\textnormal{E},\textnormal{GL}\\}$. If $R$
satisfies $T_{\textnormal{SL}}$, then the only unit of $R$ is $1$, its
identity element. Commutative rings with no units other than $1$ have been
studied by Cohn [Coh58, Theorem 3]. For every Boolean ring $B$, both $B$ and
$B[x]$ have no units other than $1$. Since $B$ and $B[x]$ are elementary
divisor rings [Sho74, Example 2], both enjoy property $T_{\textnormal{SL}}$ by
Theorem A.

Regarding the action of $\textnormal{GL}_{n}(R)$ on $\operatorname{V}_{n}(R)$,
we observe that it is essentially determined by the action of
$\textnormal{SL}_{n}(R)$. Indeed, decomposing
$\textnormal{GL}_{n}(R)\simeq\textnormal{SL}_{n}(R)\rtimes R^{\times}$, we
easily see that
$\operatorname{V}_{n}(M)/\textnormal{GL}_{n}(R)=(R^{\times})^{n}\backslash\operatorname{V}_{n}(M)/\textnormal{SL}_{n}(R)$
where $(R^{\times})^{n}$ acts on the left via componentwise multiplication. If
for instance an $R$-module $M$ has property $\Delta_{\textnormal{SL}}$, then
$\operatorname{V}_{n}(M)/\textnormal{GL}_{n}(R)$ identifies with

$R^{\times}\backslash\operatorname{V}_{1}(\det(M))\simeq(R/\operatorname{Fitt}_{\mu-1}(M))^{\times}/(R^{\times}+\operatorname{Fitt}_{\mu-1}(M)).$

If a ring $R$ has the property $T_{\textnormal{GL}}$, then obviously every
surjective ring homomorphism $R\twoheadrightarrow\overline{R}$ must induce a
surjective homomorphism $R^{\times}\twoheadrightarrow(\overline{R})^{\times}$
on unit groups. The latter condition is equivalent to say that $R$ has stable
rank $1$, see [EO67, Lemma 6.1] or [Guy20, Proposition 4.2]. Suppose $R$ has
property $\Delta_{\textnormal{SL}}$, e.g., $R$ lies in $\mathcal{D}$. Then it
is also clear that $R$ has property $T_{\textnormal{GL}}$ if $R$ has moreover
stable rank $1$. For instance, the ring in [Che17, Example 7] is a Jacobson
principal ideal domain of stable rank $1$ and thus enjoys property
$T_{\textnormal{GL}}$.

Besides the action of a subgroup $G\subseteq\textnormal{GL}_{n}(R)$ on
$\operatorname{V}_{n}(M)$, it is also natural to consider the action of
$\textnormal{Aut}_{R}(M)$, the group of the $R$-automorphisms of $M$, acting
diagonally on $M^{n}$ from the left. This action commutes with the action of
$G$ from the right, so that $\textnormal{Aut}_{R}(M)\times G$ acts also on
$\operatorname{V}_{n}(M)$ from the left. The following is an immediate
consequence of Theorem A and of the invariant decompositions of the modules at
play (see Theorems 2.9 and 2.12 below):

######  Corollary D.

Let $R$ be a ring in $\mathcal{D}$. Let $M$ be a finitely presented
$R$-module. Then $\textnormal{Aut}_{R}(M)\times\textnormal{SL}_{n}(R)$ acts
transitively on $\operatorname{V}_{n}(M)$ for every $n\geq\mu(M)$.

We finally consider the action of $\textnormal{E}_{n}(R)$ on generating
vectors. Evidently, Theorem A provides us with a description of
$\operatorname{V}_{n}(M)/\textnormal{E}_{n}(R)$ for $n\geq\mu(M)$ if
$\textnormal{SL}_{n}(R)=\textnormal{E}_{n}(R)$. If the latter identity holds,
the ring $R$ is termed a _$\textnormal{GE}_{n}$ -ring_. If $R$ is
$\textnormal{GE}_{n}$ ring for every $n\geq 2$, then $R$ is termed a _GE
-ring_. The class of GE-rings includes quasi-Euclidean rings, rings of stable
rank $1$ and Dedekind domains of arithmetical type with infinitely many units
[Vas72]. By taking direct products of GE-rings, we can obtain non-GE-rings for
which $\textnormal{E}_{n}(R)$ can replace $\textnormal{SL}_{n}(R)$ in Theorem
A. It is well-known that the class of $\textnormal{GE}_{n}$-ring is stable
under the formation of direct products with finitely many factors [Coh66,
Theorem 3.1], but not for infinitely many ones. A typical counter-example is
$\prod_{x}\mathbb{Z}$ where $x$ ranges in an infinite set (use, e.g., [Coh66,
Lemma 5.1]). Although the latter ring is not $\textnormal{GE}_{2}$, it is a
$\textnormal{GE}_{n}$-ring for every $n>2$. This follows from [CK83, Main
Theorem] which implies more generally that an arbitrary direct product of
rings of integers or their localizations, is a $\textnormal{GE}_{n}$-ring for
every $n>2$. For such products, combining Theorem A with Proposition C yields
a description of $\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ provided
that $\mu=\mu(M)>2$.

If $R$ is not a $\textnormal{GE}_{n}$-ring, describing
$\operatorname{V}_{n}(M)/\textnormal{E}_{n}(R)$ is already challenging when we
restrict to $M=R$ and $n=2$. In this case,
$\operatorname{V}_{2}(R)/\textnormal{E}_{2}(R)$ identifies with
$\textnormal{E}_{12}(R)\backslash\textnormal{SL}_{2}(R)/\textnormal{E}_{2}(R)$
where $\textnormal{E}_{12}(R)$ is the group of matrices of the form
$\begin{pmatrix}1&r\\\ 0&1\end{pmatrix}$ with $r\in R$ (see Lemma 4.4). If for
instance $R$ is the ring of integers of an imaginary quadratic field
$\mathbb{Q}(\sqrt{-d})$ that is not $\textnormal{GE}_{2}$, then it follows
from [FF88, Theorem 2.4] together with the Normal Form Theorem for Free
Products with Amalgamation [LS77, Theorem IV.2.6] that
$\operatorname{V}_{2}(R)/\textnormal{E}_{2}(R)$ is infinite; see [Nic11,
Theorem 1.5] and [She16] for an elementary proof of this fact. The quotient
$\operatorname{V}_{2}(R)/\textnormal{E}_{2}(R)$ clearly surjects on the
quotient of $\textnormal{SL}_{2}(R)$ by the subgroup of unipotent matrices.
For infinitely many $d>0$, this quotient surjects in turn on a non-Abelian
free group [GMV94, Proof of Theorem 1.2]. With our last result we depart from
such intricate situations, assuming that $\mu(M)>2$ and that $M$ has a non-
trivial torsion submodule. Under these assumptions, we can describe
$\operatorname{V}_{n}(M)/\textnormal{E}_{n}(R)$ by means of the determinant
map and the special Whitehead group $\operatorname{SK}_{1}(R)$ of $R$ (see
Section 5 for the definition of $\operatorname{SK}_{1}$). Let us denote by
$\tau(M)$ the _torsion submodule_ of $M$, that is,

$\tau(M)\Doteq\\{m\in M\,|\,rm=0\text{ for some }r\in R\text{ which is not a
zero divisor}\\}.$

A ring $R$ is a _Bézout ring_ if its finitely generated ideals are principal.

######  Theorem E.

Assume that $R$ is a Bézout ring whose proper quotients have stable rank $1$
or that $R$ is an almost local-global coherent Prüfer ring (see Section 2.3
for definitions). Let $M$ be a finitely presented $R$-module. Let $e$ be an
idempotent of $R$ such that $\mu(eM)=\mu(M)$ and set
$\mu^{\prime}=\mu(\tau(M))$, $\mu=\mu(M)$. Then the following hold.

  * $(i)$

If $\mu^{\prime}>1$ or $\mu=1$, then the determinant map induces a bijection
from from $\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ onto
$\operatorname{V}_{1}(\det(M))$.

  * $(ii)$

If $\mu^{\prime}=1$ and $\mu>2$, then
$\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ is equipotent with
$\operatorname{V}_{1}(\det(M))\times\operatorname{SK}_{1}(R)$.

  * $(iii)$

If either $\mu^{\prime}\neq 0$, or $\mu^{\prime}=0$ and $eM$ surjects onto a
non-principal invertible ideal of $Re$, then $\textnormal{E}_{n}(R)$ acts
transitively on $\operatorname{V}_{n}(M)$ for every $n>\mu$. This holds in
particular if $R$ is an almost local-global Prüfer domain and $M$ is torsion-
free but not free.

  * $(iv)$

The group $\textnormal{E}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(M)$ for every $n>\mu+1$.

A Bézout ring $R$ such that the proper quotients of $R/\mathcal{J}(R)$ have
stable rank $1$ is an elementary divisor ring [McG08, Proposition 3.5 and
Theorem 3.7]. To the best of our knowledge, all known examples of elementary
divisor rings satisfy this property (this is corroborated by [McG08, Remark
4.7]). Combining Theorem E with Proposition 2.7 below, we gain thus insight on
all such examples.

#### Final Remarks

The proof Theorem A hinges on the fact that ideals are two-generated for rings
in $\mathcal{D}$ (see specifically the results of Section 3). The class of
Bass rings is another well-studied class of rings whose ideals are two-
generated [Lam00, §4] and whose finitely generated torsion-free modules are
well-understood [LW85]. The Bass rings constitute therefore a class of choice
to pursue the study of equivalent generating vectors in the spirit of Theorems
A and E.

#### Layout

This paper is organized as follows. Section 2 introduces notation and
background results. It encloses some of the fundamental properties of the
determinant map, of the Bass stable rank and of the class $\mathcal{D}$.
Section 3 is dedicated to the study of the action of $\textnormal{SL}_{n}(R)$
on $\operatorname{V}_{n}(I)$ for $I$ a two-generated ideal of $R$. Section 4
addresses the finitely generated modules which admit an invariant
decomposition in the sense of Theorems 2.9 and 2.12. However, the idempotents
appearing in the decomposition of Theorem 2.12 are only handled in Section 5
where Proposition C is proven. Section 6 is dedicated to the proofs of the
Theorems A and E.

#### Acknowledgments

We are grateful to François Couchot and Jean Fasel for their encouragements.
We are thankful to Justin Chen, Henri Lombardi, Bogdan Nica and Wilberd van
der Kallen for helpful comments and references.

##  2\. Notation and background

We assume throughout that $R$ is a commutative unital ring and $M$ is a
finitely generated $R$-module. We denote by $\mathcal{J}(R)$ the _Jacobson
radical_ of $R$, that is, the intersection of the maximal ideals of $R$. If
$\mathfrak{p}$ is a prime ideal of $R$, we denote by $R_{\mathfrak{p}}$ the
localization of $R$ at $\mathfrak{p}$. Similarly, we denote by
$M_{\mathfrak{p}}$ its localization at $\mathfrak{p}$, which we identify with
$M\otimes_{R}R_{\mathfrak{p}}$. Given an element $m$ in $M$, we abuse notation
in denoting also by $m$ its image $m\otimes_{R}1\in M_{\mathfrak{p}}$. If $M$
and $N$ are two submodules of a given $R$-module, we denote by $(M:N)$ the
ideal consisting of all elements $r\in R$ such that $rN\subseteq M$. We define
the _annihilator $\operatorname{ann}_{R}(M)$ of $M$_ as $(0:M)$. If $Rm$ and
$Rm^{\prime}$ are two cyclic submodules of $M$, we simply write
$(m:m^{\prime})$ for $(Rm:Rm^{\prime})$.

#### Regular elements

An element $r$ of a ring $R$ is a _zero divisor_ if
$\operatorname{ann}_{R}(r)\Doteq(0:r)\neq\\{0\\}$. An element of $R$ is
_regular_ if it is not a zero divisor. An ideal of $R$ is _regular_ if it
contains a regular element. The _total ring of quotients $K(R)$_ of $R$ is the
localization $R[S^{-1}]$ where $S$ is the set of regular elements of $R$. The
above definition of a regular element should not be confused with the
definition of a _von Neumann regular element_ of $R$, that is an element for
which there exists $a\in R$ satisfying $r=ara$. A ring whose elements are von
Neumann regular is called a _von Neumann regular ring_.

#### Matrices over $R$ and $M$

Given two $1\times n$ row vectors $\mathbf{r}=(r_{1},\dots,r_{n})\in R^{n}$
and $\mathbf{m}=(m_{1},\dots,m_{n})\in M^{n}$, we denote by
$\mathbf{r}^{\top}$ and $\mathbf{m}^{\top}$ the $n\times 1$ column vectors
obtained by transposition and we define their product
$\mathbf{r}\mathbf{m}^{\top}=\mathbf{m}\mathbf{r}^{\top}\Doteq\sum_{i}r_{i}m_{i}$.
Based on these identities, the product of any two matrices over $R$ and $M$,
with compatible numbers of rows and columns, is uniquely defined. We denote
respectively by $0_{m\times n}$ and by ${1_{n}}$ the $m\times n$ zero matrix
and the $n\times n$ identity matrix over $R$.

###  2.1. Fitting ideals and the determinant module

The invariant on which our study of the action of $\textnormal{SL}_{\mu}(R)$
on $\operatorname{V}_{\mu}(M)$ hinges is the determinant map. This map can be
effectively computed thanks to the $\mu$th-Fitting ideal.

#### Fitting ideals

Let

$F\xrightarrow[]{\varphi}G\rightarrow M\rightarrow 0$

be an exact sequence where $F$ and $G$ are free $R$-modules and $G$ is
finitely generated. Let $I_{i}(\varphi)$ the ideal of $R$ generated by the
$i\times i$ minors of the matrix of $\varphi:F\rightarrow G$, agreeing that
$I_{0}(\varphi)=R$. Then the _$(i+1)$ -th Fitting ideal of $M$_ is
$\operatorname{Fitt}_{i}(M)\Doteq I_{\mu(G)-i}(\varphi)$. The Fitting ideals
are independent of $\varphi$ by Fitting’s lemma [Eis95, Corollary 20.4]. They
commute with _base change_ , i.e,
$\operatorname{Fitt}_{i}(M\otimes_{R}S)=\operatorname{Fitt}_{i}(M)\otimes_{R}S$
for any $R$-algebra $S$ [Eis95, Corollary 20.5], and hence with localization.

#### The determinant of a finitely generated module

Let $\mu=\mu(M)$. Recall that the determinant $\det(M)$ of $M$ is the exterior
product $\bigwedge^{\mu}M$. This is a cyclic $R$-module whose annihilator is
$\operatorname{Fitt}_{\mu-1}(M)$ [Eis95, Exercise 20.9.i]. Given
$\mathbf{m}=(m_{1},\dots,m_{\mu})\in\operatorname{V}_{\mu}(M)$, we denote by
$\phi_{\mathbf{m}}:\det(M)\rightarrow R/\operatorname{Fitt}_{\mu-1}(M)$ the
isomorphism induced by the map $m_{1}\wedge\cdots\wedge m_{\mu}\rightarrow
1+\operatorname{Fitt}_{\mu-1}(M)$. Given another generating pair
$\mathbf{m}^{\prime}=(m_{1}^{\prime},\dots,m_{\mu}^{\prime})\in M^{\mu}$, we
define

$\text{det}_{\mathbf{m}}(\mathbf{m}^{\prime})\Doteq\phi_{\mathbf{m}}(m_{1}^{\prime}\wedge\dots\wedge
m_{\mu}^{\prime}).$

It is easy to check that $\det_{\mathbf{m}}(\operatorname{V}_{\mu}(M))$ is a
subgroup of $(R/\operatorname{Fitt}_{\mu-1}(M))^{\times}$ which doesn’t depend
on $\mathbf{m}$. The following lemma is straightforward.

######  Lemma 2.1.

Let
$\mathbf{m},\mathbf{m}^{\prime},\mathbf{m}^{\prime\prime}\in\operatorname{V}_{\mu}(M)$.
Then the following assertions hold.

  * $(i)$

$\det_{\mathbf{m}}(\mathbf{m}^{\prime}A)=\det_{\mathbf{m}}(\mathbf{m}^{\prime})\det(A)$,
for every $\mu\times\mu$ matrix $A$ over $R$.

  * $(ii)$

$\det_{\mathbf{m}}(\mathbf{m}^{\prime\prime})=\det_{\mathbf{m}}(\mathbf{m}^{\prime})\det_{\mathbf{m}^{\prime}}(\mathbf{m}^{\prime\prime})$.

  * $(iii)$

$\det_{\mathbf{m}}(\mathbf{m}^{\prime})=1+\operatorname{Fitt}_{\mu-1}(M)$, if
and only if,
$\det_{\mathbf{m}}(\mathbf{m}^{\prime})=1+\operatorname{Fitt}_{\mu-1}(M_{\mathfrak{p}})$
for every maximal ideal $\mathfrak{p}$ of $R$, where $\mathbf{m}$ and
$\mathbf{m}^{\prime}$ denote (abusively) their natural images in
$\operatorname{V}_{\mu-1}(M_{\mathfrak{p}})$ in the left-hand side of the
identity.

$\square$

When there is no risk of ambiguity, we simply denote by $1$ the identity
element of $R/\operatorname{Fitt}_{\mu-1}(M)$. The ideal
$\operatorname{Fitt}_{\mu-1}(M)$ has a convenient description which makes the
computation of $\det_{\mathbf{m}}$ effective when a workable presentation of
$M$ is given. Let $\mathbf{m}\in\operatorname{V}_{\mu}(M)$. We say that an
element $r\in R$ is _involved in a relation of $\mathbf{m}$_ if there is
$(r_{1},\dots,r_{\mu})\in R^{\mu}$ such that $\sum_{i=1}^{\mu}r_{i}m_{i}=0$
and $r=r_{i}$ for some $i$.

######  Lemma 2.2.

Let $\mathbf{m}\in\operatorname{V}_{\mu}(M)$. Then
$\operatorname{Fitt}_{\mu-1}(M)$ is the set of the elements of $R$ involved in
a relation of $\mathbf{m}$.

$\square$

Let $\overline{\mathbf{m}}=\pi(\mathbf{m})$ be the image of $\mathbf{m}\in
V_{\mu}(M)$ by the natural map $\pi:M\rightarrow
M/\operatorname{Fitt}_{\mu-1}(M)M$ and let $\mathbf{e}$ be the canonical basis
of $(R/\operatorname{Fitt}_{\mu-1}(M))^{\mu}$. Then the map
$\overline{\mathbf{m}}\mapsto\mathbf{e}$ induces an isomorphism
$\varphi_{\mathbf{m}}$ from $M/\operatorname{Fitt}_{\mu-1}(M)M$ onto
$(R/\operatorname{Fitt}_{\mu-1}(M))^{\mu}$. The next lemma shows how the map
$\pi$ can be used to compute $\det_{\mathbf{m}}$.

######  Lemma 2.3.

Let $\mathbf{m},\mathbf{m}^{\prime}\in\operatorname{V}_{\mu}(M)$. Then
$\det_{\mathbf{m}}(\mathbf{m}^{\prime})$ is the determinant of the unique
$\mu\times\mu$ matrix $\overline{A}$ over $R/\operatorname{Fitt}_{\mu-1}(M)$
satisfying $\pi(\mathbf{m}^{\prime})=\pi(\mathbf{m})\overline{A}$, that is the
matrix of $\varphi_{\mathbf{m}}\circ\pi(\mathbf{m}^{\prime})$ with respect to
$\mathbf{e}$.

$\square$

###  2.2. Ranks

Following [Rie83, Section 9] and [MR87, Section 6.7.2], we define the Bass
stable rank of a finitely generated $R$-module $M$. An integer $n>0$ lies in
the _stable range of_ $M$ if for every
$\mathbf{m}=(m_{1},\dots,m_{n+1})\in\operatorname{V}_{n+1}(M)$, there is
$(r_{1},\dots,r_{n})\in R^{n}$ such that
$(m_{1}+r_{1}m_{n+1},\dots,m_{n}+r_{n}m_{n+1})$ belongs to
$\operatorname{V}_{n}(M)$. If $n$ lies in the stable range of $M$, then so
does $k$ for every $k>n$ [MR87, Lemma 11.3.3]. The _stable rank
$\textnormal{sr}(M)$ of $M$_ is the least integer in the stable range of $M$.

The Bass stable rank can be characterized in terms of a lifting property for
the generating vectors of quotient modules.

######  Proposition 2.4.

Let $M$ be a finitely generated $R$-module and let $n\geq\mu(M)$. Then the
following are equivalent:

  * $(i)$

$\textnormal{sr}(M)\leq n$.

  * $(ii)$

For every $R$-submodule $N\subseteq M$, the map
$\operatorname{V}_{n}(M)\rightarrow\operatorname{V}_{n}(M/N)$ sending
$(m_{1},\dots,m_{n})$ to $(m_{1}+N,\dots,m_{n}+N)$ is surjective. In other
words, every generating $n$-vector of $M/N$ lifts to a generating $n$-vector
of $M$.

###### Proof.

Straightforward, see the proof of [Guy20, Proposition 4.2]. ∎

Extending naturally the ranks introduced in [MR87, Section 11.3.3] to finitely
generated modules, we define the _linear rank $\textnormal{glr}_{R}(M)$_ and
the _elementary rank $\textnormal{er}_{R}(M)$_ of $M$ as the least integer
$n\geq\mu(M)$ such that $\textnormal{GL}_{k}(R)$, respectively
$\textnormal{E}_{k}(R)$, acts transitively on $\operatorname{V}_{k}(M)$ for
every $k>n$. Note that the analogous definition based on
$\textnormal{SL}_{k}(R)$ yields the rank $\textnormal{glr}_{R}(M)$. When there
is no risk of ambiguity, we simply write $\textnormal{glr}(M)$ and
$\textnormal{er}(M)$ instead of $\textnormal{glr}_{R}(M)$ and
$\textnormal{er}_{R}(M)$.

It is easy to check that $\textnormal{SL}_{2}(R)$ acts transitively on
$\operatorname{V}_{2}(R)$. Hence $\textnormal{glr}(R)=2$ is equivalent to
$\textnormal{glr}(R)=1$. A ring $R$ is said to be a _Hermite ring_ if
$\textnormal{glr}(R)=1$. This is [Lam06, Definition I.4.6]; Lam’s Hermite
rings are sometimes called _completable_ rings.

Following [Coh66], we call $R$ a _$\textnormal{GE}_{n}$ -ring_ if
$\textnormal{SL}_{n}(R)=\textnormal{E}_{n}(R)$ and $R$ is said to be a
_generalized Euclidean ring_ , or simply a _GE -ring_, if it is a
$\textnormal{GE}_{n}$-ring for all $n>1$.

######  Remark 2.5.

The ring $R$ is a $\textnormal{GE}_{2}$-ring if and only if
$\textnormal{E}_{2}(R)$ acts transitively on $\operatorname{V}_{2}(R)$. If $R$
is Hermite, then $R$ is a GE-ring if and only if $\textnormal{er}(R)=1$.

We denote by $\mathcal{J}(M)$, the _Jacobson radical of $M$_, that is, the
intersection of the maximal submodules of $M$ (see [Lam01, §24] for properties
and examples).

######  Proposition 2.6.

[MR87, Proposition 11.3.11 and Lemma 11.4.6] Let $M$ be a finitely generated
$R$-module and let $N$ be a submodule of $M$. Then the following hold.

  * $(i)$

$\textnormal{er}(R)\leq\textnormal{sr}(R)$.

  * $(ii)$

$\mu(M)\leq\textnormal{glr}(M)\leq\textnormal{er}(M)$ and
$\textnormal{sr}(M)\leq\mu(M)+\textnormal{sr}(R)-1$.

  * $(iii)$

$\textnormal{sr}(M/N)\leq\textnormal{sr}(M)$ and equality holds if
$N\subseteq\mathcal{J}(M)$.

######  Proposition 2.7.

Let $M$ be a finitely generated $R$-module. Then the following hold.

  * $(i)$

$\textnormal{er}(M)=\textnormal{er}_{R/I}(M)$ for every ideal
$I\subseteq\operatorname{ann}(M)$.

  * $(ii)$

$\textnormal{glr}(M)=\textnormal{glr}_{R/I}(M)$ for every ideal
$I\subseteq\operatorname{ann}(M)\cap\mathcal{J}(R)$.

  * $(iii)$

If $M\neq\\{0\\}$, then $\textnormal{er}(M)\leq\mu(M)+\textnormal{sr}(M)-1$.

  * $(iv)$

$\textnormal{glr}(M/N)=\textnormal{glr}(M)$ and
$\textnormal{er}(M/N)=\textnormal{er}(M)$ for every $R$-submodule
$N\subseteq\mathcal{J}(M)$.

  * $(v)$

If $M\neq\\{0\\}$ and $\textnormal{sr}(R)=1$ then
$\textnormal{er}(M)=\textnormal{sr}(M)=\mu(M)$.

###### Proof.

$(i)$. Observe that the natural map
$\textnormal{E}_{n}(R)\rightarrow\textnormal{E}_{n}(R/I)$ is surjective for
every ideal $I$ of $R$ and every $n>0$. The result follows immediately.
$(ii)$. Observe that the natural map
$\textnormal{GL}_{n}(R)\rightarrow\textnormal{GL}_{n}(R/I)$ is surjective for
every ideal $I\subseteq\mathcal{J}(R)$ and every $n>0$ [Wei13, Exercise
I.1.12.iv].

$(iii)$. We can assume, without loss of generality, that
$\textnormal{sr}(M)<\infty$. Let $n>\mu(M)+\textnormal{sr}(M)-1$ and let
$\mathbf{m}\in\operatorname{V}_{n}(M)$,
$\mathbf{m}^{\prime}=(m_{1}^{\prime},\dots,m_{\mu}^{\prime},0,\dots,0)\in\operatorname{V}_{n}(M)$
with $\mu\Doteq\mu(M)$. Let $s\Doteq\textnormal{sr}(M)$. It readily follows
from the definition of $s$ that
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(m_{1}^{\prime\prime},\dots,m_{s}^{\prime\prime},0,\dots,0)$
for some
$(m_{1}^{\prime\prime},\dots,m_{s}^{\prime\prime})\in\operatorname{V}_{s}(M)$.
The following equivalences are then immediate.

$\displaystyle\mathbf{m}$ $\displaystyle\sim_{\textnormal{E}_{n}(R)}$
$\displaystyle(m_{1}^{\prime\prime},\dots,m_{s}^{\prime\prime},m_{1}^{\prime},\dots,m_{\mu}^{\prime},0,\dots,0),$
$\displaystyle\sim_{\textnormal{E}_{n}(R)}$
$\displaystyle(0,\dots,0,m_{1}^{\prime},\dots,m_{\mu}^{\prime},0,\dots,0),$
$\displaystyle\sim_{\textnormal{E}_{n}(R)}$
$\displaystyle\mathbf{m}^{\prime}.$

This shows that $\textnormal{er}(M)<n$ and hence
$\textnormal{er}(M)\leq\mu+s-1$.

$(iv)$. The inequality $\textnormal{glr}(M)\geq\textnormal{glr}(M/N)$ is
trivial. To prove the reverse inequality, let $\mu\Doteq\mu(M)=\mu(M/N)$ and
$n>\textnormal{glr}(M/N)$. Let $\mathbf{m}\in\operatorname{V}_{n}(M)$ and
$\mathbf{m}^{\prime}=(m_{1}^{\prime},\dots,m_{\mu}^{\prime},0,\dots,0)\in\operatorname{V}_{n}(M)$.
By hypothesis, we can find $\mathbf{n}=(\nu_{1},\dots,\nu_{n})\in N^{n}$ such
that $\mathbf{m}\sim_{\textnormal{GL}_{n}(R)}\mathbf{m}^{\prime}+\mathbf{n}$.
As $(m_{1}^{\prime}+\nu_{1},\dots,m_{\mu}^{\prime}+\nu_{\mu})$ generates $M$,
we easily derive that
$\mathbf{m}^{\prime}+\mathbf{n}\sim_{\textnormal{GL}_{n}(R)}\mathbf{m}^{\prime}$.
Thus we established that
$\mathbf{m}\sim_{\textnormal{GL}_{n}(R)}\mathbf{m}^{\prime}$ for every
$\mathbf{m}\in\operatorname{V}_{n}(R)$, which proves that
$\textnormal{glr}(M)\leq\textnormal{glr}(M/N)$. The proof of the analog result
for the elementary rank is identical.

$(v)$. The equality $\textnormal{sr}(M)=\mu(M)$ is given by Proposition
2.6.$ii$. We shall prove that $\textnormal{er}(M)=\mu(M)$ by induction on
$\mu\Doteq\mu(M)>0$. If $\mu=1$, then $M\simeq R/\operatorname{ann}(M)$. As
$\textnormal{er}(M)=\textnormal{er}_{R/\operatorname{ann}(M)}(M)$ by $(i)$, we
deduce from Proposition 2.6 that
$\textnormal{er}(M)\leq\textnormal{sr}(R/\operatorname{ann}(M))\leq\textnormal{sr}(R)$.
Therefore $\textnormal{er}(M)=1$. Assume now that $\mu>1$ and let
$n>\mu,\mathbf{m}\in\operatorname{V}_{n}(M)$ and
$\mathbf{m}^{\prime}=(m_{1}^{\prime},\dots,m_{\mu}^{\prime},0,\dots,0)\in\operatorname{V}_{n}(M)$.
Since $n>\mu=\textnormal{sr}(M)$, we have
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(m_{1},\dots,m_{\mu},m_{1}^{\prime},0,\dots,0)$
for some $(m_{1},\dots,m_{\mu})\in\operatorname{V}_{\mu}(M)$. The induction
hypothesis applied to $M/Rm_{1}^{\prime}$ yields
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(m_{1}^{\prime}+\lambda_{1}m_{1}^{\prime},\dots,m_{\mu}^{\prime}+\lambda_{\mu}m_{1}^{\prime},m_{1}^{\prime},0,\dots,0)$
for some $\lambda_{1},\dots,\lambda_{\mu}\in R$. Therefore
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}\mathbf{m}^{\prime}$. ∎

A ring $R$ is _semi-local_ if it has only finitely many maximal ideals. A ring
$R$ is _K-Hermite_ if for every $n\geq 2$ and every $\mathbf{r}\in R^{n}$,
there is $\gamma\in\textnormal{GL}_{n}(R)$ such that
$\mathbf{r}\gamma=(0,\dots,0,d)$ for some $d\in R$. It follows immediately
from the definition that a K-Hermite ring is both a Bézout ring and a Hermite
ring.

######  Proposition 2.8.

The following hold:

  * $(i)$

$\textnormal{sr}(R)=1$ if $R$ is semi-local [Bas64, Corollary 10.5].

  * $(ii)$

$\textnormal{sr}(R)\leq 2$ if $R$ is a K-Hermite ring [MM82, Proposition
8.$i$].

  * $(iii)$

$\textnormal{sr}(R)\leq 2$ if every proper quotient of $R/\mathcal{J}(R)$ has
stable rank $1$ [McG08, Theorem 3.6].

  * $(iv)$

$\textnormal{sr}(R)\leq\dim_{\text{Krull}}(R)+1$ [Hei84, Corollary 2.3] or
[CLQ04, Theorem 2.4].

###  2.3. Elementary divisor rings and coherent Prüfer rings

In this section, we underscore the properties of the rings in $\mathcal{D}$
which are instrumental in the proofs of Theorems A and E. We detail in
particular the structure of finitely presented modules over these rings by
stating the corresponding invariant decomposition theorems. Complete
definitions are provided and some remarkable examples and results are briefly
outlined. Recall that the class $\mathcal{D}$ is the union of two classes: the
elementary divisor rings and the almost local-global coherent Prüfer rings ;
the latter are the almost local-global semihereditary rings of Couchot
[Cou07].

A ring $R$ is an _elementary divisor ring_ if every matrix over $R$ with
finitely many rows and columns admits a Smith Normal Form, that is, for every
$m\times n$ matrix $A$ over $R$, we can find
$B\in\textnormal{GL}_{m}(R),C\in\textnormal{GL}_{n}(R)$ such that $BAC$ is a
diagonal matrix whose diagonal coefficients $d_{1},\dots,d_{k}$ satisfy the
divisibility condition $d_{i+1}\,|\,d_{i}$ for every $1\leq i\leq k-1$.

The class of elementary divisor rings is easily seen to be stable under the
formation of direct products and quotients. An elementary divisor ring is
clearly a K-Hermite ring (see definition right before Proposition 2.8), hence
a Hermite ring. A finitely presented module over an elementary divisor ring
decomposes into a direct sum of finitely many cyclic modules whose annihilator
forms a descending chain.

######  Theorem 2.9.

[Kap49, Theorem 9.1] Let $R$ be an elementary divisor ring. Then $R$ satisfies
the Invariant Factor Theorem, that is, every finitely presented $R$-module $M$
decomposes as

$R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}\,(k\geq 0)$

where the ideals $\mathfrak{a}_{i}\subseteq R$ form a descending chain
$R\supsetneq\mathfrak{a}_{1}\supseteq\cdots\supseteq\mathfrak{a}_{k}$.

Conversely, a ring whose finitely presented modules have a decomposition into
a direct sum of cyclic modules is an elementary divisor rings [LLS74, Theorem
3.8].

######  Examples 2.10.

Principal ideal rings, i.e., rings whose ideals are principal, are elementary
divisor rings [Kap49, Theorem 12.3] . The ring of entire functions and the
ring of algebraic integers are two emblematic non-Noetherian elementary
divisor domains [DB72, Examples 1 and 2 of Section 4]. A Boolean ring and more
generally a von Neumann ring $R$ [GH56, Remark 12] as well as $R[X]$ [Sho74,
Example 2] are elementary divisor rings.

We turn now to the definition of an _almost local-global coherent Prüfer
ring_.

A ring is _reduced_ if it has no non-zero nilpotent element. A ring is
_arithmetic_ if its finitely generated ideals are locally principal. A
_Prüfer_ ring in the sense of Hermida and Sànchez-Giralda [HSG86, Definition
4] is a reduced arithmetical ring [LQ15, Proposition and definition VIII.4.4].
A ring is _coherent_ if its finitely generated ideals are finitely presented.
A Prüfer coherent ring is characterized by the property that its finitely
generated ideals are projective [LQ15, Theorem XII.4.1]. A Prüfer coherent
ring is often called a _semihereditary_ ring, a terminology we shall
discontinue (following Lombardi and Quitté [LQ15]) as we find it not
suggestive enough.

A ring $R$ is _local-global_ , or _LG_ for short, if every (possibly
multivariate) polynomial over $R$ whose values generate $R$, represents a
unit. The LG rings [MW81, EG82] are a natural generalization of the rings
satisfying van der Kallen’s primitive criterion [vdK77, Definition 1.10], see
[MW81, Proposition on page 456]. Let $f(X)=aX+b$ with
$(a,b)\in\operatorname{V}_{2}(R)$. Clearly, the ideal generated by the values
$f(R)$ of $f$ is $R$. If $R$ is LG, then there is $x\in R$ such that $ax+b$ is
a unit. Therefore LG rings have stable rank $1$. A ring $R$ is _almost local-
global_ in the sense of Couchot [Cou07], or _almost-LG_ for short, if $R/Rr$
is LG for every regular element $r\in R$.

######  Examples 2.11.

Semi-local rings and rings of Krull dimension $0$ (e.g. von Neumann regular
rings) are LG rings [FS01, Examples 4.1 and 4.2 of Chapter V], [LQ15, Fact
IX.6.2]. The Nagata ring $R(X)$ of a ring $R$ is an LG ring [LQ15, Fact
IX.6.7].

A _Dedekind domain_ is a Noetherian Prüfer domain. Dedekind domains are one-
dimensional [Mat89, Theorem 11.6], hence almost-LG. The classical structure
theorem of Steinitz for finitely generated modules over Dedekind rings [Mag02,
Theorem 7.48] generalizes to almost-LG coherent Prüfer rings in the following
way.

######  Theorem 2.12.

[Cou07, Theorem 2] Let $R$ be an almost-LG coherent Prüfer ring. Then $R$
satisfies the Invariant Factor Theorem, that is, every finitely presented
$R$-module $M$ decomposes as

$M\simeq\prod_{i=1}^{m}R/I_{i}\times\prod_{j=1}^{n}\left(J_{j,1}e_{j}\times\cdots\times
J_{j,k_{j}}e_{j}\right).$

where $R\supsetneq I_{1}\supseteq\cdots\supseteq I_{m}\neq\\{0\\}$ are
invertible ideals of $R$, $(e_{1},\dots,e_{n})$ is a sequence of orthogonal
idempotents, $(k_{1},...,k_{n})$ a strictly increasing sequence of positive
integers and each ideal $J_{j,k}e_{j}$ is an invertible ideal of $Re_{j}$. In
addition, the isomorphism class of $M$ is completely determined by the
following invariants:

  * •

the ideals $I_{1},\dots,I_{m}$,

  * •

the idempotents $e_{1},\dots,e_{n}$,

  * •

the integers $k_{1},\dots,k_{n}$ and

  * •

the isomorphism class of $(J_{1,1}\cdots
J_{1,k_{1}}e_{1})\times\cdots\times(J_{n,1}\cdots J_{n,k_{n}}e_{n})$.

######  Remark 2.13.

Theorem 2.12 implies in particular that $J_{j,1}e_{j}\times\cdots\times
J_{j,k_{j}}e_{j}\simeq(J_{1,1}\cdots
J_{1,k_{j}}e_{j})\times(Re_{j})^{k_{j}-1}$ for every $1\leq j\leq n$.

A ring $R$ is said to be _of finite character_ if $R/I$ is semi-local for
every non-zero ideal $I$ of $R$. Using Examples 2.11, it is immediate to check
that a Prüfer domain which is one-dimensional or of finite character is an
almost-LG coherent Prüfer ring but not necessarily Noetherian.

######  Examples 2.14.

The ring of algebraic integers is a non-Noetherian Bézout domain (see Examples
2.10) which is one-dimensional by the going-up and going-down theorems [Mat89,
Theorem 9.4]. Let $p$ be a prime number, let $\mathbb{Z}_{p}$ be the ring of
the $p$-adic integers and let $\mathbb{Q}_{p}$ be its fraction field. Then the
integral closure of $\mathbb{Z}_{p}$ in an algebraic closure of
$\mathbb{Q}_{p}$ is a one-dimensional valuation domain, with value group
$\mathbb{Q}$, hence not Noetherian [Mat89, Theorem 11.1].

We conclude this section with two properties of importance for the proofs of
our theorems.

######  Proposition 2.15.

Let $R$ be a ring in $\mathcal{D}$. Then the two following hold.

  * $(i)$

Every finitely generated ideal of $R$ can be generated by two elements.

  * $(ii)$

The stable rank of $R$ is at most $2$.

###### Proof.

Assertion $(i)$ is obvious for a Bézout ring, hence for any elementary divisor
ring. If $R$ is an almost-LG coherent Prüfer ring, then $(i)$ follows from
[Cou07, Theorem 2 and Lemma 10]. Assertion $(ii)$ for an elementary divisor
ring is given by Proposition 2.8.$ii$. For an almost-LG coherent Prüfer ring,
combine Proposition 3.8 below and [Cou07, Lemma 10]. ∎

##  3\. The quotient $\operatorname{V}_{n}(I)/\textnormal{SL}_{n}(R)$ for
$n\geq 2$

In this section, $I$ denotes a two-generated ideal of $R$. Our results assume
that $I$ enjoys at least one of the following two properties.

######  Definition 3.1.

Let $I$ be an ideal of $R$ and let $I^{-1}\Doteq\\{x\in K(R)\,|\,xI\subseteq
R\\}$ where $K(R)$ is the total ring of quotients of $R$.

  * •

The ideal $I$ is said to be _invertible_ if $II^{-1}=R$.

  * •

The ideal $I$ is said to be _$1\nicefrac{{1}}{{2}}$ -generated_ if $I/Rx$ is
generated by one element for every $x\in R\setminus\\{0\\}$.

It is well-known that an invertible ideal contains a regular element and that
it is projective of constant rank $1$ (see, e.g., [Eis95, Theorem 11.6], where
this claim is proved in 11.6.c and 11.6.d without actually assuming that $R$
is Noetherian). A $1\nicefrac{{1}}{{2}}$-generated ideal which contains a
regular element is invertible [LM88, Theorem 1]. In a one-dimensional domain
[Hei76, Theorem 3.1] or in a ring of finite character [GH70, Theorem 3],
invertible ideals are $1\nicefrac{{1}}{{2}}$-generated.

######  Lemma 3.2.

Let $I$ and $\mathfrak{a}$ be ideals of $R$ with $I$ invertible. If
$I/I\mathfrak{a}$ is cyclic then $I/I\mathfrak{a}\simeq R/\mathfrak{a}$.

###### Proof.

Since $I/I\mathfrak{a}$ is cyclic, it suffices to show that
$\operatorname{ann}(I/I\mathfrak{a})=\mathfrak{a}$. Clearly
$\operatorname{ann}(I/I\mathfrak{a})=(I\mathfrak{a}:I)$ and the inclusions
$(I\mathfrak{a}:I)\supseteq\mathfrak{a}$ and
$(I\mathfrak{a}:I)\subseteq(II^{-1}\mathfrak{a}:II^{-1})$ are immediate. To
conclude, we observe that $(II^{-1}\mathfrak{a}:II^{-1})=\mathfrak{a}$ holds
because $I$ is invertible. ∎

######  Lemma 3.3.

Let $I$ be a $1\nicefrac{{1}}{{2}}$-generated ideal of $R$. Assume that at
least one of the following holds.

  * $(i)$

Every proper quotient of $R$ has stable rank $1$.

  * $(ii)$

Every proper quotient of $R/\mathcal{J}(R)$ has stable rank $1$ and $I$ is
invertible.

Then $\textnormal{sr}(I)\leq 2$. In particular $\textnormal{sr}(R)\leq 2$ and
$R$ is Hermite.

###### Proof.

Let $(a,b,c)\in\operatorname{V}_{3}(I)$. We need to prove the existence of
$(r_{1},r_{2})\in R^{2}$ such that
$(a+r_{1}c,b+r_{2}c)\in\operatorname{V}_{2}(I)$.

We assume first that $(i)$ holds. The proof follows closely the lines of
[McG08, Theorem 3.6]. If $a=0$, then we can take $(r_{1},r_{2})=(1,0)$. Thus
we can suppose that $a\neq 0$. If $I=Ra$, then we can take
$(r_{1},r_{2})=(0,0)$. Thus we can suppose that $Ra\neq I$. As $I$ is
$1\nicefrac{{1}}{{2}}$-generated, the non-empty set $X$ of the ideals
$I^{\prime}$ verifying $Ra\subseteq I^{\prime}\subsetneq I$ is isomorphic to
the set of proper ideals of $R$ containing $(a:I)$ as a set partially ordered
by inclusion. Hence we can consider the intersection $J$ of the maximal
elements of $X$. Since $J$ contains $a$, it is a non-zero ideal. Thus
$\textnormal{sr}(I/J)=1$, so that we can find $r\in R$ satisfying $b+rc+J=I$.
Reasoning by contradiction, we assume that $Ra+R(b+rc)\subsetneq I$. Then
$Ra+R(b+rc)$ is contained in a maximal element $K\in X$. As a result, we have
$J\subseteq K\subsetneq I$. We obtain a contradiction by observing that
$K\supseteq Ra+R(b+rc)+J=I$. Therefore we can take $(r_{1},r_{2})=(0,r)$.

Let us assume now that $(ii)$ holds. If $I\mathcal{J}(R)=\\{0\\}$, then
$\mathcal{J}(R)=\\{0\\}$ so that $(i)$ is satisfied. Otherwise, we have
$I/\mathcal{J}(R)I\simeq R/\mathcal{J}(R)$ by Lemma 3.2. Hence
$\textnormal{sr}(I/\mathcal{J}(R)I)\leq 2$ by $(i)$. The result follows then
from Proposition 2.6.$iii$. ∎

######  Proposition 3.4.

Let $R$ and $I$ be as in Lemma 3.3. Then $\textnormal{er}(I)\leq 2$.

###### Proof.

Let $(a_{1},\dots,a_{n})\in\operatorname{V}_{n}(I)$ with $n>2$. If any of
$(i)$ or $(ii)$ holds, then $\textnormal{sr}(I)=2$ by Lemma 3.3. Hence we can
assume, without loss of generality, that $n=3$ and $a_{1}\neq 0$.

We assume first that $(i)$ holds. Let $b\in I$ be such that $I=Ra_{1}+Rb$.
Then the following equivalence holds

(1) $(a_{1},a_{2},a_{3})\sim_{\textnormal{E}_{3}(R)}(a_{1},b,0).$

Indeed, since $\textnormal{sr}(I/Ra_{1})=\textnormal{sr}(R/(a_{1}:I))=1$, we
have
$(a_{2}+Ra_{1},a_{3}+Ra_{1})\sim_{\textnormal{E}_{2}(R)}(b+Ra_{1},Ra_{1})$
from which we easily infer the equivalence (1). Fix now
$(a_{1}^{\prime},a_{2}^{\prime},0)\in\operatorname{V}_{3}(I)$ with
$a_{1}^{\prime}\neq 0$ and let $b^{\prime}\in I$ be such that
$I=Ra_{1}^{\prime}+Rb^{\prime}$. Then we have

(2) $(a_{1},b,0)\sim_{\textnormal{E}_{3}(R)}(a_{1}^{\prime},a_{1},b)$

while the two equivalences

(3)
$(a_{1}^{\prime},a_{1},b)\sim_{\textnormal{E}_{3}(R)}(a_{1}^{\prime},b^{\prime},0),\,(a_{1}^{\prime},a_{2}^{\prime},0)\sim_{\textnormal{E}_{3}(R)}(a_{1}^{\prime},b^{\prime},0)$

can be derived from (1) using obvious substitutions. Combining (1), (2) and
(3) yields
$(a_{1},a_{2},a_{3})\sim_{\textnormal{E}_{3}(R)}(a_{1}^{\prime},a_{2}^{\prime},0)$,
which proves the result.

Let us now assume that $(ii)$ holds. We can also assume, without loss of
generality, that $I\mathcal{J}(R)\neq\\{0\\}$, since otherwise $(i)$ is
satisfied. Thus $I/I\mathcal{J}(R)\simeq R/\mathcal{J}(R)$ by Lemma 3.2. As we
have
$\textnormal{er}(I)=\textnormal{er}(I/I\mathcal{J}(R))=\textnormal{er}_{R/\mathcal{J}(R)}(I/I\mathcal{J}(R))$
by Proposition 2.7, the result follows from $(i)$. ∎

######  Lemma 3.5.

Let $I$ be an invertible ideal of $R$. Suppose there is $a\in I$ such that
$a+\mathcal{J}(R)$ is contained in only finitely maximal ideals of $R$. Then
there is $b\in I$ such that $I=Ra+Rb$.

Lemma 3.5 implies that an invertible ideal $I$ of $R$ is generated by
$1\nicefrac{{1}}{{2}}$ element if $R/\mathcal{J}(R)$ is of finite character.
This lemma is a straightforward generalization of [GH70, Theorem 3]. We
provide a proof for the convenience of the reader.

###### Proof.

If $a=0$, then $R$ is semi-local so the result holds by [GH70, Corollary 2].
If $I=Ra$, the result is obvious. Otherwise $\mathfrak{b}\Doteq(a:I)$ is a
proper ideal of $R$ and satisfies $Ra=I\mathfrak{b}$. Since
$Ra+\mathcal{J}(R)\subseteq\mathfrak{b}+\mathcal{J}(R)$, there are only
finitely many maximal ideals $\mathfrak{p}_{1},\dots,\mathfrak{p}_{n}$ of $R$
containing $\mathfrak{b}+\mathcal{J}(R)$. By [GH70, Theorem 2], we can find
$b\in I\setminus\bigcup_{i=1}^{n}I\mathfrak{p}_{i}$. We have
$Rb=I\mathfrak{c}$ for $\mathfrak{c}=(b:I)$ and $\mathfrak{c}+\mathcal{J}(R)$
is not contained in any of the ideals $\mathfrak{p}_{i}$. Therefore
$Ra+Rb=I(\mathfrak{b}+\mathfrak{c})$ and
$R=\mathfrak{b}+\mathfrak{c}+\mathcal{J}(R)$ as the latter ideal is not
contained in any maximal ideal of $R$. Thus $I=Ra+Rb+\mathcal{J}(R)I$ and the
result follows from Nakayama’s Lemma [Eis95, Corollary 4.8]. ∎

######  Corollary 3.6.

Assume that every proper quotient of $R/\mathcal{J}(R)$ has stable rank $1$.
Let $I$ be an invertible $1\nicefrac{{1}}{{2}}$-generated ideal of $R$. Then
$\textnormal{SL}_{n}(R)$ acts transitively on $\operatorname{V}_{n}(I)$ for
every $n\geq 2$.

###### Proof.

The group $\textnormal{E}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(I)$ for every $n\geq 3$ by Proposition 3.4. Since $I$ is
invertible, it is a faithful projective module of constant rank $1$. By
[Guy20, Corollary 3.2], the group $\textnormal{SL}_{2}(R)$ acts transitively
on $\operatorname{V}_{2}(I)$. ∎

######  Corollary 3.7.

Assume that $R/\mathcal{J}(R)$ is of finite character. Let $I$ be an
invertible ideal of $R$. Then $\textnormal{SL}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(I)$ for every $n\geq 2$.

###### Proof.

Combine Lemma 3.3, Lemma 3.5 and Corollary 3.6. ∎

A ring $R$ is a _$PP$ -ring_ if every principal ideal of $R$ is projective. A
coherent Prüfer ring is, for instance, a PP-ring. Coherent Prüfer rings can be
characterized as the arithmetic PP-rings [LQ15, Theorem XII.4.1].

######  Proposition 3.8.

Assume $R$ is a $PP$-ring and let $I$ be an invertible ideal of $R$. Suppose
moreover that the two following hold:

  * (i)

$\textnormal{sr}(R/Rr)=1$ for every regular element $r\in R$.

  * (ii)

$\mu(I/Rr)=1$ for every regular element $r\in I$.

Then $I$ is generated by two regular elements and we have:

  * •

$\textnormal{sr}(I)\leq 2$,

  * •

$\textnormal{er}(I)\leq 2$.

In particular, we have $\textnormal{sr}(R)\leq 2$ and $R$ is Hermite. In
addition, the group $\textnormal{SL}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(I)$ for every $n\geq 2$.

###### Proof.

Since $I$ is invertible, it contains a regular element $x$. Using hypothesis
$(ii)$, we infer that $I=Rx+Rx^{\prime}$ for some $x^{\prime}\in R$. As $R$ is
a $PP$-ring, it is additively regular [Mat83, Propositions 1.4 and 2.8]. Hence
we can find $\lambda\in R$ such that $y\Doteq x+\lambda x^{\prime}$ is
regular. Thus $I$ is generated by the two regular elements $x$ and $y$.

Let us prove first that $\textnormal{sr}(I)\leq 2$. To do so, consider
$(a,b,c)\in\operatorname{V}_{3}(I)$. If $a=0$, we can replace, without loss of
generality, $a$ with $x$. In this case, we can suppose $a$ regular so that
$\textnormal{sr}(I/Ra)=\textnormal{sr}(R/(a:I))=1$. Consequently, there is
$r\in R$ such that $b+rc+Ra$ generates $I/Ra$, i.e.,
$(a,b+rc)\in\operatorname{V}_{2}(I)$. If $a$ is not regular, then $(0:a)=Re$
for some idempotent $e\in R$. Applying the previous reasoning to $Ie$ and
$I(1-e)$ in the PP-rings $Re$ and $R(1-e)$ yields $(r_{1},r_{2})\in R^{2}$
such that $(a+r_{1}c,b+r_{2}c)\in\operatorname{V}_{2}(I)$.

Let us prove now that $\textnormal{er}(I)\leq 2$. Let
$(a,b,c,a_{1}\dots,a_{n-3})\in\operatorname{V}_{n}(I)$ with $n>2$. Because
$\textnormal{sr}(I)\leq 2$, we can assume without loss of generality that
$n=3$ and $c=x$. Using the fact that
$\textnormal{sr}(I/Rx)=\textnormal{sr}(R/(x:I))=1$, we deduce that
$(a+Rx,b+Rx)\sim_{\textnormal{E}_{2}(R)}(0,y+Rx)$ and hence
$(a,b,c)\sim_{\textnormal{E}_{3}(R)}(x,y,0)$. Therefore
$\textnormal{er}(I)\leq 2$.

As $\textnormal{SL}_{2}(R)$ acts transitively on $\operatorname{V}_{2}(I)$ by
[Guy20, Corollary 3.2], we deduce that $\textnormal{SL}_{n}(R)$ acts
transitively on $\operatorname{V}_{n}(I)$ for every $n\geq 2$. ∎

##  4\. Direct products of cyclic modules and ideals

By Theorems 2.9 and 2.12, an $R$-module $M$ as in Theorems A and E decomposes
into a direct product of cyclic modules and ideals. In this section, we
describe the action of
$G\in\\{\textnormal{SL}_{n}(R),\textnormal{E}_{n}(R)\\}$ on the generating
vectors of such a direct product under assumptions which are less restrictive
than those of Theorems A and E.

Let $\mathfrak{a}_{1},\dots,\mathfrak{a}_{k}$ be ideals of $R$. We denote by
$e_{i}$ ($1\leq i\leq k$) the element of the direct product
$R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}$ whose $i$-th
component is the identity of $R/\mathfrak{a}_{i}$ and whose other components
are zero. If $\mathbf{m}=(m_{1},\dots,m_{n})\in(M\times N)^{n}$ with
$M=R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}$ and $N$ a finitely
generated module, we let

$\operatorname{Mat}(\mathbf{m})\Doteq((m_{j})_{i})$

be the $(k+1)\times n$ matrix whose columns are the elements
$m_{j}=((m_{j})_{i})$ with $(m_{j})_{k+1}\in N$ and $(m_{j})_{i}\in
R/\mathfrak{a}_{i}$ for $1\leq i\leq k$ and $1\leq j\leq n$. Clearly, we have
$\operatorname{Mat}(\mathbf{m}A)=\operatorname{Mat}(\mathbf{m})A$ for every
$n\times n$ matrix $A$ over $R$.

###  4.1. Products with two factors

This section is dedicated to modules of the form $R/\mathfrak{a}\times I$
($\mathfrak{a},I\subseteq R$). These modules are used as bases in each
induction proof of Section 4.2.

######  Lemma 4.1.

Let $M$ be a finitely generated $R$-module and let $\mathfrak{b}$ be an ideal
of $R$ contained in $\operatorname{ann}(M)$. Let
$\mathbf{m}=((m_{1},0),\dots,(m_{n-1},0),(m_{n},1))\in\operatorname{V}_{n}(M\times
R/\mathfrak{b})$ with $n\geq\mu(M)$. Then
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}((m_{1},0),\dots,(m_{n-1},0),(0,1))$.

###### Proof.

It suffices to show that $m_{n}$ lies in the submodule of $M$ generated by
$m_{1},\dots,m_{n-1}$. Since $\mathbf{m}$ generates $M\times R/\mathfrak{b}$,
we can find $\mathbf{r}=(r_{1},\dots,r_{n})\in R^{n}$ such that
$\mathbf{m}\mathbf{r}^{\top}=(m_{n},0)$. We infer from the previous identity
that $r_{n}\in\mathfrak{b}$ and hence $m_{n}=\sum_{i=1}^{n-1}r_{i}m_{i}$. ∎

######  Lemma 4.2.

Let $M=R/\mathfrak{a}\times R$ with $\mathfrak{a}$ an ideal of $R$. Let
$\mathbf{m}\in\operatorname{V}_{2}(M)$. Then
$\mathfrak{a}=\operatorname{Fitt}_{1}(M)$ and
$\mathbf{m}\sim_{\textnormal{SL}_{2}(R)}(\operatorname{det}_{\mathbf{e}}(\mathbf{m})e_{1},e_{2})$
where $\mathbf{e}=(e_{1},e_{2})$ with $e_{1}=(1,0)$ and $e_{2}=(0,1)$. In
particular, the map $\det_{\mathbf{e}}$ induces a bijection from
$\operatorname{V}_{2}(M)/\textnormal{SL}_{2}(R)$ onto
$(R/\mathfrak{a})^{\times}$.

###### Proof.

The fact that $\mathfrak{a}=\operatorname{Fitt}_{1}(M)$ is a direct
consequence of Lemma 2.2. By definition, there is a $2\times 2$ matrix $A$
over $R$ such that $\mathbf{m}=\mathbf{e}A$. Since $\textnormal{SL}_{2}(R)$
acts transitively on $\operatorname{V}_{2}(R)$, we can find
$\sigma\in\textnormal{SL}_{2}(R)$ such that $A\sigma=\begin{pmatrix}u&v\\\
0&1\end{pmatrix}$ for some $u,v\in R$. As $\mathbf{m}\sigma$ generates $M$,
Lemma 4.1 applies with $\mathfrak{b}=\\{0\\}$ so that
$\mathbf{m}\sim_{\textnormal{SL}_{2}(R)}(ue_{1},e_{2})$. Therefore
$\det_{\mathbf{e}}(\mathbf{m})=u+\operatorname{Fitt}_{1}(M)=u+\mathfrak{a}$,
which completes the proof. ∎

######  Proposition 4.3.

Let $M$ be a two-generated faithful $R$-module such that $M_{\mathfrak{p}}$ is
isomorphic to $R_{\mathfrak{p}}/\mathfrak{a}_{\mathfrak{p}}\times
R_{\mathfrak{p}}$, for every maximal ideal $\mathfrak{p}$ of $R$, with
$\mathfrak{a}_{\mathfrak{p}}$ some ideal of $R_{\mathfrak{p}}$. Let
$\mathbf{m},\mathbf{m}^{\prime}\in\operatorname{V}_{2}(M)$. Then the following
are equivalent.

  * $(i)$

$\mathbf{m}\sim_{\textnormal{SL}_{2}(R)}\mathbf{m}^{\prime}$,

  * $(ii)$

$\det_{\mathbf{m}}(\mathbf{m}^{\prime})=1$.

###### Proof.

The implication $(i)\Rightarrow(ii)$ follows from Lemma 2.1.$i$. Let us prove
$(ii)\Rightarrow(i)$. By [Guy20, Proposition 3.1] and Lemma 2.1.$iii$, we can
assume that $R$ is local. Applying Lemma 4.2 and Lemma 2.1.$ii$ yields the
result. ∎

Given an ideal $\mathfrak{a}$ of $R$, we denote by
$\textnormal{E}_{12}(\mathfrak{a})$ the group of matrices of the form
$\begin{pmatrix}1&r\\\ 0&1\end{pmatrix}$ with $r\in\mathfrak{a}$.

######  Lemma 4.4.

Let $M=R/\mathfrak{a}\times R$ with $\mathfrak{a}$ an ideal of $R$ and let
$\mathbf{e}\Doteq(e_{1},e_{2})$. For every
$\mathbf{m}\in\operatorname{V}_{2}(M)$, fix
$\sigma_{\mathbf{m}}\in\textnormal{SL}_{2}(R)$ such that
$\mathbf{m}=(\operatorname{det}_{\mathbf{e}}(\mathbf{m})e_{1},e_{2})\sigma_{\mathbf{m}}$.
Then the two following hold.

  * $(i)$

The map
$\Delta_{\mathbf{e}}:\mathbf{m}\textnormal{E}_{2}(R)\mapsto\left(\rm{det}_{\mathbf{e}}(\mathbf{m}),\textnormal{E}_{12}(\mathfrak{a})\sigma_{\mathbf{m}}\textnormal{E}_{2}(R)\right)$
defines a bijection from $\operatorname{V}_{2}(M)/\textnormal{E}_{2}(R)$ onto
$(R/\mathfrak{a})^{\times}\times\left(\textnormal{E}_{12}(\mathfrak{a})\backslash\textnormal{SL}_{2}(R)/\textnormal{E}_{2}(R)\right)$.

  * $(ii)$

If $G$ is a normal subgroup of $\textnormal{SL}_{2}(R)$ containing
$\textnormal{E}_{2}(R)$, then $\operatorname{V}_{2}(M)/G$ is equipotent with
$(R/\mathfrak{a})^{\times}\times\textnormal{SL}_{2}(R)/G$.

In general $\textnormal{E}_{2}(R)$ is not a normal subgroup of
$\textnormal{SL}_{2}(R)$, see e.g., [Nic11, Theorem 1.5] or [Lam06, Claim
I.8.16]. More tractable subgroups $G$ could be considered instead in Lemma
4.4.$ii$, e.g., the normal closure of $\textnormal{E}_{2}(R)$ in
$\textnormal{SL}_{2}(R)$ or the group of $2\times 2$ unipotent matrices over
$R$.

######  Example 4.5.

Let $R$ be the ring of integers of a totally imaginary quadratic field. If $R$
is not a $GE$-ring, then it follows from [FF88, Theorem 2.4] and [LS77,
Theorem IV.2.6] that
$\textnormal{E}_{12}(R)\backslash\textnormal{SL}_{2}(R)/\textnormal{E}_{2}(R)$
is infinite (alternatively use [Nic11, Theorem 1.5] and [She16]). In
particular, $\operatorname{V}_{2}(R/\mathfrak{a}\times
R)/\textnormal{SL}_{2}(R)$ is infinite for every ideal $\mathfrak{a}\subseteq
R$. Note that [Sta18, Proof of Proposition 7.5] offers a geometric insight on
$\textnormal{SL}_{2}(R)/\textnormal{E}_{2}(R)$.

###### Proof of Lemma 4.4.

$(i)$. Given $\mathbf{m}\in\operatorname{V}_{2}(M)$, the existence of
$\sigma_{\mathbf{m}}$ is provided by Lemma 4.2. In order to see that
$\Delta_{\mathbf{e}}$ is well-defined, observe that $\det_{\mathbf{e}}$ is
$\textnormal{E}_{2}(R)$-invariant and that the stabilizer of
$\mathbf{m}=(\delta e_{1},e_{2})$ in $\textnormal{SL}_{2}(R)$ is
$\textnormal{E}_{12}(\mathfrak{a})$ for every
$\delta\in(R/\mathfrak{a})^{\times}$. Checking that $\Delta_{\mathbf{e}}$ is
bijection is straightforward. $(ii)$. Check that
$\mathbf{m}G\mapsto\left(\rm{det}_{\mathbf{e}}(\mathbf{m}),\sigma_{\mathbf{m}}\it
G\right)$ is well-defined and argue as in $(i)$. ∎

######  Lemma 4.6.

Let $I$ be a two-generated ideal of $R$. Let $\mathfrak{a}$ be an ideal of $R$
such that $\textnormal{glr}_{R}(R/\mathfrak{a})=1$. Then $R/\mathfrak{a}\times
I$ is two-generated if and only if there is $(a,b)\in\operatorname{V}_{2}(I)$
such that $(1,a)$ and $(0,b)$ generate $R/\mathfrak{a}\times I$. Moreover, for
every $(a,b)\in\operatorname{V}_{2}(I)$, the following are equivalent:

  * $(i)$

$R/\mathfrak{a}\times I$ is generated by $(1,a)$ and $(0,b)$.

  * $(ii)$

$(b:a)+\mathfrak{a}=R$.

If in addition we suppose $I$ invertible, then each of the above assertions is
equivalent to:

  * $(iii)$

$I=I\mathfrak{a}+Rb$.

###### Proof.

Assuming that $\mu(R/\mathfrak{a}\times I)\leq 2$, we pick an arbitrary
$\mathbf{m}\in\operatorname{V}_{2}(R/\mathfrak{a}\times I)$. By hypothesis, we
can find $\sigma\in\textnormal{SL}_{2}(R)$ such that
$\operatorname{Mat}(\mathbf{m})\sigma=\begin{pmatrix}1&0\\\ a&b\end{pmatrix}$
for some $(a,b)\in\operatorname{V}_{2}(I)$, which proves the first part of the
statement.

$(i)\Rightarrow(ii)$. By hypothesis, there is $(\lambda,\mu)\in R^{2}$ such
that $\lambda(1,a)+\mu(0,b)=(0,a)$, which implies $\lambda-1\in(b:a)$ and
$\lambda\in\mathfrak{a}$. $(ii)\Rightarrow(i)$. By hypothesis, there is
$(a,b)\in\operatorname{V}_{2}(I)$, $\lambda,\nu\in R$ and $\mu\in\mathfrak{a}$
such that $\lambda+\mu=1$ and $\lambda a=\nu b$. As $(0,a)=\mu(1,a)+\nu(0,b)$,
we deduce that $(1,a)$ and $(0,b)$ generate $R/\mathfrak{a}\times I$.

To complete the proof, it suffices to show $(ii)\Leftrightarrow(iii)$ under
the additional hypothesis. Multiplying both sides of $(ii)$ by $I$, we obtain
$I=I\mathfrak{a}+I(b:a)$. Clearly $(b:a)=(b:I)$. As $I$ is assumed to be
invertible, we have $(b:I)=bI^{-1}$, hence $I=I\mathfrak{a}+Rb$. Conversely,
multiplying both sides of $(iii)$ by $I^{-1}$ yields $R=\mathfrak{a}+(b:a)$. ∎

######  Lemma 4.7.

Let $I,\mathfrak{a}$ be two ideals of $R$ with $I$ invertible and $\mu(I)\leq
2$. Let $M\Doteq R/\mathfrak{a}\times I\subseteq R/\mathfrak{a}\times R$. Then
$\operatorname{Fitt}_{1}(M)=\mathfrak{a}$.

###### Proof.

Let us consider two presentations

$\displaystyle F_{1}\xrightarrow[]{\varphi_{1}}R\rightarrow
R/\mathfrak{a}\rightarrow 0$ $\displaystyle
F_{2}\xrightarrow[]{\varphi_{2}}R^{2}\rightarrow I\rightarrow 0.$

where $F_{1}$ and $F_{2}$ are free $R$-modules. This gives rise to a
presentation

$F\xrightarrow[]{\varphi}R^{3}\rightarrow M\rightarrow 0.$

with $F=F_{1}\times F_{2}$ and $\varphi=\varphi_{1}\times\varphi_{2}$. By
[Eis95, Corollary 20.4], we have
$\operatorname{Fitt}_{1}(M)=I_{2}(\varphi)=I_{1}(\varphi_{2})I_{1}(\varphi_{1})+I_{2}(\varphi_{2})=\operatorname{Fitt}_{1}(I)\mathfrak{a}+\operatorname{Fitt}_{0}(I)$.
By [Eis95, Propositions 20.5 and 20.6], we have
$\operatorname{Fitt}_{0}(I)=\\{0\\}$ and $\operatorname{Fitt}_{1}(I)=R$.
Therefore $\operatorname{Fitt}_{1}(M)=\mathfrak{a}$. ∎

######  Proposition 4.8.

Let $\mathfrak{a}\neq 0$ be an ideal of $R$ such that
$\textnormal{glr}_{R}(R/\mathfrak{a})=1$. Let $I$ be an invertible ideal of
$R$ and let $M\Doteq R/\mathfrak{a}\times I\subseteq R/\mathfrak{a}\times R$.
We suppose moreover that at least one of the following holds:

  * $(i)$

$I$ is $1\nicefrac{{1}}{{2}}$-generated.

  * $(ii)$

$\mathfrak{a}$ is regular and $I/Rr$ is cyclic for every regular element $r\in
I$.

Then there is $(a,b)\in\operatorname{V}_{2}(I)$ with $a\in
I\mathfrak{a}\setminus\\{0\\}$ such that $R/\mathfrak{a}\times I$ is generated
by $\left(e_{1}+ae_{2},be_{2}\right)$ where $e_{1}=(1,0)$ and $e_{2}=(0,1)$.
If $\mathbf{m}$ is a generating pair of this form and
$\mathbf{m}^{\prime}\in\operatorname{V}_{2}(M)$, then

$\mathbf{m}^{\prime}\sim_{\textnormal{SL}_{2}(R)}(\operatorname{det}_{\mathbf{m}}(\mathbf{m}^{\prime})e_{1}+ae_{2},be_{2}).$

In particular, the map $\det_{\mathbf{m}}$ induces a bijection from
$\operatorname{V}_{2}(M)/\textnormal{SL}_{2}(R)$ onto
$(R/\mathfrak{a})^{\times}$.

###### Proof.

If $(i)$ holds, we can pick $a\in I\mathfrak{a}\setminus\\{0\\}$ such that
$I=Ra+Rb$ for some $b\in I$. If $(ii)$ holds, we can find two regular elements
$x\in\mathfrak{a},y\in I$. Setting $a\Doteq xy$, we also obtain that $I=Ra+Rb$
for some $b\in I$. In particular, we have $I=I\mathfrak{a}+Rb$ so that
$\mathbf{m}\Doteq(e_{1}+ae_{2},be_{2})$ generates $M$ by Lemma 4.6. Let
$\mathbf{m}^{\prime}\in\operatorname{V}_{2}(M)$ and let
$\delta\Doteq\det_{\mathbf{m}^{\prime}}(\mathbf{m})$. As $\mathbf{m}$
generates $M$, so does $\mathbf{m}^{\prime\prime}\Doteq(\delta
e_{1}+ae_{2},be_{2})$. We claim that
$\det_{\mathbf{m}}(\mathbf{m}^{\prime\prime})=\delta$. The equivalence
$\mathbf{m}^{\prime}\sim_{\textnormal{SL}_{2}(R)}\mathbf{m}^{\prime\prime}$
follows from our claim. Indeed, we have $M_{\mathfrak{p}}\simeq
R_{\mathfrak{p}}/\mathfrak{a}R_{\mathfrak{p}}\times R_{\mathfrak{p}}$ for
every maximal ideal $\mathfrak{p}$ of $R$ because $I$ is invertible, so that
Proposition 4.3 applies. In order to establish our claim, we consider the
natural map $\pi:M\twoheadrightarrow M/\operatorname{Fitt}_{1}(M)M$. Since
$a\in\mathfrak{a}=\operatorname{Fitt}_{1}(M)$ by Lemma 4.7, we have
$\pi(\mathbf{m}^{\prime\prime})=\pi(\mathbf{m})\begin{pmatrix}\delta&0\\\
0&1\end{pmatrix}$ and Lemma 2.3 yields
$\det_{\mathbf{m}}(\mathbf{m}^{\prime\prime})=\delta$. This concludes the
proof. ∎

######  Proposition 4.9.

Let $I,\mathfrak{a}$ and $M$ be as in Proposition 4.8. Assume moreover that
$\textnormal{er}_{R}(R/\mathfrak{a})=1$ and that at least one of the following
holds:

  * $(i)$

$I$ is $1\nicefrac{{1}}{{2}}$-generated and
$\textnormal{er}_{R}(R/\mathfrak{b})=1$ for every non-zero
$\mathfrak{b}\subseteq I$.

  * $(ii)$

$R$ is a PP-ring, $\mathfrak{a}$ is regular, $I/\mathfrak{b}$ is cyclic and
$\textnormal{er}_{R}(R/\mathfrak{b})=1$ for every regular ideal
$\mathfrak{b}\subseteq I$.

Then $\textnormal{er}(M)=2$.

###### Proof.

We assume first that $(i)$ holds and let
$n>2,\mathbf{m}\in\operatorname{V}_{n}(M)$. We begin by proving that
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(e_{1}+ae_{2},be_{2},0,\dots,0)$ for
some $(a,b)\in\operatorname{V}_{2}(I)$ with $a\in
I\mathfrak{a}\setminus\\{0\\}$. Since $\textnormal{er}_{R}(R/\mathfrak{a})=1$,
we deduce that

(4) $\operatorname{Mat}(\mathbf{m}E)=\begin{pmatrix}1&0&\mathbf{0}\\\
a&b&\mathbf{c}\end{pmatrix}$

for some $E\in\textnormal{SL}_{2}(R)$, some $(a,b)\in I^{2}$ with $a\neq 0$
and some $\mathbf{c}=(c_{1},\dots,c_{n-2})\in I^{n-2}$. Since $\mathbf{m}E$
generates $M$, there is a row vector $\mathbf{r}=(r,s,\mathbf{t})\in R^{n}$
with $\mathbf{t}\in R^{n-2}$ such that
$\operatorname{Mat}(\mathbf{m}E)\mathbf{r}^{\top}=\begin{pmatrix}0\\\
a\end{pmatrix}$. As a result, we can write
$a=ar+bs+\mathbf{c}\mathbf{t}^{\top}$ with $r\in\mathfrak{a}$. Multiplying
$\mathbf{m}E$ on the right by an elementary matrix if needed, we can therefore
assume that $a\in I\mathfrak{a}$. Since $I$ is
$1\nicefrac{{1}}{{2}}$-generated, we have $\mu(I/Ra)=1$ and hence
$\textnormal{er}_{R}(I/Ra)=1$, so that
$(b+Ra,c_{1}+Ra,\dots,c_{n-2}+Ra)\sim_{\textnormal{E}_{n-1}(R)}(d+Ra,0,\dots,0)$
for some $d\in I$. Thus we can assume, without loss of generality, that
$c_{i}\in Ra$ for every $i\geq 1$ and $I=I\mathfrak{a}+Rb$. This shows in
particular that $(a,b)\in\operatorname{V}_{2}(I)$ and we infer from Lemma 4.6
that $(e_{1}+ae_{2},be_{2})$ generates $M$. It follows immediately that
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(e_{1}+ae_{2},be_{2},0,\dots,0)$, as
claimed.

Let $\mathbf{m}^{\prime}=(e_{1}+a^{\prime}e_{2},b^{\prime}e_{2},0,\dots,0)$
with $(a^{\prime},b^{\prime})\in\operatorname{V}_{2}(I)$ and $a^{\prime}\in
I\mathfrak{a}\setminus\\{0\\}$. We shall conclude our proof by showing that
$\mathbf{m}^{\prime}\sim_{\textnormal{E}_{n}(R)}\mathbf{m}$. We begin by
observing that

$\operatorname{Mat}(\mathbf{m}^{\prime})\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}1&0&0&\mathbf{0}\\\
a^{\prime}&b^{\prime}&a-a^{\prime}&\mathbf{0}\end{pmatrix}\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}1&0&0&\mathbf{0}\\\
a&b^{\prime}&a-a^{\prime}&\mathbf{0}\end{pmatrix}.$

Using again the fact that $\textnormal{er}_{R}(I/Ra)=1$, we deduce that
$(b^{\prime},a-a^{\prime})\sim_{\textnormal{E}_{2}(R)}(b+\lambda a,\mu a)$ for
some $(\lambda,\mu)\in R^{2}$. Since $(e_{1}+ae_{2},(b+\lambda a)e_{2})$
generates $M$ by Lemma 4.6, we have

$\displaystyle\operatorname{Mat}(\mathbf{m}^{\prime})\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}1&0&0&\mathbf{0}\\\
a&b+\lambda a&\mu
a&\mathbf{0}\end{pmatrix}\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}1&0&0&\mathbf{0}\\\
a&b+\lambda a&a&\mathbf{0}\end{pmatrix}$

from which $\mathbf{m}^{\prime}\sim_{\textnormal{E}_{n}(R)}\mathbf{m}$ easily
follows.

Let us assume now that $(ii)$ holds. Proceeding as before, we only need to
show that given the identity (4) with $a\in I\mathfrak{a}$, we can further
suppose that $a$ is regular. Since $R$ is a PP-ring, there is an idempotent
$e\in R$ such that $Re=(0:a)$. Let $r$ be a regular element in
$I\mathfrak{a}$. Then there are $\lambda\in R,\bm{\mu}\in R^{n-2}$ such that
$re=\lambda be+\bm{\mu}\mathbf{c}^{\top}e$. As $a+re$ is regular, we can
therefore assume, without loss of generality, that so is $a$. From there on,
the proof is the same. ∎

###  4.2. Products of more than two factors

In this section, we consider $R$-modules $M$ of one of the two following forms

(5) $\displaystyle R/\mathfrak{a}_{1}\times\cdots\times
R/\mathfrak{a}_{k}\,(k\geq 1),$ (6) $\displaystyle R/I_{1}\times\cdots\times
R/I_{k}\times I\times R^{l}\,(k\geq 0,l\geq 0)$

where $R\supsetneq\mathfrak{a}_{1}\supseteq\cdots\supseteq\mathfrak{a}_{k}$ is
a descending chain of ideals, $R\supsetneq I_{1}\supseteq\cdots\supseteq
I_{k}$ is a descending chain of invertible ideals, $I$ is either $\\{0\\}$ or
an invertible ideal and $l>0$ if $I=\\{0\\}$. We show that the conclusions of
Theorems A and E hold for such modules with less restrictive assumptions on
$R$. This is achieved with Proposition 4.10 ($M$ as in (6), torsion free but
not free), Propositions 4.12 and 4.14 ($M$ as in (5)), Proposition 4.16 ($M$
as in (6) and $k\geq 1$) and Proposition 4.18 ($M$ is free) below.

We denote by $\textnormal{SL}_{k}(R)\cap\textnormal{E}_{n}(R)\,(k\leq n)$ the
subgroup of $\textnormal{SL}_{k}(R)$ consisting of the matrices $\sigma$ such
that $\begin{pmatrix}\sigma&\mathbf{0}\\\
\mathbf{0}&1_{n-k}\end{pmatrix}\in\textnormal{E}_{n}(R)$.

######  Proposition 4.10.

Assume that $R$ is Hermite. Let $I$ be an ideal of $R$ such that $\mu(I)=2$.
Let $M=I\times R^{k}\subset R^{k+1}\,(k\geq 1)$. Then the following hold.

  * $(i)$

$\mu(M)=k+2$ and $\operatorname{Fitt}_{k+1}(M)=\operatorname{Fitt}_{1}(I)$.

  * $(ii)$

The map $(a,b)\mapsto(ae_{1},be_{1},e_{2},\dots,e_{k+1})$ induces a bijection
from
$\operatorname{V}_{2}(I)/\textnormal{SL}_{2}(R)$ onto
$\operatorname{V}_{k+2}(M)/G$ with $G=\textnormal{SL}_{k+2}(R)$. If
$\textnormal{er}(R)\leq 2$, we can take $G=\textnormal{E}_{\mu}(R)$ and the
above map also induces a bijection from
$\operatorname{V}_{2}(I)/(\textnormal{SL}_{2}(R)\cap\textnormal{E}_{k+2}(R))$
onto $\operatorname{V}_{k+2}(M)/\textnormal{E}_{k+2}(R)$.

  * $(iii)$

For every $n>k+2$, the map

$(a_{1},\dots,a_{n-k})\mapsto(a_{1}e_{1},\dots,a_{n-k}e_{n-k},e_{n-k+1},\dots,e_{n})$

induces a bijection from
$\operatorname{V}_{n-k}(I)/(\textnormal{SL}_{n-k}(R)\cap G)$ onto
$\operatorname{V}_{n}(M)/G$ where $G=\textnormal{SL}_{n}(R)$. If
$\textnormal{er}(R)\leq 2$, we can take $G=\textnormal{E}_{n}(R)$.

  * $(iv)$

If $R$ and $I$ are as in Proposition 3.4 or as in Proposition 3.8, then
$\textnormal{er}(M)=k+2$.

###### Proof.

We begin by proving $(i)$ and $(ii)$ simultaneously under the unique
assumption $\textnormal{glr}(R)=1$. Let
$\mathbf{m}\in\operatorname{V}_{\mu}(M)$. First, we shall prove by induction
on $k\geq 0$ that $\mu=k+2$ and that
$\mathbf{m}\sim_{\textnormal{SL}_{\mu}(R)}(ae_{1},be_{1},e_{2},\dots,e_{k+1})$
for some $(a,b)\in\operatorname{V}_{2}(I)$ depending on $\mathbf{m}$, which is
obvious for $k=0$. Suppose now that $k\geq 1$. As $\textnormal{glr}(R)=1$, we
deduce that

(7)
$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{SL}_{\mu}(R)}\begin{pmatrix}A&\mathbf{b}\\\
\mathbf{0}&1\end{pmatrix}$

with $A$ a $k\times(\mu-1)$ matrix and $\mathbf{b}$ a column vector of size
$k$. It follows from Lemma 4.1 that

$\begin{pmatrix}A&\mathbf{b}\\\
\mathbf{0}&1\end{pmatrix}\sim_{\textnormal{E}_{\mu}(R)}\begin{pmatrix}A&\mathbf{0}\\\
\mathbf{0}&1\end{pmatrix}.$

Our claim is then proven by applying the induction hypothesis to the
generating vector $\mathbf{m}^{\prime}\in\operatorname{V}_{\mu-1}(M^{\prime})$
defined by $\operatorname{Mat}(\mathbf{m}^{\prime})=A$.

The identity $\operatorname{Fitt}_{k+1}(M)=\operatorname{Fitt}_{1}(I)$ follows
from Lemma 2.2 applied to $\mathbf{m}=(ae_{1},be_{1},e_{2},\dots,e_{k+1})$. We
have already established that the map

$(a,b)\mapsto(ae_{1},be_{1},e_{2},\dots,e_{k+1})$

induces a surjection from $\operatorname{V}_{2}(I)/\textnormal{SL}_{2}(R)$
onto $\operatorname{V}_{k+2}(M)/\textnormal{SL}_{k+2}(R)$. In order to show
that the induced map is also injective, let us consider
$(a,b),(a^{\prime},b^{\prime})\in\operatorname{V}_{2}(I)$ and
$\sigma\in\textnormal{SL}_{k+2}(R)$ such that

$(a^{\prime}e_{1},b^{\prime}e_{1},e_{2},e_{3},\dots,e_{k+1})=(ae_{1},be_{1},e_{2},e_{3},\dots,e_{k+1})\sigma.$

Then $\sigma$ must be of the form
$\begin{pmatrix}\sigma^{\prime}&\mathbf{0}\\\ \mathbf{0}&1_{k-1}\end{pmatrix}$
with $\sigma^{\prime}\in\textnormal{SL}_{2}(R)$. Thus
$(a^{\prime},b^{\prime})\sim_{\textnormal{SL}_{2}(R)}(a,b)$.

If in addition $\textnormal{er}(R)\leq 2$, we show likewise that
$\mathbf{m}\sim_{\textnormal{E}_{k+2}(R)}(ae_{1},be_{1},e_{2},\dots,e_{k+1})$
for some $(a,b)\in\operatorname{V}_{2}(I)$ by induction on $k$. Since
$\textnormal{er}(R)\leq 2<k+2$, we can use $\textnormal{E}_{k+2}(R)$ to reduce
$\operatorname{Mat}(\mathbf{m})$ in (7). From there, the proof follows the
same lines.

$(iii)$. Proceed by induction on $k$ as in the proof of assertion $(ii)$.

Assertion $(iv)$ follows from $(iii)$, since $\textnormal{E}_{n-k}(R)$ acts
then transitively on $\operatorname{V}_{n-k}(I)$ by Propositions 3.4 and 3.8.
∎

######  Lemma 4.11.

Let $M=R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}\,(k\geq 1)$
where the ideals $\mathfrak{a}_{i}\subseteq R$ form an descending chain
$R\supsetneq\mathfrak{a}_{1}\supseteq\cdots\supseteq\mathfrak{a}_{k}$. Then
$\mu(M)=k$ and $\operatorname{Fitt}_{k-1}(M)=\mathfrak{a}_{1}$.

###### Proof.

Clearly $\mu(M)\leq k$. Let $\mathfrak{p}$ be a maximal ideal of $R$
containing $\mathfrak{a}_{1}$. Then $M/\mathfrak{p}M$ surjects onto
$(R/\mathfrak{p})^{k}$ which is vector space of dimension $k$ over
$R/\mathfrak{p}$. Therefore $\mu(M)\geq k$ and hence $\mu(M)=k$. The identity
$\operatorname{Fitt}_{k-1}(M)=\mathfrak{a}_{1}$ follows from Lemma 2.2 applied
to $\mathbf{m}=(e_{1},\dots,e_{k})$. ∎

######  Proposition 4.12.

Let $M$ be as in Lemma 4.11. Let
$G\in\\{\textnormal{SL}_{n}(R),\textnormal{E}_{n}(R)\\}$ with $n\geq k$ and
let $\mathbf{e}=(e_{1},\dots,e_{k})$. Assume that one of the following holds:

  * $(i)$

$G=\textnormal{SL}_{n}(R)$ and $\textnormal{glr}_{R}(R/\mathfrak{a}_{i})\leq
i$ for every $i\geq 1$.

  * $(ii)$

$G=\textnormal{E}_{n}(R)$ and
$\textnormal{er}_{R}(R/\mathfrak{a}_{i})\leq\max(1,i-1)$ for every $i\geq 1$.

Then for every $\mathbf{m}\in\operatorname{V}_{n}(M)$, we have:

$\mathbf{m}\sim_{G}\left\\{\begin{array}[]{cc}(\det_{\mathbf{e}}(\mathbf{m})e_{1},e_{2},\dots,e_{k}),&\text{
if }n=k,\\\ (e_{1},e_{2},\dots,e_{k},0,\dots,0),&\text{ if
}n>k.\end{array}\right.$

In particular $\det_{\mathbf{e}}$ induces a bijection from
$\operatorname{V}_{k}(M)/G$ onto $(R/\mathfrak{a}_{1})^{\times}$.

######  Remark 4.13.

The hypothesis $(i)$ of Proposition 4.12 is clearly satisfied if $R$ is a
K-Hermite ring. It is also satisfied if at least one of the following holds:

  * $(a)$

Every proper quotient of $R$ has stable rank $1$.

  * $(b)$

The ring $R$ is as in Proposition 3.8 and each ideal $\mathfrak{a}_{i}$ is
either $\\{0\\}$ or regular.

Hypothesis $(ii)$ is satisfied if any of $(a)$ or $(b)$ holds and
$\mathfrak{a}_{2}\neq\\{0\\}$.

###### Proof of Proposition 4.12.

We shall proceed by induction on $k\geq 1$. If $k=1$, the result holds
trivially under any of the assumptions. We assume throughout that $k\geq 2$
and hypothesis $(ii)$ holds, in particular $G=\textnormal{E}_{n}(R)$. The
proof can be trivially adapted to hypothesis $(i)$.

Consider $\mathbf{m}\in\operatorname{V}_{n}(M)$ with $n\geq k$. Since
$n>\textnormal{er}_{R}(R/\mathfrak{a}_{k})$, we have

$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}A&\mathbf{b}\\\
\mathbf{0}&1\end{pmatrix}$

with $A$ a $k\times(n-1)$ matrix and $\mathbf{b}$ a column vector of size $k$.
By Lemma 4.1, we obtain

$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}A&\mathbf{0}\\\
\mathbf{0}&1\end{pmatrix}.$

If $n=k$, we use the induction hypothesis for the generating vector
$\mathbf{m}^{\prime}\in\operatorname{V}_{k-1}(M^{\prime})$ satisfying
$A=\operatorname{Mat}(\mathbf{m}^{\prime})$ to infer that
$\mathbf{m}\sim_{\textnormal{E}_{k}(R)}(\delta e_{1},e_{2},\dots,e_{k})$ for
some $\delta\in(R/\mathfrak{a}_{1})^{\times}$. It follows from Lemmas 2.2 and
4.11 that $\delta=\det_{\mathbf{e}}(\mathbf{m})$. If $n>k$, we infer that
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(e_{1},\dots,e_{k-1},0,\dots,0,e_{k})$
and using one further elementary matrix, we obtain
$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(e_{1},e_{2},\dots,e_{k},0,\dots,0)$. ∎

######  Proposition 4.14.

Let $M$ be as in Lemma 4.11. Let $d\geq 0$ and $n>k+d$. Assume that
$\textnormal{er}_{R}(R/\mathfrak{a}_{i})\leq i+d$ for every $i\geq 1$. Then
for every $\mathbf{m}\in\operatorname{V}_{n}(M)$, we have
$\mathbf{m}\sim_{\textnormal{E}_{n}}(e_{1},e_{2},\dots,e_{k},0,\dots,0)$.

###### Proof.

Proceed by induction on $k\geq 1$ as in the proof of Proposition 4.12. ∎

######  Lemma 4.15.

Let $M$ be a finitely generated $R$-module and let
$\mathfrak{a}_{1},\dots,\mathfrak{a}_{k}$ be ideals of $R$ such that
$\mathfrak{a}_{1}+\cdots+\mathfrak{a}_{k}\subseteq\operatorname{ann}(M)$. Let
$\mathbf{m}\in\operatorname{V}_{n}(M\times
R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k})$ with
$n\geq\mu(M\times R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k})$ be
such that

$\operatorname{Mat}(\mathbf{m})=\begin{pmatrix}\mathbf{m}_{0}&\mathbf{m}_{1}\\\
\mathbf{0}&1_{k}\end{pmatrix}$

where $\mathbf{m}_{0}\in M^{n-k}$ and $\mathbf{m}_{1}\in M^{k}$. Then every
component of $\mathbf{m}_{1}$ is an $R$-linear combination of the components
of $\mathbf{m}_{0}$. In particular, we have

$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{n}(R)}\begin{pmatrix}\mathbf{m}_{0}&\mathbf{0}\\\
\mathbf{0}&1_{k}\end{pmatrix}.$

###### Proof.

Let us write $\mathbf{m}_{0}=(m_{1},\dots,m_{n-k})$,
$\mathbf{m}_{1}=(m_{n-k+1},\dots,m_{n})$ and fix a component $m$ of
$\mathbf{m}_{1}$. Since $\mathbf{m}$ generates $M\times
R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}$, there is
$\mathbf{r}=(r_{1},\dots,r_{n})\in R^{n}$ such that
$\operatorname{Mat}(\mathbf{m})\mathbf{r}^{\top}=\begin{pmatrix}m\\\
\mathbf{0}\\\ \end{pmatrix}$. Hence we have $m=\sum_{i=1}^{n}r_{i}m_{i}$ and
$r_{n-k+i}\in\mathfrak{a}_{i}$ for $1\leq i\leq k$. Since
$\mathfrak{a}_{i}\subseteq\operatorname{ann}(M)$ for every $i$ by hypothesis,
we actually have $m=\sum_{i=1}^{n-k}r_{i}m_{i}$. ∎

######  Proposition 4.16.

Let $I$ be a non-principal invertible ideal of $R$.

Let

$M=R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}\times I\times
R^{l}\,(k\geq 1,l\geq 0)$

where the ideals $\mathfrak{a}_{i}\subseteq R$ form an descending chain
$R\supsetneq\mathfrak{a}_{1}\supseteq\cdots\supseteq\mathfrak{a}_{k}$ and
$\mathfrak{a}_{k}\neq 0$.

Assume moreover that at least one of the following holds:

  * $(a)$

Every proper quotient of $R$ has stable rank $1$ and $I$ is
$1\nicefrac{{1}}{{2}}$-generated.

  * $(b)$

$R$ and $I$ are as in Proposition 3.8 and each ideal $\mathfrak{a}_{i}$ is
regular.

Let $\mathbf{e}\Doteq(e_{1},\dots,e_{k+l+1})$,
$(a,b)\in\operatorname{V}_{2}(I)$ with $a\in I\mathfrak{a}_{k}$.

Then the following hold:

  * $(i)$

$\mu(M)=k+l+1$ and $\operatorname{Fitt}_{k+l}(M)=\mathfrak{a}_{1}$.

  * $(ii)$

For every $\mathbf{m}\in\operatorname{V}_{k+l+1}(M)$, we have

$\mathbf{m}\sim_{G}(\textnormal{det}_{\mathbf{e}}(\mathbf{m})e_{1},e_{2},\dots,e_{k},e_{k}+ae_{k+1},be_{k+1},e_{k+2},\dots,e_{k+l+1})$

for $G=\textnormal{SL}_{k+l+1}(R)$. This holds also for
$G=\textnormal{E}_{k+l+1}(R)$ if $k>1$.

  * $(iii)$

For every $\mathbf{m}\in\operatorname{V}_{n}(M)$, we have

$\mathbf{m}\sim_{\textnormal{E}_{n}(R)}(e_{1},e_{2},\dots,e_{k},e_{k}+ae_{k+1},be_{k+1},e_{k+2},\dots,e_{k+l+1},0,\dots,0)$

if $n>k+l+1$.

In particular, $\det_{\mathbf{e}}$ induces a bijection from
$\operatorname{V}_{k+l+1}(M)/\textnormal{SL}_{k+l+1}(R)$ onto
$(R/\mathfrak{a}_{1})^{\times}$.

###### Proof.

$(i)$. Let $\mathfrak{p}$ be a maximal ideal of $R$ containing
$\mathfrak{a}_{1}$. Then $M$ surjects onto $(R/\mathfrak{p})^{k}\times
I/I\mathfrak{p}\times(R/\mathfrak{p})^{l}$. By hypothesis $I/I\mathfrak{p}$ is
cyclic so that $I/I\mathfrak{p}\simeq R/\mathfrak{p}$ by Lemma 3.2. Therefore
$M$ surjects onto $(R/\mathfrak{p})^{k+l+1}$ and hence cannot be generated by
less than $k+l+1$ elements. Since $R/\mathfrak{a}_{k}\times I$ can be
generated by two elements thanks to Proposition 4.8, $M$ can be generated by
$k+l+1$ elements. Thus $\mu(M)=k+l+1$. Let
$\mathbf{m}=(e_{1},e_{2},\dots,e_{k},e_{k}+ae_{k+1},be_{k+1},e_{k+2},\dots,e_{k+l+1})$.
Then $\mathbf{m}$ generates $M$ and it follows from Lemma 2.2 that
$\operatorname{Fitt}_{k+l}(M)=\mathfrak{a}_{1}+\operatorname{Fitt}_{1}(R/\mathfrak{a}_{k}\times
I)$. Since $\operatorname{Fitt}_{1}(R/\mathfrak{a}_{k}\times
I)=\mathfrak{a}_{k}$ by Lemma 4.7, we obtain that
$\operatorname{Fitt}_{k+l}(M)=\mathfrak{a}_{1}$.

$(ii)$. We shall prove by induction on $l\geq 0$ that

$\mathbf{m}\sim_{G}(\delta
e_{1},e_{2},\dots,e_{k},e_{k}+ae_{k+1},be_{k+1},e_{k+2},\dots,e_{k+l+1})$

for some $\delta=\delta(\mathbf{m})\in(R/\mathfrak{a}_{1})^{\times}$. The
identity $\delta=\det_{\mathbf{e}}(\mathbf{m})$ is a straightforward
consequence of assertion $(i)$ and Lemma 2.2.

If $l=0$ and $k=1$, assertion $(ii)$ is given by Proposition 4.8.

If $l=0$ and $k>1$, Proposition 4.9 yields

$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{k+l+1}(R)}\begin{pmatrix}A&\mathbf{b}&\mathbf{c}\\\
\mathbf{0}&1&0\\\ \mathbf{0}&a&b\end{pmatrix}$

where $A$ is a $(k+l-1)\times(n-1)$ matrix, $\mathbf{b}$ and $\mathbf{c}$ are
column vectors of size $k+l-1$. By Lemma 3.2, the map $b\mapsto 1$ induces an
isomorphism from $I/I\mathfrak{a}_{k}$ onto $R/\mathfrak{a}_{k}$. This
isomorphism allows us to identify $M/\mathfrak{a}_{k}M$ with
$R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}\times
R/\mathfrak{a}_{k}$. Under this identification the image
$\overline{\mathbf{m}}$ of $\mathbf{m}$ in $(M/\mathfrak{a}_{k}M)^{k+l+1}$
satisfies
$\operatorname{Mat}(\overline{\mathbf{m}})=\begin{pmatrix}A&\mathbf{b}&\mathbf{c}\\\
\mathbf{0}&1&0\\\ \mathbf{0}&0&1\end{pmatrix}$. By Lemma 4.15, the vectors
$\mathbf{b}$ and $\mathbf{c}$ are in the $R$-linear span of the columns of
$A$. Thus
$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{k+l+1}(R)}\begin{pmatrix}A&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&1&0\\\ \mathbf{0}&a&b\end{pmatrix}$. We complete the proof of the
induction basis by applying Proposition 4.12 to the generating vector
$\mathbf{m}^{\prime}\in\operatorname{V}_{k}(R/\mathfrak{a}_{1}\times\cdots\times
R/\mathfrak{a}_{k})$ defined by $\operatorname{Mat}(\mathbf{m}^{\prime})=A$.

Assume now that $l>0$. We have $\textnormal{er}(R)\leq 2$ by Lemma 3.3 under
assumption $(a)$ and by Proposition 3.8 under assumption $(b)$. Thus

$\operatorname{Mat}(\mathbf{m})\sim_{\textnormal{E}_{k+l+1}(R)}\begin{pmatrix}A&\mathbf{b}\\\
\mathbf{0}&1\end{pmatrix}\sim_{\textnormal{E}_{k+l+1}(R)}\begin{pmatrix}A&\mathbf{0}\\\
\mathbf{0}&1\end{pmatrix}$

where $A$ is a $(k+l)\times(n-1)$ matrix, $\mathbf{b}$ a column vector of size
$k+l$ and the second equivalence is given by Lemma 4.1. Our claim follows now
from the induction hypothesis applied to the generating vector
$\mathbf{m}^{\prime}\in\operatorname{V}_{k+l}(R/\mathfrak{a}_{1}\times\cdots\times
R/\mathfrak{a}_{k}\times I\times R^{l-1})$ defined by
$\operatorname{Mat}(\mathbf{m}^{\prime})=A$.

$(iii)$. We proceed again by induction on $l\geq 0$. Assume that $l=0$. If
moreover $k=1$, then assertion $(iii)$ is given by Proposition 4.9. If $k>1$,
reduce $\operatorname{Mat}(\mathbf{m})$ as in $(ii)$ and use Proposition 4.12
to complete the proof of this induction basis. If $l>0$, reduce
$\operatorname{Mat}(\mathbf{m})$ as in $(ii)$ and apply the induction
hypothesis. ∎

######  Proposition 4.17.

Let $R$ be a Hermite ring. Then the following hold.

(8) $\displaystyle\operatorname{V}_{2}(R)/\textnormal{E}_{2}(R)$
$\displaystyle\simeq$
$\displaystyle\textnormal{E}_{12}(R)\backslash\textnormal{SL}_{2}(R)/\textnormal{E}_{2}(R),$
(9) $\displaystyle\operatorname{V}_{n}(R)/\textnormal{E}_{n}(R)$
$\displaystyle\simeq$
$\displaystyle\textnormal{SL}_{n-1}(R)\textnormal{E}_{n}(R)\backslash\textnormal{SL}_{n}(R)\text{
for }n>2,$ (10)
$\displaystyle\operatorname{V}_{n}(R^{n})/\textnormal{E}_{n}(R)$
$\displaystyle\simeq$ $\displaystyle
R^{\times}\times\left(\textnormal{SL}_{n}(R)/\textnormal{E}_{n}(R)\right)\text{
for }n\geq 2,$ (11)
$\displaystyle\operatorname{V}_{n}(R^{l})/\textnormal{E}_{n}(R)$
$\displaystyle\simeq$
$\displaystyle\textnormal{SL}_{n-l}(R)\textnormal{E}_{n}(R)\backslash\textnormal{SL}_{n}(R)\text{
for }n>l\geq 2.$

###### Proof.

(8) Apply Lemma 4.4 with $\mathfrak{a}=R$. (9) We have
$\operatorname{V}_{n}(R)=e_{1}\textnormal{SL}_{n}(R)$. The stabilizer of
$e_{1}$ for the action of $\textnormal{SL}_{n}(R)$ is
$\textnormal{SL}_{n-1}(R)$. As $\textnormal{E}_{n}(R)$ is normal in
$\textnormal{SL}_{n}(R)$ by Suslin’s Normality Theorem [Mag02, Theorem 10.8],
the result follows. (10) By Proposition 4.12, we have
$\operatorname{V}_{n}(R)=\bigsqcup_{u\in
R^{\times}}(ue_{1},e_{2},\dots,e_{n})\textnormal{SL}_{n}(R)$. As the
stabilizer of $(ue_{1},e_{2},\dots,e_{n})$ for the action of
$\textnormal{SL}_{n}(R)$ is trivial, the result follows. (11) By Proposition
4.12, we have

$\operatorname{V}_{n}(R)=(e_{1},e_{2},\dots,e_{l},0,\dots,0)\textnormal{SL}_{n}(R).$

The stabilizer $S$ of $(e_{1},e_{2},\dots,e_{l},0,\dots,0)$ for the action of
$\textnormal{SL}_{n}(R)$ satisfies $\textnormal{SL}_{n-l}(R)\subset
S\subset\textnormal{E}_{n}(R)\textnormal{SL}_{n-l}(R)$. Since
$\textnormal{E}_{n}(R)$ is normal in $\textnormal{SL}_{n}(R)$, the result
follows. ∎

######  Proposition 4.18.

Let $R$ be a Hermite ring.

  * $(i)$

If $n>l+\textnormal{sr}(R)-1$, then $\textnormal{E}_{n}(R)$ acts transitively
on $\operatorname{V}_{n}(R^{l})$.

  * $(ii)$

If $\textnormal{sr}(R)\leq 2$, then we have

(12) $\displaystyle\operatorname{V}_{n}(R^{n})/\textnormal{E}_{n}(R)$
$\displaystyle\simeq$ $\displaystyle
R^{\times}\times\operatorname{SK}_{1}(R)\text{ for }n>2,$ (13)
$\displaystyle\operatorname{V}_{n+1}(R^{n})/\textnormal{E}_{n+1}(R)$
$\displaystyle\simeq$ $\displaystyle\operatorname{SK}_{1}(R)\text{ for }n\geq
2.$

###### Proof.

$(i)$. This is an immediate consequence of Proposition 4.14. $(ii)$. Combine
the identifications (10) and (11) of Proposition 4.17 with
$\operatorname{SK}_{1}$ Stability Theorem [Mag02, Corollary 11.19.$v$]. ∎

######  Lemma 4.19.

Let $\mathfrak{a}$ be an ideal of $R$ and let $M$ be a finitely generated
$R$-module. Assume that at least one of the following holds.

  * $(i)$

$R$ is Hermite and $M=R/\mathfrak{a}\times R^{l}$ with $l>1$.

  * $(ii)$

$R$ and $I$ are as in Proposition 3.8, $\mathfrak{a}$ is regular and
$M=R/\mathfrak{a}\times I\times R^{l}$ with $l>0$.

Then $\mathfrak{a}=\operatorname{Fitt}_{\mu-1}(M)$ and
$\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ is equipotent with

$(R/\mathfrak{a})^{\times}\times\textnormal{SL}_{\mu}(R)/\textnormal{E}_{\mu}(R)$

where $\mu=\mu(M)$. If in addition $\textnormal{sr}(R)\leq 2$, e.g., if $(ii)$
holds, then $\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ can be further
identified with $(R/\mathfrak{a})^{\times}\times\operatorname{SK}_{1}(R)$.

###### Proof.

The identity $\mathfrak{a}=\operatorname{Fitt}_{\mu-1}(M)$ is given by Lemma
4.11 and Proposition 4.16. If $(i)$, respectively $(ii)$ holds, then
$\operatorname{V}_{\mu}(M)=\bigsqcup_{u\in(R/\mathfrak{a})^{\times}}\mathbf{m}_{u}\textnormal{SL}_{\mu}(R)$
where $\mathbf{m}_{u}=(ue_{1},e_{2},\dots,e_{\mu})$, respectively
$\mathbf{m}_{u}=(ue_{1},e_{2}+ae_{3},be_{3},e_{4},\dots,e_{\mu})$ and $\mu=l$,
respectively $\mu=l+2$, by Propositions 4.12 and 4.16. In each case, it is
easily checked that the stabilizer $S_{u}$ of $\mathbf{m}_{u}$ for the action
of $\textnormal{SL}_{\mu}(R)$ consists in the matrices of the form

$\begin{pmatrix}1&\mathbf{r}\\\ \mathbf{0}&1_{\mu-1}\\\ \end{pmatrix}$

where $\mathbf{r}$ is a row vector of length $\mu-1$ whose components lie in
$\mathfrak{a}$. Therefore $S_{u}\subset\textnormal{E}_{\mu}(R)$. Since
$\textnormal{E}_{\mu}(R)$ is normal in $\textnormal{SL}_{\mu}(R)$ by Suslin’s
Normality Theorem [Mag02, Theorem 10.8], we deduce that
$\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)$ is equipotent with
$(R/\mathfrak{a})^{\times}\times\textnormal{SL}_{\mu}(R)/\textnormal{E}_{\mu}(R)$.
If $\textnormal{sr}(R)\leq 2$, then
$\textnormal{SL}_{\mu}(R)/\textnormal{E}_{\mu}(R)\simeq\operatorname{SK}_{1}(R)$
by the $\operatorname{SK}_{1}$ Stability Theorem [Mag02, Corollary 11.19.$v$].
Hence
$\operatorname{V}_{\mu}(M)/\textnormal{E}_{\mu}(R)\simeq(R/\mathfrak{a})^{\times}\times\operatorname{SK}_{1}(R)$.
∎

##  5\. Direct products of rings and quotient rings

The proofs of Theorems A and E rely on the existence of invariant
decompositions defined in Theorems 2.9 and 2.12. The results in Section 4.2
don’t address the decomposition of a module $M$ as in Theorem 2.12 when non-
trivial idempotents $e_{i}$ split the torsion-free part of $M$. They are
however sufficient to prove Theorems A and E in full generality. Indeed,
thanks to the splittings $R\simeq\prod_{i}Re_{i}$ and
$M\simeq\prod_{i}e_{i}M$, Proposition 5.3 below allows us to handle the
submodules $e_{i}M$ independently. We also prove Proposition C by establishing
Propositions 5.3 and 5.4 below.

Given a ring $R$, let
$\textnormal{E}(R)\Doteq\bigcup_{n}\textnormal{E}_{n}(R)$ and
$\textnormal{SL}(R)\Doteq\bigcup_{n}\textnormal{SL}_{n}(R)$ be the ascending
unions for which the embeddings
$\textnormal{E}_{n}(R)\rightarrow\textnormal{E}_{n+1}(R)$ and
$\textnormal{SL}_{n}(R)\rightarrow\textnormal{SL}_{n+1}(R)$ are defined
through $A\mapsto\begin{pmatrix}A&0\\\ 0&1\end{pmatrix}$. The _special
Whitehead group_ $\operatorname{SK}_{1}(R)$ of $R$ is the quotient group
$\textnormal{SL}(R)/\textnormal{E}(R)$.

######  Lemma 5.1.

Let $\\{R_{x}\\}$ be a family of rings where $x$ ranges in a set $X$ and let
$R=\prod_{x}R_{x}$. Then the following hold.

  * $(i)$

The natural maps
$\textnormal{SL}_{n}(R)\rightarrow\prod_{x}\textnormal{SL}_{n}(R_{x})$ and
$\textnormal{GL}_{n}(R)\rightarrow\prod_{x}\textnormal{GL}_{n}(R_{x})$ are
isomorphisms for every $n\geq 1$.

  * $(ii)$

If $X$ is finite, then the natural map
$\textnormal{E}_{n}(R)\rightarrow\prod_{x}\textnormal{E}_{n}(R_{x})$ is an
isomorphism for every $n\geq 1$.

  * $(iii)$

Let $n\geq 2$. If every matrix in $\textnormal{E}_{n}(R_{x})$ is the product
of at most $\nu_{n}$ elementary matrices where $\nu_{n}<\infty$ is independent
on $x$, then the natural map
$\textnormal{E}_{n}(R)\rightarrow\prod_{x}\textnormal{E}_{n}(R_{x})$ is an
isomorphism.

  * $(iv)$

If $X$ is finite, or if $\textnormal{sr}(R_{x})$ is bounded and the natural
map $\textnormal{E}_{n}(R)\rightarrow\prod_{x}\textnormal{E}_{n}(R_{x})$ is an
isomorphism, then the natural map

$\operatorname{SK}_{1}(R)\rightarrow\prod_{x}\operatorname{SK}_{1}(R_{x})$

is an isomorphism.

###### Proof.

We first observe that the natural maps considered in the assertions $(i)$ to
$(iii)$ are clearly injective. Assertion $(i)$ is obvious. For $(ii)$, see
[Coh66, Proof of Theorem 3.1] or reason as in $(iii)$.

$(iii)$. Let us prove that the natural map
$\varphi:\textnormal{E}_{n}(R)\rightarrow\prod_{x}\textnormal{E}_{n}(R_{x})$
is surjective. Let $(E_{x})\in\prod_{x}\textnormal{E}_{n}(R_{x})$. Write
$E_{x}=E_{x,1}\cdots E_{x,\nu_{n}}$ with
$E_{x,k}=E_{i(x)j(x)}(r_{x,k})\,(r_{x,k}\in R_{x})$ where $E_{ij}(r)$ denotes
the elementary matrix whose $(i,j)$-coefficient is $r$ and whose other off-
diagonal coefficients are zero. Then we have
$E_{x}=\prod_{k=1}^{\nu_{n}}\prod_{1\leq i\neq j\leq n}E_{ij}(r_{x,k}(i,j))$
where the factors indexed by the pairs $(i,j)$ are ordered lexicographically
and

$r_{x,k}(i,j)=\left\\{\begin{array}[]{cc}r_{x,k}&\text{ if
}(i,j)=(i(x),j(x)),\\\ 0&\text{ else.}\end{array}\right.$

Set $E\Doteq\prod_{k=1}^{\nu_{n}}\prod_{1\leq i\neq j\leq n}E_{ij}(r_{kij})$,
with $r_{kij}=(r_{k,x}(i,j))\in R$. Then $\varphi(E)=(\gamma_{x})$, which
proves that $\varphi$ is an isomorphism.

$(iv)$. It is trivial to check that the natural map
$\psi:\operatorname{SK}_{1}(R)\rightarrow\prod_{x}\operatorname{SK}_{1}(R_{x})$
is always surjective. In order to show that $\psi$ is injective, consider
$\sigma\textnormal{E}(R)\in\ker(\psi)$ with
$\sigma=(\sigma_{x})\in\textnormal{SL}_{n}(R)$ for some $n\geq 1$. Then
$\sigma_{x}\in\textnormal{E}(R_{x})$ for every $x\in X$. If $X$ is finite,
then we can find $N\geq 1$, such $\sigma_{x}\in\textnormal{E}_{N}(R_{x})$ for
every $x$. Therefore $(\sigma_{x})\in\prod_{x}\textnormal{E}_{N}(R_{x})$ and
the result follows from $(iii)$. If $\textnormal{sr}(R_{x})<N$ for some $N\geq
2$, then $\sigma_{x}\in\textnormal{E}_{N}(R_{x})$ for every $x$ by the
$\operatorname{SK}_{1}$ Stability Theorem [Mag02, Corollary 11.19.$v$]. Thus
$(\sigma_{x})\in\textnormal{E}(R)$, provided that
$\textnormal{E}_{N}(R)\rightarrow\prod_{x}\textnormal{E}_{N}(R_{x})$ is an
isomorphism. ∎

######  Lemma 5.2.

Let $\\{R_{x}\\}$ be a family of rings where $x$ ranges in a set $X$. Let $M$
be module over $R\Doteq\prod_{x}R_{x}$. Denote by $e_{x}$ the element of $R$
whose component in $R_{x}$ is $1$ and whose other components are zero. Let
$\varphi:M\rightarrow\prod_{x}e_{x}M$ be the $R$-homomorphism defined by
$\varphi(m)=(e_{x}m)$. Then the following hold.

  * $(i)$

If $X$ is finite, then $\varphi$ is an isomorphism.

  * $(ii)$

If $M$ is finitely generated, then $\varphi$ is surjective.

  * $(iii)$

If $M$ is finitely presented, then $\varphi$ is an isomorphism.

###### Proof.

$(i)$. The identity of $R$ decomposes as $1=\sum_{x}e_{x}$. Thus, for every
$m\in M$, we have $m=1\cdot m=\sum_{x}e_{x}m$, which shows that $\varphi$ is
injective. If $(m_{x})\in\prod_{x}e_{x}M$, we can set
$m\Doteq\sum_{x}e_{x}m_{x}$ so that $\varphi(m)=(m_{x})$. Hence $\varphi$ is
surjective.

$(ii)$. Let $(m_{1},\dots,m_{n})$ be a generating vector of $M$. Let
$(m_{x})\in\prod_{x}e_{x}M$. Since $(e_{x}m_{1},\dots,e_{x}m_{n})$ generates
$e_{x}M$, we can find $\lambda_{i,x}\in R_{x}\,(1\leq i\leq n)$ such that
$m_{x}=\sum_{i}\lambda_{i,x}e_{x}m_{i}$. Setting
$\lambda_{i}\Doteq(\lambda_{i,x})_{x}\in R$, we observe that
$(m_{x})=\varphi(\sum_{i}\lambda_{i}m_{i})$. Therefore $\varphi$ is
surjective.

$(iii)$. By hypothesis $M$ is the cokernel of an $n\times m$ matrix $A$
associated to a generating vector $(m_{1},\dots,m_{n})\in M^{n}$. Let
$\mathbf{c}_{1},\dots,\mathbf{c}_{m}\in R^{n}$ be the columns of $A$. Given
$\bm{\lambda}=(\lambda_{1},\dots,\lambda_{n})\in R^{n}$, the element
$\sum_{i}\lambda_{i}m_{i}$ is zero if and only if $\bm{\lambda}$ lies in the
$R$-linear span of the columns $\mathbf{c}_{j}$. Let
$m=\sum_{i}\lambda_{i}m_{i}\in M$ be such that $\varphi(m)=0$. Since for every
$x\in X$ we have $\sum_{i}e_{x}\lambda_{i}m_{i}=0$, it follows that
$e_{x}\bm{\lambda}=\sum_{j}\mu_{j,x}\mathbf{c}_{j}$ for some $\mu_{j,x}\in R$
with $1\leq j\leq m$. Setting $\mu_{j}=(\mu_{j,x})_{x}$, we observe that
$\bm{\lambda}=\sum_{j}\mu_{j}\mathbf{c}_{j}$. Therefore $m=0$, which shows
that $\varphi$ is injective. As $\varphi$ is also surjective by $(ii)$, it is
an isomorphism. ∎

######  Proposition 5.3.

Let $\\{R_{x}\\}$ be a family of rings where $x$ ranges in a set $X$. Let $M$
be a finitely presented module over $R\Doteq\prod_{x}R_{x}$.

  * $(i)$

If the $R_{x}$-module $e_{x}M$ has property $\Delta_{\textnormal{SL}}$ for
every $x\in X$, then $M$ has property $\Delta_{\textnormal{SL}}$.

  * $(ii)$

Assume that the natural map
$\textnormal{E}_{n}(R)\rightarrow\prod_{x}\textnormal{E}_{n}(R_{x})$ is an
isomorphism for every $n\geq\mu(M)$. If in addition the $R_{x}$-module
$e_{x}M$ has property $\Delta_{\textnormal{E}}$ for every $x\in X$, then $M$
has property $\Delta_{\textnormal{E}}$.

###### Proof.

We shall only consider the property $\Delta_{\textnormal{SL}}$. The
corresponding result for $\Delta_{\textnormal{E}}$ can be derived in the same
way. Assume that $(i)$ holds. Let $n\geq\mu(M)$. Using the natural isomorphism
$\textnormal{SL}_{\mu}(R)\simeq\prod_{x}\textnormal{SL}_{\mu}(R_{x})$ and
Lemma 5.2.$iii$, we obtain a bijection
$\operatorname{V}_{n}(M)/\textnormal{SL}_{n}(R)\simeq\prod_{x}\operatorname{V}_{n}(e_{x}M)/\textnormal{SL}_{n}(R_{x})$.
By hypothesis the determinant map induces a bijection
$\operatorname{V}_{n}(e_{x}M)/\textnormal{SL}_{n}(R_{x})\simeq\operatorname{V}_{1}(\bigwedge^{n}e_{x}M)$
for every $x$. It is easily checked that
$\prod_{x}\operatorname{V}_{1}(\bigwedge^{n}e_{x}M)\simeq\operatorname{V}_{1}(\bigwedge^{n}M)$
by means of the natural $R$-isomorphism
$\prod_{x}\bigwedge^{n}e_{x}M\simeq\bigwedge^{n}M$. The result follows. ∎

######  Proposition 5.4.

Let $R$ be a ring and let $\overline{R}$ be a quotient of $R$.

  * $(i)$

If $R$ satisfies $\Delta_{\textnormal{E}}$, then $\overline{R}$ satisfies
$\Delta_{\textnormal{E}}$.

  * $(ii)$

If $R$ satisfies $\Delta_{\textnormal{SL}}$ and if the natural map
$\textnormal{SL}_{n}(R)\rightarrow\textnormal{SL}_{n}(\overline{R})$ is
surjective for every $n$, then $\overline{R}$ satisfies
$\Delta_{\textnormal{SL}}$.

######  Remark 5.5.

The natural map
$\textnormal{SL}_{n}(R)\rightarrow\textnormal{SL}_{n}(\overline{R})$ is
surjective for every $n$ and every quotient $\overline{R}$ of $R$ if
$\textnormal{sr}(R)\leq 2$ [EO67, Corollary 8.3]. See also the partial
stability result [EO67, Theorem 8.3] when $R$ is a ring of univariate
polynomials over a principal ideal domain.

###### Proof.

$(i)$. The quotient map $R\twoheadrightarrow\overline{R}$ endows
$\overline{R}$ with an $R$-algebra structure. Let $\overline{M}$ be an
$\overline{R}$-module and let $M$ be the $R$-module with underlying set
$\overline{M}$ that is obtained from $\overline{M}$ by restricting scalars. We
have the following commutative diagram

${\operatorname{V}_{n}(M)/\textnormal{E}_{n}(R)}$${\operatorname{V}_{1}(\bigwedge^{n}M)}$${\operatorname{V}_{n}(\overline{M})/\textnormal{E}_{n}(\overline{R})}$${\operatorname{V}_{1}(\bigwedge^{n}\overline{M})}$$\scriptstyle{\det}$$\scriptstyle{\det}$

where the left vertical arrow is given by the surjectivity of the natural map
$\textnormal{E}_{n}(R)\rightarrow\textnormal{E}_{n}(\overline{R})$ and the
right vertical arrow is given by the base change isomorphism [Eis95,
Proposition A2.2.b]. The bottom horizontal arrow is a bijection as the other
arrows are bijections. $(ii)$. Proceed as in $(i)$. ∎

##  6\. Proofs

#### Proof of Theorem A

If $R$ is an elementary divisor ring, then $R$ is K-Hermite and $M$ has an
invariant decomposition in the sense of Theorem 2.9. The result follows by
combining Propositions 4.10 and 4.12.

Let us suppose that $R$ is an almost-LG coherent Prüfer ring. Let
$(e_{i})_{i}$ be the idempotents of $R$ appearing in the invariant
decomposition of $M$ in Theorem 2.12. Then each $Re_{i}$ is an almost-LG
coherent Prüfer ring and $M$ is isomorphic to $\prod_{i}e_{i}M$ by Lemma
5.2.$i$. By Proposition 5.3.$i$, we can assume, without loss of generality,
that $M$ is of the form

$R/I_{1}\times\cdots\times R/I_{k}\times I\times R^{l}\,(k\geq 0,l\geq 0)$

where $R\supsetneq I_{1}\supseteq\cdots\supseteq I_{k}$ is a descending chain
of invertible ideals and $I$ is either $\\{0\\}$ or an invertible ideal. As
$R$ is a coherent Prüfer ring, it is certainly a PP-ring. If $I$ is
invertible, then $R$ and $I$ satisfy the hypotheses of Proposition 3.8 by
[Cou07, Lemma 10]. In particular $\textnormal{er}(R)\leq 2$ and $R$ is
Hermite. The result follows by combining Propositions 4.10, 4.12 and 4.16.
$\square$

#### Proof of Theorem E

Let us assume first that $R$ is a Bézout ring whose proper quotients have
stable rank $1$. By [McG08, Theorem 3.7], the ring $R$ is an elementary
divisor ring of stable rank at most $2$.

$(i)$. If $\mu=1$, the result is trivial. Let us suppose that
$\mu^{\prime}>1$. By Theorem 2.9, the module $M$ is of the form
$R/\mathfrak{a}_{1}\times\cdots\times R/\mathfrak{a}_{k}\,(k>1)$ where
$\mathfrak{a}_{1}\supseteq\cdots\supseteq\mathfrak{a}_{k}$ is a descending
chain of ideals such that $R\supsetneq\mathfrak{a}_{1}$ and
$\mathfrak{a}_{2}\neq\\{0\\}$. Then the result follows from Proposition 4.12
and Remark 4.13.

$(ii)$. As $M$ is of the form $R/\mathfrak{a}_{1}\times R^{l}\,(l>1)$ with
$R\supsetneq\mathfrak{a}_{1}\neq\\{0\\}$, the result follows from Lemma 4.19.

$(iii)$. As $\textnormal{sr}(R)\leq 2$, Proposition 4.14 applies with $d=0$ if
$\mu^{\prime}>0$. Otherwise, there is nothing to show since $R$ is a Bézout
ring.

$(iv)$. Proposition 4.14 applies with $d=1$.

Let us assume now that $R$ is an almost-LG coherent Prüfer ring.

$(i)$. If $\mu=1$, the result is trivial. Let us suppose that
$\mu^{\prime}>1$. Because of Theorem 2.12, Proposition 5.3.$ii$ and Lemma
5.1.$ii$, we can assume, without loss of generality, that $M$ is of the form
$R/I_{1}\times\cdots\times R/I_{k}\times I\times R^{l}\,(k>1,l\geq 0)$ where
$R\supsetneq I_{1}\supseteq\cdots\supseteq I_{k}\neq\\{0\\}$ is a descending
chain of invertible ideals and $I$ is either $\\{0\\}$ or an invertible ideal.
Then the result follows from Proposition 4.12, Remark 4.13 and Proposition
4.16.$ii$

$(ii)$. We can assume, without loss of generality, that $M$ is of the form
$R/I_{1}\times I\times R^{l}\,(l>0)$ where $I_{1}$ is an invertible ideal, $I$
is either invertible or $\\{0\\}$ and $l>1$ if $I=\\{0\\}$. Then the result
follows from Lemma 4.19.

$(iii)$. Let us assume first that $\mu^{\prime}>0$. We can assume without loss
of generality that $M$ is of the form $R/I_{1}\times\cdots\times R/I_{k}\times
I\times R^{l}\,(k>0,l\geq 0)$ where $R\supsetneq I_{1}\supseteq\cdots\supseteq
I_{k}\neq\\{0\\}$ is a descending chain of invertible ideals and $I$ is either
$\\{0\\}$ or an invertible ideal. If $\mu=1$, then $M=R/I_{1}$. As
$\textnormal{sr}(R/I_{1})=1$, Proposition 2.7.$iii$ yields the result. If
$\mu>1$ then the result follows from Proposition 4.9, Proposition 4.12 and
Proposition 4.16.$iii$.

Suppose now that $\mu^{\prime}=0$ and that $eM$ surjects onto an invertible
non-principal ideal of $Re$ where $e$ is an idempotent of $R$ satisfying
$\mu(eM)=\mu(M)$. We can assume, without loss of generality, that $e$ is the
last idempotent in the decomposition of Theorem 2.12. We can also assume that
$M$ is of the form $I\times R^{l}\,(l\geq 0)$ with $I$ invertible, in which
case the result is given by Proposition 3.8 and Proposition 4.10.$iv$. Indeed,
if $e^{\prime}$ is an idempotent element of $R$ such that
$\mu(e^{\prime}M)<\mu(M)$, then $\textnormal{E}_{n}(R)$ acts transitively on
$\operatorname{V}_{n}(e^{\prime}M)$ for every $n>\mu(M)$ by Proposition 3.8,
Proposition 4.10.$iv$ and Proposition 4.18.$i$. $(iv)$. By $(iii)$, we can
assume that $M$ is free, so that the result follows from Proposition 4.18.$i$.
$\square$

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