Ring Elements of Stable Range One

Dinesh Khurana and T. Y. Lam

###### Abstract

A ring element $\,a\in R\,$ is said to be of right stable range one if, for
any $\,t\in R$, $\,aR+tR=R\,$ implies that $\,a+t\,b\,$ is a unit in $\,R\,$
for some $\,b\in R$. Similarly, $\,a\in R\,$ is said to be of left stable
range one if $\,R\,a+R\,t=R\,$ implies that $\,a+b^{\prime}t\,$ is a unit in
$\,R\,$ for some $\,b^{\prime}\in R$. In the last two decades, it has often
been speculated that these two notions are actually the same for any $\,a\in
R$. In §3 of this paper, we will prove that this is indeed the case. The key
to the proof of this new symmetry result is a certain “Super Jacobson’s
Lemma”, which generalizes Jacobson’s classical lemma stating that, for any
$\,a,b\in R$, $\,1-ab\,$ is a unit in $\,R\,$ iff so is $\,1-ba$. Our proof
for the symmetry result above has led to a new generalization of a classical
determinantal identity of Sylvester, which will be published separately in
[KL3]. In §§4-5, a detailed study is offered for stable range one ring
elements that are unit-regular or nilpotent, while §6 examines the behavior of
stable range one elements via their classical Peirce decompositions. The paper
ends with a more concrete §7 on integral matrices of stable range one,
followed by a final §8 with a few open questions.

§1. Introduction

The concept and the basic theory of the stable range of unital rings were
first introduced by H. Bass [Ba] in 1964 in his original study of the
stability properties of the general linear groups in algebraic K-theory. Over
the years, many papers have been written on the stable range of rings, while
the notion of rings of stable range one has often stood out as a particularly
important special case. In 2005, inspired by Bass’s ground-breaking paper, the
authors introduced in [KL2: Def. (3.1)] an element-wise notion of stable range
one, as follows.

Definition 1.1. We say that a ring element $\,a\in R\,$ has right stable range
one (written $\,{\rm sr}_{R}(a)=1$, or just $\,{\rm sr}\,(a)=1\,$ if $\,R\,$
is understood) if, for any $\,t\in R$, $\,aR+tR=R\,$ implies that
$\,a+t\,b\in{\rm U}(R)\,$ (the group of units in $\,R\,$) for some $\,b\in R$.
Similarly, we say that $\,a\in R\,$ has left stable range one (written $\,{\rm
sr}\,^{\prime}(a)={\rm sr}\,^{\prime}_{R}(a)=1$) if, for any $\,t\in R$,
$\,R\,a+R\,t=R\,$ implies that $\,a+b^{\prime}\,t\in{\rm U}(R)\,$ for some
$\,b^{\prime}\in R$. Finally, the ring $\,R\,$ itself is said to be of stable
range one if $\,{\rm sr}\,(a)=1\,$ for all $\,a\in R$.

From the abstract definitions above, it is of course not clear at all whether,
for any element $\,a\in R$, $\,{\rm sr}\,(a)=1\,$ is equivalent to $\,{\rm
sr}\,^{\prime}(a)=1\,$ (over any ring $\,R$). Quite pleasantly, this long-
suspected statement did turn out to be true: we will give a nontrivial proof
for it in §3 after covering some general preparatory material in §2. After
this, of course, the distinction between the two conditions $\,{\rm
sr}\,(a)=1\,$ and $\,{\rm sr}\,^{\prime}(a)=1\,$ will become unnecessary. On
the other hand, if $\,a\in S\,$ where $\,S\,$ is a subring of $\,R$, there
does not seem to be much correlation between the condition $\,{\rm
sr}_{S}(a)=1\,$ and the condition $\,{\rm sr}_{R}(a)=1$. Nevertheless, if
$\,f:R\to S\,$ is a surjective ring homomorphism, then $\,{\rm sr}_{R}(a)=1\,$
does imply that $\,{\rm sr}_{S}(f(a))=1$; see Theorem 2.4(D) below.

This paper is intended to be an introduction to the notion of “element-wise
stable range one” essentially from first principles. In particular, a couple
of results in the literature that are more or less known to experts (e.g. the
Product Theorem 2.8) are re-proved in §2 and §5 for the reader’s convenience.
For a sampling of old and new papers in the literature written on the theme of
stable range one for rings and for ring elements, we can cite for instance
[EO], [Ka], [Ev], [Va1, Va2], [Wa], [GM], [CY1, CY2], [Yu1, Yu2], [Can], [Ar],
[Ch3, Ch4], [We], [CN], [Khu], [KhM], [AO1, AO2], [HN], [CLM], [DGK], and
[CP1, CP2]. As we have mentioned in the previous paragraph, the first main
result of our paper is the following.

Theorem 1.2. For any ring element $\,a\in R$, $\,{\rm sr}\,(a)=1\,$ iff
$\,{\rm sr}\,^{\prime}(a)=1$.

The key tool to be used for proving this important symmetry result is Super
Jacobson’s Lemma 3.2, which in its most primitive form states that, for any
three elements $\,a,b,x\,$ in a ring $\,R\,$ with unit group $\,{\rm U}(R)\,$:

$a+b-axb\in{\rm U}(R)\;\,\Longleftrightarrow\;\,a+b-bxa\in{\rm U}(R).$ $None$

In fact, in the full version of Super Jacobson’s Lemma 3.2, the equivalence in
(1.3) above is shown to hold if $\,{\rm U}(R)\,$ is replaced by the sets of
regular or unit-regular elements in any ring $\,R$. (Later in Theorem 6.10,
Super Jacobson’s Lemma will also be proved if $\,{\rm U}(R)\,$ in (1.3) is
replaced by the set of stable range one elements in $\,R$.) The detailed
proofs of these results will be given in §3, after some basic facts on
“element-wise stable range one” (such as multiplicative closure) are recalled
or proved in §2.

In §4, we study in detail the relationship between stable range one elements
and unit-regular elements in any ring $\,R$. While it is essentially well
known (say from [Go: Proposition 4.12] or [KL2: Theorem 3.5]) that unit-
regular elements in $\,R\,$ are exactly the (von Neumann) regular elements in
$\,R\,$ with stable range one, a few more characterizations of unit-regular
elements in terms of the stable range one notion are given in Theorem 4.3. As
an application, it is shown for instance in Corollary 4.7 that a ring $\,R\,$
is an IC ring (“ring with internal cancellation”) in the sense of [KL2] iff
every regular element in $\,R\,$ has stable range one, or equivalently, is a
product of unit-regular elements.

Instead of working with regular and unit-regular elements, §5 is devoted to
the study of nilpotent elements in a ring $\,R$. If $\,a\in R\,$ is a strongly
nilpotent element (in the sense of Levitzki [Lv]), or a central quasi-
nilpotent element (in the sense of Harte [Ha]), it is shown in Theorem 5.1
that we must have $\,{\rm sr}\,(a)=1$. In the case of an exchange ring $\,R\,$
(in the sense of Warfield), we show in Theorem 5.2 that every regular
nilpotent element $\,a\in R\,$ is unit-regular, with the property that $\,{\rm
sr}\,(a^{i})=1\,$ for all $\,i\geq 1$. For any pair of orthogonal idempotents
$\,e,f\,$ in any ring $\,R$, Theorem 5.5 shows that $\,{\rm sr}\,(erf)=1\,$
for all $\,r\in R$, and also that $\,{\rm sr}\,(erfse)=1\,$ for all $\,r,s\in
R$. For some concrete examples, it is shown as a part of Theorem 5.8 that, if
$\,R\,$ is a matrix ring $\,{\mathbb{M}}_{n}(S)\,$ where $\,n\geq 2$, then
$\,{\rm sr}\,\bigl{(}aE_{kj}\bigr{)}=1\,$ for all $\,a\in S\,$ and all
$\,k,\,j\,$, where $\,\\{E_{kj}\\}\,$ are the matrix units in $\,R$.

As is to be expected, the element-wise notion of stable range one behaves well
with respect to the passage from a ring $\,R\,$ to its Peirce corner rings
$\,eRe\,$ for any idempotent $\,e\in R$. In §6, we prove a Suspension Theorem
6.2 to the effect that, if $\,f\\!:=1-e\,$ and $\,a\in eR\,e$, $\,p\in fR\,e$,
then $\,{\rm sr}_{eR\,e}(a)=1\,$ iff $\,{\rm sr}_{R}(a+p+f)=1$, in which case
we have automatically $\,{\rm sr}_{R}(a)=1$. In the case where $\,R\,$ is a
matrix ring $\,{\mathbb{M}}_{n}(S)\,$ and $\,A=(a_{ij})\in R\,$ is a lower
triangular matrix with $\,{\rm sr}_{S}(a_{ii})=1\,$ for all $\,i\,$, Theorem
6.8 shows that $\,{\rm sr}_{R}(A)=1$, while Theorem 7.1 shows that if $\,A\,$
is a diagonal matrix with at least one diagonal entry $\,a_{ii}=0$, then again
$\,{\rm sr}_{R}(A)=1$. Finally, if $\,S\,$ is assumed to be a commutative
elementary divisor domain, Theorem 7.2 shows that $\,{\rm sr}_{R}(B)=1\,$ for
any matrix $\,B\in{\mathbb{M}}_{n}(S)\,$ such that $\,{\rm det}\,(B)\,$ is
either zero or a unit in $\,S$. After §7, the paper concludes with a final
section §8 which lists several open problems on stable range one elements in
rings.

The terminology and notations introduced so far in this Introduction will be
used freely throughout the paper. In addition, for any ring $\,R$, we shall
use the standard notations $\,{\rm idem}\,(R)$, $\,{\rm nil}\,(R)$, $\,{\rm
reg}\,(R)$, $\,{\rm ureg}\,(R)\,$ and $\,{\rm sreg}\,(R)\,$ respectively for
the sets of idempotents, nilpotent elements, (von Neumann) regular elements,
unit-regular elements, and strongly regular elements in any ring $\,R$. Also,
as we have mentioned before, we will use $\,E_{ij}\,$ to denote the matrix
units in any matrix ring $\,{\mathbb{M}}_{n}(S)\,$ over a given ring $\,S$.
Other standard terminology and conventions in ring theory follow mainly those
in [Go], [La1] and [La2].

§2. Basic Facts On Element-Wise Stable Range One

Throughout this section (and all subsequent sections), the term “stable range
one” shall mean “right stable range one”, and the notation $\,{\rm
sr}\,(a)={\rm sr}_{R}(a)=1\,$ will be taken to mean that the element $\,a\,$
has right stable range one in $\,R$. We begin by mentioning some of the most
basic examples of such elements.

Example 2.1. To begin with, if $\,a\in{\rm U}(R)$, clearly $\,{\rm
sr}\,(a)=1\,$ since we can pick $\,b=0\,$ in Definition 1.1. (A more general
statement on 1-sided invertible elements will be given in Theorem 2.4(C)
below.) Next, for any $\,e\in{\rm idem}\,(R)\,$ with complementary idempotent
$\,f\\!:=1-e$, we have $\,{\rm sr}\,(e)=1$. Indeed, if $\,e\,x+ty=1\,$ for
some elements $\,t,x,y\in R$, then $\,exf+tyf=f$. Adding $\,e\,$ gives
$\,e+tyf=1-exf\in{\rm U}(R)\,$ (with inverse $\,1+exf$), so we have checked
that $\,{\rm sr}\,(e)=1$.

The following result offers a number of alternative characterizations for
elements of stable range one in any ring $\,R$.

Theorem 2.2. For any element $\,a\in R$, the following statements are
equivalent:

(1) $\,{\rm sr}\,(a)=1$.
(2) $\,aR+K=R\,$ for a right ideal $\,K\subseteq R\,\Rightarrow\,a+k\in{\rm
U}(R)\,$ for some $\,k\in K$.
(3) For any $\,x\in R$, there exists $\,b\in R\,$ such that $\,a+b-axb\in{\rm
U}(R)$.
(4) For any $\,x,c\in R$, $\,ax+c\in{\rm U}(R)\,\Rightarrow\,a+c\,b\in{\rm
U}(R)\,$ for some $\,b\in R$.
(5) For any $\,x,c\in R$, $\,ax+c=1\,\Rightarrow\,a+c\,b\in{\rm U}(R)\,$ for
some $\,b\in R$.

Proof. $(1)\Rightarrow(2)\Rightarrow(3)\,$ are trivial implications.

$(3)\Rightarrow(4)$. Assume (3), and consider any $\,x,c\in R\,$ such that
$\,u\\!:=ax+c\in{\rm U}(R)$. Then $\,axu^{-1}+cu^{-1}=1$. Applying (3), we
have $\,a+(1-axu^{-1})\,b_{0}\in{\rm U}(R)\,$ for some $\,b_{0}\in R$. Thus,
$\,a+c\,(u^{-1}b_{0})\in{\rm U}(R)$, which proves (4).

$(4)\Rightarrow(5)\,$ is a tautology.

$(5)\Rightarrow(1)$. Assume (5), and consider any $\,t\in R\,$ such that
$\,aR+tR=R$. Fixing an equation $\,ax+ty=1\,$ and applying (5) with $\,c=ty$,
we see that $\,a+(ty)\,b\in{\rm U}(R)\,$ for some $\,b\in R$. According to
Definition 1.1, this checks that $\,{\rm sr}\,(a)=1$. ∎

In the special case where the right ideal $\,aR\,$ is subject to some
“suitability condition” in the sense of Nicholson ([Ni1], [Ni2]), Theorem 2.2
can be transformed in some way into the corollary below, which is directly
inspired by the work of Camillo-Yu (in [CY1: Lemma 2] and [Yu1: Theorem 9]) on
the criterion for an exchange ring to have stable range one.

Corollary 2.3. Let $\,a\in R\,$ be such that, whenever $\,aR+tR=R\,$ for some
$\,t\in R$, there exists an idempotent $\,f\in aR\,$ such that $\,1-f\in tR$.
(This would be the case, for instance, if all elements in $\,aR\,$ are
suitable in $\,R\,$; see [DKN: Example 3.2].) Then the following three
statements are equivalent :

(1) $\,{\rm sr}\,(a)=1$.
(6) For any $\,x\in R\,$ such that $\,ax=(ax)^{2}$, there exists $\,b\in R$
such that $\,a+(1-ax)\,b\in{\rm U}(R)$.
(7) For any $\,x\in R\,$ and $\,e\in{\rm idem}\,(R)$,
$\,ax+e=1\,\Rightarrow\,a+e\,b\in{\rm U}(R)\,$ for some $\,b\in R$.

Proof. Referring back to condition (5) in Theorem 2.2, it suffices to check
that $(5)\Rightarrow(7)\Rightarrow(6)\Rightarrow(1)$. Here, the first two
implications are obvious. To prove the last implication, assume that $(6)$
holds. To check (1), we consider any equation $\,aR+tR=R$. By our hypothesis
on $\,a$, there exists an idempotent $\,f=ax\,$ for some $\,x\in R\,$ such
that $\,e\\!:=1-f=ty\,$ for some $\,y\in R$. Since $\,ax+e=f+e=1$, $(6)$
guarantees the existence of some $\,b\in R\,$ such that
$\,a+e\,b=a+t\,(y\,b)\in{\rm U}(R)$. This checks that $\,{\rm sr}\,(a)=1$.∎

Coming back to Theorem 2.2 itself, we collect in the next result a number of
other basic facts on the notion of “element-wise right stable range one”.

Theorem 2.4. (A) For any ring element $\,a\in R$, $\,{\rm sr}\,(a)=1\,$ iff,
for any $\,t\in R$, $\,aR+tR=R\,$ implies that $\,{\rm sr}\,(a+ty)=1\,$ for
some $\,y\in R$.
(B) For any $\,a\in R\,$ and $\,u,v\in{\rm U}(R)$, $\,{\rm
sr}\,(a)=1\,\Rightarrow\,{\rm sr}\,(uav)=1$.
(C) If $\,a\,$ has a one-sided inverse in $\,R$, then $\,{\rm sr}\,(a)=1\,$
iff $\,a\in{\rm U}(R)$.
(D) For any ideal $\,J\subseteq R$, let $\,\overline{R}=R/J$. For any $\,a\in
R$, $\,{\rm sr}\,(a)=1\,\Rightarrow\,{\rm
sr}\,\bigl{(}\overline{a}\bigr{)}=1$. The converse holds if the quotient map
$\,f:R\rightarrow R/J\,$ “reflects units”, in the sense that $\,f(a)\in{\rm
U}(R/J)\Rightarrow a\in{\rm U}(R)$. In particular, if $\,J\subseteq{\rm
rad}\,(R)\,$ (the Jacobson radical of $\,R\,$), then $\,{\rm sr}_{R}(a)=1\,$
iff $\,{\rm sr}_{R/J}\bigl{(}\overline{a}\bigr{)}=1$.

Proof. (A) The “only if” part is trivial as we can choose $\,y=0$. To prove
the “if” part, consider any equation $\,aR+tR=R$. By assumption, there exists
$\,y\in R\,$ such that $\,a^{\prime}=a+ty\,$ has $\,{\rm sr}\,(a^{\prime})=1$.
Then $\,a^{\prime}R+tR\supseteq aR+tR=R\,$ implies that there exists
$\,y_{0}\in R\,$ such that $\,a^{\prime}+ty_{0}\in{\rm U}(R)$. As
$\,a^{\prime}+ty_{0}=a+t\,(y+y_{0})$, this checks that $\,{\rm sr}\,(a)=1$.

(B) After replacing $\,uav\,$ by its conjugate $\,u^{-1}(uav)\,u=avu$, we need
only show that $\,{\rm sr}\,(av)=1$. If $\,avR+tR=R$, we have $\,aR+tR=R$, so
$\,a+tb\in{\rm U}(R)\,$ for some $\,b\in R$. This implies that
$\,av+t\,(bv)\in{\rm U}(R)$, which checks the claim that $\,{\rm sr}\,(av)=1$.

(C) The result in (C) here should be regarded as an “element-wise version” of
the well known fact that rings of stable range one (see (2.5A) below) are
always Dedekind-finite. In view of what we said at the beginning of Example
2.1, we need only prove the “only if” part in (C), so assume that $\,{\rm
sr}\,(a)=1$. If $\,ax=1\,$ for some $\,x\in R$, applying (2.2)(5) for
$\,c=0\,$ shows directly that $\,a\in{\rm U}(R)$. If $\,ya=1\,$ for some
$\,y\in R\,$ instead, applying (2.2)(4) yields some $\,b\in R\,$ such that
$\,u\\!:=a+(1-ay)\,b\in{\rm U}(R)$. Left multiplying this equation by $\,y\,$
shows that $\,yu=ya+(y-yay)\,b=1$. Thus, $\,y\in{\rm U}(R)$, and hence
$\,a\in{\rm U}(R)$.

(D) The two main conclusions in (D) follow from a routine application of the
criterion for stable range one elements given in Theorem 2.2(3). Given these
two conclusions, the last part of (D) follows from the well known fact that
the quotient map $\,R\rightarrow R/J\,$ reflects units if $\,J\subseteq{\rm
rad}\,(R)$.∎

Example 2.5A. Recall that a ring $\,R\,$ has stable range one iff $\,{\rm
sr}_{R}(a)=1\,$ for every $\,a\in R$. Well known classes of such rings are:
(1) semilocal rings (including local rings and one-sided artinian rings
according to [La1: (20.3)] and [La3: (2.10)]); (2) unit-regular rings
(including all abelian regular rings) according to [Go: (4.12)]; (3) strongly
$\,\pi$-regular rings according to [Ar: Theorem 4] (see also [Khu]); (4) left-
and-right self-injective rings according to [La3: Cor. 7.17]; and (5) the ring
of all algebraic integers according to [Va2: (1.2)]. Many other interesting
examples of commutative and noncommutative rings of stable range one were
given in [Va1, Va2], [EO], [GM], [CY1, CY2], [Yu1, Yu2], [Ch3, Ch4], [La3],
etc.

Example 2.5B. Using (2.2) and (2.4), it is easy to determine which integers
$\,n\,$ have stable range one in the ring $\,{\mathbb{Z}}$. The answer is,
$\,{\rm sr}\,(a)=1\,$ iff $\,a\in\\{0,\,{\pm 1}\\}$. The “if” part is clear
(say from Example 2.1). For the “only if” part, start with $\,{\rm
sr}\,(n)=1\,$ for some $\,n\in{\mathbb{Z}}$. By Theorem 2.4(B), we may replace
$\,n\,$ by $\,-n\,$ if necessary to assume that $\,n\geq 0$. If $\,n\geq 2$,
taking $\,x=-n\,$ in Theorem 2.2(3) shows that there exists
$\,b\in{\mathbb{Z}}\,$ such that $\,n+(1+n^{2})\,b\in\\{{\pm 1}\\}$. This is
clearly impossible, so we must have $\,n\in\\{0,1\\}$. Note that this example
shows that the set of stable range one elements in a ring is in general not
closed under the map $\,a\mapsto 1-a\,$. It shows also that a “clean element”
(to be defined in Example 2.5F below) such as $\,2\in R={\mathbb{Z}}\,$ need
not have stable range one in $\,R$.

Example 2.5C. Since units and idempotents in any ring always have stable range
one, it will follow from the forthcoming Product Theorem 2.8 that unit-regular
elements also have stable range one. Somewhat surprisingly, however, in any
matrix ring $\,R={\mathbb{M}}_{n}(S)\,$ with $\,n\geq 2$, all matrices
$\,aE_{ij}\,$ with $\,a\in S\,$ (i.e. matrices with at most one nonzero entry)
happen to have also stable range one. This interesting fact will be proved
later in Theorem 5.8.

Example 2.5D. Some fairly large rings may have “very few” elements of stable
range one. For instance, let $\,R\,$ be a polynomial ring $\,S\,[x]\,$ over a
commutative domain $\,S$. Then the set of stable range one elements in $\,R\,$
is just $\,\\{0\\}\cup{\rm U}(S)\,$ (although $\,S\,$ itself may have many
other elements of stable range one). Indeed, if a degree $\,n\,$ polynomial
$\,a(x)\in R\setminus\\{0\\}\,$ has stable range one, then by Theorem 2.2,
there exists $\,b(x)\in R\,$ such that

$f(x)\\!:=a(x)+(1-a(x)\,x)\,b(x)\in{\rm U}(R)={\rm U}(S).$

If $\,b(x)\neq 0$, we have $\,{\rm deg}\,f(x)\geq n+1$, which is impossible.
Thus, we must have $\,b(x)=0$, and so $\,f(x)=a(x)\in{\rm U}(S)$.

Example 2.5E. For an example of a ring with a relatively large set of stable
range one elements, let $\,R=C(X)\,$ be the ring of continuous real-valued
functions on a topological space $\,X$. For any $\,a\in R\,$ such that
$\,a\,(X)\geq 0$, Azarpanah, Farokhpay and Ghashghaei have observed in [AFG:
(3.2)] that $\,{\rm sr}\,(a)=1$. Indeed, if $\,b\in R\,$ is such that
$\,aR+bR=R$, then $\,a,\,b\in C(X)\,$ cannot have a common zero. This implies
that the continuous function $\,a+b^{2}\,$ takes only positive values on
$\,X$, and so $\,a+b^{2}\in{\rm U}(R)$. In the case where $\,X\,$ is a
completely regular space, a precise description of the set of stable range one
elements in $\,C(X)\,$ has been given by Ghashghaei [Gh].

Example 2.5F. There are certainly some connections between stable range one
elements and Nicholson’s notion of clean elements in rings. In Nicholson’s
paper [Ni1], a ring element $\,a\in R\,$ is called clean if $\,a\,$ is the sum
of an idempotent and a unit in $\,R$. In view of Example 2.1, such an element
$\,a\,$ is the sum of two elements of stable range one in $\,R$, although
$\,a\,$ itself may not have stable range one. Going in the other direction,
consider for instance any two elements $\,a,b\in R\,$ such that $\,aR+bR=R$.
If either $\,{\rm sr}_{R}(a)=1\,$ or $\,{\rm sr}_{R}(b)=1$, it turns out that
$\,\footnotesize{\begin{pmatrix}a&b\\\ 0&0\end{pmatrix}}$ is a clean matrix in
$\,{\mathbb{M}}_{2}(R)$. For a proof of this interesting fact, based on the
ideas developed in [KL1: §3], see “Example (11)” listed under (17.2) in [La4].

In the next three theorems, we shall give some more applications of Theorem
2.4. The first one is the following basic result relating the Jacobson radical
$\,{\rm rad}\,(R)\,$ of a ring $\,R\,$ to the element-wise notion of (right)
stable range one.

Theorem 2.6. (1) For any $\,a\in R\,$ and $\,b\in{\rm rad}\,(R)$, we have
$\,{\rm sr}\,(a)=1\,$ iff $\,{\rm sr}\,(a+b)=1$. (In particular, $\,{\rm
sr}\,(b)=1\,$ always holds.)
(2) For any $\,b\in R$, we have $\,b\in{\rm rad}\,(R)\,$ iff $\,{\rm
sr}\,(b)=1\,$ and $\,b+{\rm U}(R)\subseteq{\rm U}(R)$.

Proof. (1) This part follows quickly from Theorem 2.4(D) since the natural
surjection $\,R\rightarrow R/{\rm rad}\,(R)\,$ reflects units, and both
conditions in (1) are equivalent to $\,{\rm
sr}\,\bigl{(}\overline{a}\bigr{)}=1\,$ in the factor ring
$\,\overline{R}=R/{\rm rad}\,(R)$.

(2) The “only if” part follows from (1) by setting $\,a=0$, and from a
familiar property of the Jacobson radical. The “if” part is basically folklore
after the appearance of [La2: Exercise 20.10B]. For more details, assume that
$\,{\rm sr}\,(b)=1\,$ and $\,b+{\rm U}(R)\subseteq{\rm U}(R)$. To get the
desired conclusion that $\,b\in{\rm rad}\,(R)$, it suffices to check that, for
any $\,x\in R$, $\,1-bx\,$ has a right inverse in $\,R$. (Here, we are using a
classical characterization result for elements in $\,{\rm rad}\,(R)$, as given
for example in [La1: Lemma 4.1].) By Theorem 2.2, there exists $\,y\in R\,$
such that $\,b+(1-bx)\,y\in{\rm U}(R)$. Thus, $\,(1-bx)\,(-y)\in b+{\rm
U}(R)\subseteq{\rm U}(R)$, which shows that $\,1-bx\,$ has a right inverse, as
desired.∎

Our next application of Theorem 2.4 is largely inspired by the work of
Godefroid [Gd] and Goodearl-Menal [GM].

Theorem 2.7. Suppose $\,a\in R\,$ has the property that, for any $\,x\in R$,
there exists $\,u\in{\rm U}(R)\,$ such that $\,x-u^{-1}\in{\rm U}(R)\,$ and
$\,{\rm sr}\,(a-u)=1$. Then $\,{\rm sr}\,(a)=1$.

Proof. It suffices to check the validity of the condition (A) in Theorem 2.4,
so consider any equation $\,ax+ty=1$. By assumption, there exists $\,u\in{\rm
U}(R)\,$ such that $\,v\\!:=x-u^{-1}\in{\rm U}(R)\,$ and $\,{\rm
sr}\,(a-u)=1$. We have then

$av+ty=a\,(x-u^{-1})+ty=1-au^{-1}=(a-u)\,(-u^{-1}),$

so $\,a+t\,(yv^{-1})=(a-u)\,(-u^{-1}v^{-1})$. According to Theorem 2.4(B),
this element has stable range one, so we have checked that the condition in
Theorem 2.4(A) holds.∎

We note in passing that Theorem 2.7 applies somewhat less widely than Theorem
2.2 and Theorem 2.4. For instance, it cannot be applied to a ring $\,R\,$
unless every element in $\,R\,$ is a sum of two units.

Applying Theorem 2.4 also retrieves easily the following result that is due to
H. Chen and W. K. Nicholson; see [CN: Lemma 17]. This important result will be
used over and over again in the forthcoming sections.

Product Theorem 2.8. If $\,{\rm sr}\,(a_{i})=1\,$ for all $\,i\in[1,n]$, then
$\,{\rm sr}\,(a_{1}\cdots a_{n})=1$. (Equivalently, the set $\,\\{a\in
R:\,{\rm sr}\,(a)=1\\}\,$ is a monoid under multiplication (with identity
$\,1\,$) containing $\,{\rm U}(R)\,$ as a subgroup.)

Proof. It suffices to check that $\,{\rm sr}\,(a)={\rm
sr}\,(a^{\prime})=1\Rightarrow{\rm sr}\,(aa^{\prime})=1$. Following [CN], we
start with an equation $\,aa^{\prime}R+tR=R$. Then $\,aR+tR=R\,$ too, so
$\,{\rm sr}\,(a)=1\,$ implies that $\,u\\!:=a+t\,b\in{\rm U}(R)\,$ for some
$\,b\in R$. Thus, $\,aa^{\prime}+t\,(ba^{\prime})=ua^{\prime}$, with $\,{\rm
sr}\,(ua^{\prime})=1\,$ by Theorem 2.4(B). According to Theorem 2.4(A), this
checks that $\,{\rm sr}\,(aa^{\prime})=1$, as desired. ∎

Example 2.9. In contrast with the theorem above, the well known example of
$\,R={\mathbb{Z}}\,$ shows that the set of stable range one elements in a ring
$\,R\,$ may not be closed under addition. However, one may still ask if
$\,{\rm sr}\,(a)={\rm sr}\,(b)=1\,$ might imply $\,{\rm sr}\,(a+b)=1\,$ if we
assume further that $\,ab=ba=0$. Unfortunately, this additional condition
won’t do it either. Indeed, in the ring
$\,R={\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}$, both of the matrices
$\,A={\rm diag}\,(2,0)\,$ and $\,B={\rm diag}\,(0,2)\,$ have stable range one
by our later result (5.8) in §5, and they satisfy the “orthogonality
condition” $\,AB=BA=0$. However, $\,A+B=2I_{2}\,$ does not have stable range
one by our later result (7.5) in §7.

To conclude this section, we rcord the following easy consequence of Theorem
2.8.

Corollary 2.10. If $\,xay=1\in R\,$ where $\,{\rm sr}\,(a)=1\,$ and either
$\,{\rm sr}\,(x)=1\,$ or $\,{\rm sr}\,(y)=1$, then $\,a\in{\rm U}(R)$.

Proof. We may assume that $\,{\rm sr}\,(x)=1\,$ (since the case where $\,{\rm
sr}\,(y)=1\,$ is similar). Here, Theorem 2.8 gives $\,{\rm sr}\,(xa)=1$. Since
$\,xa\,$ has a right inverse $\,y$, Theorem 2.4(C) shows that $\,z\,(xa)=1\,$
for some $\,z\in R$. Thus, $\,a\,$ has a left inverse $\,zx$, so another
application of Theorem 2.4(C) shows that $\,a\in{\rm U}(R)$. ∎

§3. Super Jacobson’s Lemma and Its Applications

One main question in the study of element-wise stable range one in arbitrary
(noncommutative) rings is whether such a notion is always left-right symmetric
for all elements. In the literature to date, this question has been answered
positively only for a few specific classes of rings; e.g. regular rings (as in
§4), duo rings111A ring $\,R\,$ is said to be a duo ring if every 1-sided
ideal in $\,R\,$ is a 2-sided ideal. (as in [AO2]), and the ring
$\,{\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}\,$ (as in [CP1]). The first
result in this section is devoted to resolving the question above positively
for all rings.

Symmetry Theorem 3.1. For any ring element $\,a\in R$, the property $\,{\rm
sr}\,(a)=1\,$ is equivalent to the property $\,{\rm sr}\,^{\prime}(a)=1$.
Here, as in Definition 1.1, $\,{\rm sr}\,^{\prime}(a)=1\,$ refers to the
property that $\,a\in R\,$ has left stable range one, while $\,{\rm
sr}\,(a)=1\,$ means, as always, that $\,a\in R\,$ has right stable range one.

As it turned out, the main trick for the proof of the above symmetry result is
the following interesting fact in noncommutative ring theory, which we shall
fondly call by a special name in honor of the late Yale Professor Nathan
Jacobson. Quite remarkably, this result applies uniformly to all three of the
sets $\,{\rm U}(R)\,$ (the group of units), $\,{\rm reg}\,(R)\,$ (the set of
regular elements), and $\,{\rm ureg}\,(R)\,$ (the set of unit-regular
elements), although not to $\,{\rm sreg}\,(R)\,$ (the set of strongly regular
elements).

Super Jacobson’s Lemma 3.2. For any three elements $\,a,b,x\,$ in any ring
$\,R$,the following three conclusions hold :

(1) $\,a+b-axb\in{\rm U}(R)\,$ iff $\,a+b-bxa\in{\rm U}(R)$.
(2) $\,a+b-axb\in{\rm reg}(R)\,$ iff $\,a+b-bxa\in{\rm reg}(R)$.
(3) $\,a+b-axb\in{\rm ureg}(R)\,$ iff $\,a+b-bxa\in{\rm ureg}(R)$.

However, in general, $\,a+b-axb\in{\rm sreg}(R)\,$ is not equivalent to
$\,a+b-bxa\in{\rm sreg}(R)$.

After the first draft of this paper was circulated, Professor P. Ara has
kindly informed the authors that Case (1) above was first proved by Menal and
Moncasi in [MM: p. 296]. As soon as Case (1) of this lemma is granted, a quick
recall of the criterion (2) in Theorem 2.2 for $\,{\rm sr}\,(a)=1\,$ and the
corresponding criterion for $\,{\rm sr}\,^{\prime}(a)=1\,$ will show that
these two properties are equivalent, for any element $\,a\in R\,$ in any ring
$\,R$, thus proving the Symmetry Theorem 3.1. In the language introduced by
Cǎlugǎreanu and Pop in [CP1], if $\,{\rm sr}\,(a)=1\,$ for a given element
$\,a\in R$, then for any $\,x\in R$, any “right unitizer” $\,b\,$ for $\,x\,$
will automatically be a “left unitizer” for $\,x$. This may sound almost too
good to be true, but the following proof will show that it is !

Proof of Lemma 3.2. To prove this “Super Lemma”, we will first focus on Case
(1); namely, the case of the unit group $\,{\rm U}(R)$. Of course, it will be
sufficient to prove the “only if” part of the lemma. Before we proceed
formally with the proof, perhaps a short explanation on the nomenclature used
in this lemma would be helpful. Consider the special case where $\,x=1$, and
assume that $\,a+b-ab\in{\rm U}(R)$. Then

$1-(1-a)\,(1-b)=1-(1-a-b+ab)=a+b-ab\in{\rm U}(R).$

By the classical Jacobson’s Lemma (as recalled in the Abstract), this implies
that

$1-(1-b)\,(1-a)=1-(1-a-b+b\,a)=a+b-b\,a\in{\rm U}(R),$

which is what we want. Thus, to prove Lemma 3.2 in the Case (1), our job is to
come up with a generalization of the above implication when a general element
$\,x\in R\,$ is given such that $\,a+b-axb\in{\rm U}(R)$. To accomplish this
goal, we will exploit the use of $\,2\times 2\,$ matrices over our ring $\,R$.
Although we can no longer use the tool of “determinants” for such matrices, it
turns out that a suitable appeal to the basic idea behind the proof of
“Banachiewicz’s Inversion Theorem” in [Ban] comes to our rescue. In its
“modern” purely ring-theoretic incarnation as given in [La4: (10.25)], this
well known result states that, for any $\,u\in{\rm U}(R)$, a matrix
$\,A=\footnotesize{\begin{pmatrix}u&q\\\
p&r\end{pmatrix}}\in{\mathbb{M}}_{2}(R)\,$ is invertible iff
$\,r-pu^{-1}q\in{\rm U}(R)$.222The detailed expression for $\,A^{-1}\,$ in
this case is given explicitly in [La4: (10.29)]. However, this expression is
not needed for any of the results in this section. In our special setting, we
may start with the following matrix-product equation which has appeared
before, for instance, in [LN: §4]:

$\begin{pmatrix}1&-b\\\ 1-ax&a\end{pmatrix}\begin{pmatrix}1&0\\\
x&1\end{pmatrix}=\begin{pmatrix}1-b\,x&-b\\\
1&a\end{pmatrix}\in{\mathbb{M}}_{2}(R).$ $None$

If we assume that $\,a+b-axb\in{\rm U}(R)$, then Banachiewicz’s Inversion
Theorem implies that the first matrix in (3.3) above is invertible, so the
last matrix there is also invertible. After a row permutation, it follows that
$\footnotesize{\begin{pmatrix}1&a\\\ 1-b\,x&-b\end{pmatrix}}$ is invertible.
Applying Banachiewicz’s Inversion Theorem once more shows that
$\,a+b-bxa\in{\rm U}(R)$, as desired.

Having done the work above, we need to handle two more cases; namely, the
cases where $\,{\rm U}(R)\,$ is replaced by $\,{\rm reg}\,(R)\,$ or by $\,{\rm
ureg}\,(R)$. Fortunately, not much further work is needed to get the desired
conclusions in these two cases. Indeed, in the relatively new treatment of
Banachiewicz’s inversion result in [La4: Theorem 10.25], it is already proved
that this result also works in the cases where
$\,A=\footnotesize{\begin{pmatrix}u&q\\\ p&r\end{pmatrix}}$ is regular or
unit-regular in $\,{\mathbb{M}}_{2}(R)$, with the respective characterizations
that $\,r-pu^{-1}q\in{\rm reg}\,(R)\,$ or $\,r-pu^{-1}q\in{\rm ureg}\,(R)$.
This proves the truth of the desired conclusions in both Case (2) and Case
(3).

To show that the above work cannot be extended to the case of $\,{\rm
sreg}\,(R)$, we need only consider the following example. In the ring
$\,R={\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}$, let
$\,a=\footnotesize{\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}}$,
$\,b=\footnotesize{\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}}$, and
$\,x=\footnotesize{\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}}$. Then
$\,a+b-axb=\footnotesize{\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}}\in{\rm
idem}\,(R)\subseteq{\rm sreg}\,(R)$. However,
$\,c\\!:=a+b-bxa=\footnotesize{\begin{pmatrix}2&1\\\ 0&0\end{pmatrix}}$ is
clearly not in $\,R\cdot c^{2}\,$ since
$\,c^{2}=\footnotesize{\begin{pmatrix}4&2\\\ 0&0\end{pmatrix}}$. This shows
that $\,c\notin{\rm sreg}\,(R)$. (On the other hand, we do have $\,c\in{\rm
ureg}\,(R)$, so there is no contradiction to the conclusions obtained in Case
(2) or Case (3).) ∎

Example 3.4. Talking about “ternary generalizations” of Jacobson’s Lemma, we
should point out that other “more obvious” (or “better looking”) forms of such
generalizations may not hold true at all. For instance, given arbitrary
$\,a,x,b\in R$, one may wonder if $\,1-axb\in{\rm U}(R)\,$ might be equivalent
to $\,1-bxa\in{\rm U}(R)$. The answer to this question is, unfortunately, “no”
(in general), as the following easy example shows. In any nonzero matrix ring
$\,R={\mathbb{M}}_{2}(S)$, let $\,a=\footnotesize{\begin{pmatrix}1&0\\\
0&0\end{pmatrix}}$, $\,x=\footnotesize{\begin{pmatrix}0&1\\\
1&0\end{pmatrix}}$, and $\,b=\footnotesize{\begin{pmatrix}1&1\\\
0&0\end{pmatrix}}$. Then $\,1-axb=1\in{\rm U}(R)$, but
$\,1-bxa=\footnotesize{\begin{pmatrix}0&0\\\ 0&1\end{pmatrix}}\notin{\rm
U}(R)\,$! On the other hand, for the same choices of $\,a,b,x\in R$, it is
clear that neither $\,a+b-axb=\footnotesize{\begin{pmatrix}2&1\\\
0&0\end{pmatrix}}$ nor $\,a+b-bxa=\footnotesize{\begin{pmatrix}1&1\\\
0&0\end{pmatrix}}$ is in $\,{\rm U}(R)$, thus reaffirming the truth of Super
Jacobson’s Lemma in this particular case.

Historical Remark 3.5. As we have seen from the calculations given at the
beginning of the proof of (3.2), the classical form of Jacobson’s Lemma
corresponds to the special case $\,x=1\,$ of Super Jacobson’s Lemma when we
make the substitutions $\,a\mapsto 1-a\,$ and $\,b\mapsto 1-b$. The new
formulation of Super Jacobson’s Lemma 3.2 also suggests that we can generalize
another old idea of Jacobson, as follows. In Jacobson’s book [Ja2: p.8], an
associative “circle operation” for ring elements $\,a,b\in R\,$ was introduced
through the definition $\,a\circ b:=a+b-ab$. If we pick a fixed element
$\,x\in R$, we can define a new circle operation by the rule $\,a\circ
b\\!:=a+b-axb$. Quite pleasantly, this still turns out to be an associative
operation, since a quick calculation shows that $\,(a\circ b)\circ c\,$ and
$\,a\circ(b\circ c)\,$ are both equal to $\,a+b+c-axb-bxc-axc+axbxc$.
Furthermore, $0\,$ is the identity element for this new circle operation; that
is, $\,a\circ 0=a=0\circ a\,$ for every $\,a\in R$.333Note that, in general,
this new operation is not commutative. Indeed, we can check easily that it is
commutative iff $\,x\,$ is a central element of $\,R\,$ that annihilates all
“additive commutators” $\,ab-ba\,$ in $\,R$. Expressed in terms of the new
circle operation (defined via a given element $\,x\,$), Super Jacobson’s Lemma
would simply say that $\,a\circ b\in{\rm U}(R)\,$ iff $\,b\circ a\in{\rm
U}(R)$; $\,a\circ b\in{\rm reg}\,(R)\,$ iff $\,b\circ a\in{\rm reg}\,(R)$; and
$\,a\circ b\in{\rm ureg}(R)\,$ iff $\,b\circ a\in{\rm ureg}(R)$.

To further illustrate the power of Super Jacobson’s Lemma 3.2, we shall next
state and prove below four of its interesting “binary specializations” for any
ring $\,R$.

Proposition 3.6. For any $\,a,x\in R$, the following conditions are
equivalent:

(1) $\,1-ax+axa\in{\rm U}(R)$. (2) $\,1-xa+axa\in{\rm U}(R)$.
(3) $\,1-ax+a^{2}x\in{\rm U}(R)$. (4) $\,1-xa+xa^{2}\in{\rm U}(R)$.

If these conditions hold, then $\,\\{ax-xa,\;axa-
xa^{2},\;axa-a^{2}x\\}\subseteq{\rm U}(R)+{\rm U}(R)$. Also, the same result
holds if $\,{\rm U}(R)\,$ is replaced throughout by $\,{\rm reg}\,(R)$, or by
$\,{\rm ureg}\,(R)$.

Proof. It suffices to work with the case of $\,{\rm U}(R)$, since the other
two cases (for $\,{\rm reg}\,(R)\,$ and $\,{\rm ureg}\,(R)$) are completely
similar. To begin with, $(1)\Leftrightarrow(2)$ follows from Super Jacobson’s
Lemma by taking $\,b=1-a$. Next, $(1)\Leftrightarrow(4)$ follows from the
usual Jacobson’s Lemma (in all three cases) by writing
$\,1-ax+axa=1-a\,(x-xa)\,$ and $\,1-xa+xa^{2}=1-(x-xa)\,a$. With this little
trick at our disposal, $(2)\Leftrightarrow(3)$ follows similarly. Finally, the
conclusion about the three elements $\,ax-xa,\;axa-xa^{2}\,$ and
$\,axa-a^{2}x\,$ follows by subtracting (1) from (2); (4) from (2); and (3)
from (1), respectively.∎

Comparing Proposition 3.6 with what is known in the literature, we can say the
following. In the special case where we assume additionally that $\,a=axa$,
(1) and (2) were known to be both equivalent to $\,a\in{\rm sreg}\,(R)\,$
(i.e. $a\,$ is strongly regular in $\,R\,$) by a result of Puystjens and
Hartwig in [PuH], while (3) and (4) can be easily shown to be equivalent to
(1) and (2). For a detailed exposition on these four characterizations for the
elements in $\,{\rm sreg}\,(R)$, see [La4: Theorem 3.24]. The remarkable thing
about Proposition 3.6 is that it holds without any initial binding conditions
between the two elements $\,a,\,x\in R$, while it generalizes the work of
Puystjens and Hartwig on strongly regular elements in a rather unexpected way.

Upon reviewing the meaning of Super Jacobson’s Lemma in the special case where
$\,R={\mathbb{M}}_{n}(S)\,$ for some ring $\,S$, we see that, for any three
matrices $\,A,B,X\in R$, we have $\,A+B-AXB\in{\rm GL}_{n}(S)\,$ iff
$\,A+B-BXA\in{\rm GL}_{n}(S)$. In the special case where $\,S\,$ is a
commutative ring, it is well known that a matrix $\,M\in R\,$ is in $\,{\rm
GL}_{n}(S)\,$ iff $\,{\rm det}\,(M)\in{\rm U}(S)$. In this special case, Super
Jacobson’s Lemma for the matrix ring $\,R\,$ above would amount to the
statement that $\,{\rm det}\,(A+B-AXB)\in{\rm U}(S)\,$ iff $\,{\rm
det}\,(A+B-BXA)\in{\rm U}(S)$. This was the first time it occurred to us that,
for any $\,A,B,X\in R$, the two determinants above must be rather “intimately
related”. Indeed, after some additional work, the following result about
determinants was proved in [KL3: Theorem 2.1].

Theorem 3.7. For any $\,A,B,X\in R={\mathbb{M}}_{n}(S)\,$ over a commutative
ring $\,S$, we have

${\rm det}\,(A+B-AXB)={\rm det}\,(A+B-BXA)\in S.$ $None$

In particular, $\,A+B-AXB\in{\rm GL}_{n}(S)\,$ if and only if
$\,A+B-BXA\in{\rm GL}_{n}(S)$. However, we may not have $\,{\rm
tr}\,(A+B-AXB)={\rm tr}\,(A+B-BXA)$.

While we will not repeat the proof of the equation (3.8) in [KL3] here, we
should point out that, in studying the possible applications of the Symmetry
Theorem 3.1, it is perhaps to be expected that the notion of a
$\,{\ast}\,$-ring (or a “ring with involution”) would be of interest here.
Indeed, using the Symmetry Theorem 3.1, we can prove easily the following
result on “element-wise stable range one” for elements in any
$\,{\ast}\,$-ring.

Theorem 3.9. If $\,(R,\,{}^{*}\,)\,$ is a $\,{\ast}\,$-ring , then for any
$\,a\in R$, $\,{\rm sr}\,(a)=1\,$ iff $\,{\rm sr}\,(a^{*})=1$.

Proof. Thanks to the Symmetry Theorem 3.1, it suffices to prove that, if
$\,{\rm sr}\,(a)=1$, then $\,{\rm sr}\,^{\prime}(a^{*})=1$. To do this, we
take any element of $\,R\,$ (which we may conveniently write as $\,x^{*}\,$
for some $\,x\in R$), and try to show that
$\,a^{*}+b^{*}\,(1-x^{*}a^{*})\in{\rm U}(R)\,$ for some $\,b\in R$. Since
$\,{\rm sr}\,(a)=1$, there exists $\,b\in R\,$ such that $\,a+(1-ax)\,b\in{\rm
U}(R)$. Applying the involution $\,{}^{*}\,$ to this membership relation (and
noting that ∗ preserves the unit group $\,{\rm U}(R)$), we see that
$\,a^{*}+b^{*}\,(1-x^{*}a^{*})\in{\rm U}(R)$, as desired. ∎

Over a commutative ground ring, a quick application of Theorem 3.9 yields the
following nice result on the transposition of square matrices over a
commutative ring.

Corollary 3.10. If $\,R={\mathbb{M}}_{n}(S)\,$ where $\,S\,$ is any
commutative ring, then for any matrix $\,A\in R$, $\,{\rm sr}\,(A)=1\,$ iff
$\,{\rm sr}\,(A^{T})=1$, where $\,A^{T}\,$ denotes the transpose of $\,A$.

Proof. We simply note that, in this case, the transpose operation $\,A\mapsto
A^{T}\,$ is a ring involution on $\,R={\mathbb{M}}_{n}(S)$, and apply Theorem
3.9.∎

While the proof given above definitely made use of the commutativity property
of the base ring $\,S$, it is nevertheless legitimate to ask whether the
conclusion of Corollary 3.10 might remain true if $\,S\,$ is just any ring. We
will return to answer this question negatively later; see Example 6.12.

Remark 3.11. To conclude this section, we would like to make the following
further observations about the technique of applying Super Jacobson’s Lemma to
the study of the notion of stable range one. In the standard literature, it is
known that this notion is also worth studying in some of its variations, such
as unit stable range one, idempotent stable range one, regular stable range
one, and square stable range one; see, for instance, Goodearl-Menal [GM], Chen
[Ch1], Ashrafi-Nasibi [AN], and Wang et al. [WC], [KLW]. These notions arise
naturally by specializing the “element-wise stable range one” notion in
Definition 1.1 by asking that the existential element $\,b\in R\,$ be,
respectively, a unit, an idempotent, a regular element, or a perfect square in
$\,R$. While these somewhat more refined notions have again a “left version”
and a “right version” (depending on the use of left ideals or right ideals in
$\,R\,$), the very same argument used for proving the Symmetry Theorem 3.1 can
be exploited once more to show that they are all left-right symmetric notions
as well.

Example 3.12. Of course, it is entirely to be expected that the more refined
notions of stable range one mentioned in Remark 3.11 above are in general more
restrictive than the original notion of stable range one introduced in
Definition 1.1. For some quick examples to show some cases of this, consider
the matrix $\,A={\rm diag}\,(7,0)\in
R={\mathbb{M}_{2}}\bigl{(}{\mathbb{Z}}\bigr{)}$, with $\,{\rm sr}\,(A)=1\,$ to
be shown in Theorem 5.8. We claim that, in $\,R$, $\,A\,$ has neither unit
stable range one nor idempotent stable range one. To see this, let $\,B={\rm
diag}\,(2,1)\,$ and $\,C={\rm diag}\,(-3,1)$. Since $\,A+B\,C=I_{2}$, we have
$\,AR+BR=R$. If $\,A\,$ has unit stable range one, there would exist a matrix
$\,U=\footnotesize{\begin{pmatrix}a&b\\\ c&d\end{pmatrix}}\in{\rm U}(R)\,$
such that $\,A+B\,U\in{\rm U}(R)$. A quick computation shows that

${\rm det}\,(A+B\,U)=7d+2\,(ad-bc)=7d\,{\pm}\,2\notin\\{{\pm 1}\\},$

which is a contradiction. Somewhat similarly, if $\,A\,$ has idempotent stable
range one, there would exist $\,P=\footnotesize{\begin{pmatrix}p&q\\\
r&s\end{pmatrix}}\in{\rm idem}\,(R)\,$ such that $\,A+BP\in{\rm U}(R)$.
Clearly, $\,P\notin\\{0,\,I_{2}\\}$, so we must have $\,p+s=1\,$ and $\,ps-
qr=0$. A quick computation shows that $\,{\rm
det}\,(A+B\,P)=7\,(1-p)\notin\\{\pm 1\\}$, again a contradiction.

§4. Stable Range One Elements Via Unit-Regularity

To further develop the notion of ring elements of stable range one, we go back
to the basic theme of looking at regular and unit-regular elements in any
ring. The first theorem in this section is largely motivated by a classical
result of Fuchs, Kaplansky and Henriksen on regular rings that is reported in
Goodearl’s book [Go: (4.12)]. According to this result, a (von Neumann)
regular ring has stable range one iff it is unit-regular. To make this into an
“element-wise result”, we have the following theorem characterizing regular
elements with stable range one in any ring. Here, $\,(1)\Leftrightarrow(2)$
was noted in [KL2: (3.5)] (see also [HN: (3.2)]), while
$\,(2)\Leftrightarrow(4)$ was an observation of H. Chen reported in [HN:
(5.13)].

Theorem 4.1. For any element $\,a\in{\rm reg}\,(R)$, the following conditions
are equivalent :

(1) $\,{\rm sr}\,(a)=1$.
(2) $\,a\in{\rm ureg}\,(R)$.
(3) $\,a\,$ is a product of elements in $\,{\rm ureg}\,(R)$.
(4) $\;\overline{a}\,$ is unit-regular in $\,\overline{R}=R/{\rm rad}\,(R)$.

Proof. We will first prove the equivalence of (1), (2) and (3). To begin with,
$(2)\Rightarrow(3)\,$ is trivial, while $(3)\Rightarrow(1)$ follows from the
Product Theorem in view of what is said in Example 2.5C. Next, assume (1), and
write $\,a=axa\,$ for some $\,x\in R$. Then Theorem 2.2(3) gives an element
$\,b\in R\,$ such that $\,a+(1-ax)\,b\in{\rm U}(R)$. Left-multiplying this by
the idempotent $\,e=ax$, we get $\,a\in e\cdot{\rm U}(R)\subseteq{\rm
ureg}\,(R)$, which proves (2). Finally, $(2)\Rightarrow(4)\,$ is trivial, and
$(4)\Rightarrow(1)\,$ follows from Theorem 2.4(D) together with the already
known fact that (4) implies that $\,{\rm sr}\,(\overline{a})=1\,$ in
$\,\overline{R}$. ∎

Remark. In view of Theorem 2.4(A), $(2)\Leftrightarrow(1)$ in the theorem
above shows easily that, for any ring element $\,a\in R\,$ (without any
regularity assumption), $\,{\rm sr}\,(a)=1\,$ iff, for any $\,t\in R$,
$\,aR+tR=R\,\Rightarrow\,a+tb\in{\rm ureg}\,(R)\,$ for some $\,b\in R$. First
stated by Altun and Özcan in [AO1: (2.6)], this fact is an improvement of a
result of Zabavsky in [Za: (7.5)].

Example 4.2A. Over any ring $\,S$, a row vector $\,{\bf
v}=(a_{1},\dots,a_{n})\in S^{n}\,$ is called a completable row if there exists
a matrix $\,V\in{\rm GL}_{n}(S)\,$ with $\,{\bf v}\,$ as its first row. For
any such $\,{\bf v}$, the matrix $\,A\in R={\mathbb{M}}_{n}(S)\,$ with first
row $\,{\bf v}\,$ and all other rows zero has stable range one, since
$\,A=E_{11}V\in{\rm ureg}\,(R)$.

Example 4.2B. In any matrix ring $\,R={\mathbb{M}}_{2}(S)$, let
$\,U=\footnotesize{\begin{pmatrix}1&0\\\ s&1\end{pmatrix}}\in{\rm U}(R)\,$ and
$\,E=\footnotesize{\begin{pmatrix}1&t\\\ 0&0\end{pmatrix}}\in{\rm idem}\,(R)$.
Then $\,B:=UE=\footnotesize{\begin{pmatrix}1&t\\\ s&s\,t\end{pmatrix}}$ and
$\,A:=EU=\footnotesize{\begin{pmatrix}1+ts&t\\\ 0&0\end{pmatrix}}$ are both
unit-regular in $\,R$, so $\,{\rm sr}_{R}(B)={\rm sr}_{R}(A)=1$. The last
equality here also follows from Example 4.2A since the first row
$\,(1+ts,\,t)\,$ of $\,A\,$ is completable to the invertible matrix
$\,V=\footnotesize{\begin{pmatrix}1+ts&t\\\ s&1\end{pmatrix}}$. In the special
case where $\,S\,$ is a commutative ring, the fact that $\,{\rm
sr}_{R}(B)=1\,$ was first noted by Cǎlugǎreanu and Pop in [CP1].

Recall that an element $\,y\in R\,$ is called a reflexive inverse of an
element $\,a\in R\,$ if $\,a=aya\,$ and $\,y=yay$. Making use of this standard
terminology, we can produce a number of additional “inter-connected”
characterizations for unit-regular elements in any ring $\,R$, as in the
following result Theorem 4.3. Here, the two characterizations (8) and (9) are
taken from Wei’s paper [We: Lemma 2.1], but our proofs below are considerably
simpler.

Theorem 4.3. For any ring element $\,a\in R$, the following statements are
equivalent:

(1) $\;a\in{\rm ureg}\,(R)$.
(2) $\;a\in{\rm reg}\,(R)$, and every reflexive inverse $\,x\,$ of $\,a\,$ is
unit-regular.
(3) $\;a\in{\rm reg}\,(R)$, and every reflexive inverse $\,x\,$ of $\,a\,$ has
$\,{\rm sr}\,(a)=1$.
(4) $\;a\,$ has a reflexive inverse $\,x\in R\,$ such that $\,{\rm
sr}\,(x)=1$.
(5) $\;a\,$ has an inner inverse $\,x\in R\,$ such that $\,{\rm sr}\,(x)=1$.
(6) $\;a\,$ has a reflexive inverse that is unit-regular.
(7) $\;a\,$ has an inner inverse that is unit-regular.
(8)  There exist $\,y\in R\,$ and $\,u\in{\rm U}(R)\,$ such that
$\,a=aya=ayu$.
(9)  There exist $\,y\in R\,$ and $\,w\in{\rm ureg}\,(R)\,$ such that
$\,a=aya=ayw$.
(10) There exist $\,y,\,z\in R\,$ with $\,{\rm sr}\,(z)=1\,$ such that
$\,a=aya=ayz$.

Proof. We will first prove the equivalence of (1) through (5). To begin with,
$(2)\Rightarrow(3)\,$ follows from Theorem 4.1, and
$(3)\Rightarrow(4)\Rightarrow(5)\,$ are trivial.

$(5)\Rightarrow(1)$. Assume that $\,a=axa\,$ for some $\,x\in R\,$ such that
$\,{\rm sr}\,(x)=1$. According to Theorem 2.2, there exists $\,b\in R\,$ such
that $\,u\\!:=x+(1-xa)\,b\in{\rm U}(R)$. Left multiplying this by $\,a\,$
gives $\,au=ax+(a-axa)\,b=ax$, so $\,a=(ax)\,a=aua\in{\rm ureg}\,(R)$.

$(1)\Rightarrow(2)$. Let $\,x\,$ be any reflexive inverse of $\,a\in{\rm
ureg}\,(R)$. We have $\,{\rm sr}\,(a)=1\,$ by Theorem 4.1, and $\,x\,$ has
inner inverse $\,a$. Thus, applying the preceding implication
$(5)\Rightarrow(1)\,$ to $\,x\,$ shows that $\,x\in{\rm ureg}\,(R)$.

Having shown the equivalence of (1)–(5), we’ll next prove that
$(1)\Rightarrow(6)\Rightarrow(7)\Rightarrow(5)$. Here, $(6)\Rightarrow(7)$ is
a tautology, and $(7)\Rightarrow(5)$ follows from Theorem 4.1. To prove
$(1)\Rightarrow(6)$, fix a unit inner inverse $\,u\,$ for $\,a\in{\rm
ureg}\,(R)$. Then $\,a\,$ has reflexive inverse $\,uau$, which is easily seen
to be unit-regular.

To complete the proof of the theorem, we note that $(1)\Rightarrow(8)$ follows
by writing $\,a=aya\,$ for some $\,y\in{\rm U}(R)\,$ and choosing $\,u\,$ to
be $\,y^{-1}$. Next, $(8)\Rightarrow(9)\,$ is trivial, while
$(9)\Rightarrow(10)\,$ follows from Theorem 4.1. Finally, if $\,y\,$ and
$\,z\,$ exist as in (10), then $\,a=aya\in{\rm reg}\,(R)$, and $\,a=(ay)\,z\,$
(with $\,ay\in{\rm idem}\,(R)\,$) implies that $\,{\rm sr}\,(a)=1\,$ again by
Theorem 2.8. Thus, $\,a\in{\rm ureg}\,(R)\,$ by Theorem 4.1.∎

Remark 4.4. With respect to the equivalence of (1) and (6) in the above
theorem, one may wonder what would happen if we try to strengthen the
condition (6) to $\,a\,$ having a reflexive inverse that is strongly regular.
This question has been recently answered by X. Mary, who showed in [My:
Theorem 4.1] that this strengthened version of (6) is equivalent to $\,a\,$
having a “special clean decomposition” (see [KNS]) in the sense that
$\,a=e+u\,$ where $\,e=e^{2}\in R\,,$ and $\,u\in{\rm U}(R)\,$ is such that
$\,a=au^{-1}a$.

Returning to the theme of Theorem 4.1 and using once more the Product Theorem
2.8, we will next prove the interesting result below about products of unit-
regular elements in any ring $\,R$.

Theorem 4.5. (1) Any finite product of elements in $\,{\rm ureg}\,(R)\,$ has
stable range one.
(2) If $\,a_{1},\dots,a_{n}\in{\rm ureg}\,(R)\,$ are such that
$\,b\\!:=a_{1}\cdots a_{n}\in{\rm reg}\,(R)$, then $\,b\in{\rm ureg}\,(R)$.
(In particular, if $\,a\in{\rm ureg}\,(R)\,$, then for any positive integer
$\,n$, $\,a^{n}\in{\rm reg}\,(R)\,$ iff $\,a^{n}\in{\rm ureg}\,(R)$.)
(3) If the set $\,{\rm reg}\,(R)\,$ is closed under multiplication, then so is
the set $\,{\rm ureg}\,(R)$.

Proof. (1) follows from Theorem 4.1 and the Product Theorem 2.8. Since
obviously (2) implies (3), we may finish by proving (2). By (1), we have
$\,{\rm sr}\,(b)=1$. If we assume that $\,b\in{\rm reg}\,(R)$, another
application of Theorem 4.1 shows that $\,b\in{\rm ureg}\,(R)$. (Obviously,
part (3) here implies the main result of Hannah and O’Meara [HO: (3.2)], which
states that, in any regular ring $\,R$, the set $\,{\rm ureg}\,(R)\,$ is
closed under multiplication. For an independent proof of (2) in the special
case where $\,n=2$, see [La3: (6.33)].) ∎

Remark 4.6. Concerning Theorem 4.5(2) again, we should point out that, if
$\,a_{1},\dots,a_{n}\in{\rm sreg}\,(R)\,$ are such that $\,b\\!:=a_{1}\cdots
a_{n}\in{\rm reg}\,(R)$, then $\,b\,$ may not be in $\,{\rm sreg}\,(R)\,$
(although we do have $\,b\in{\rm ureg}\,(R)\,$ by Theorem 4.5(1)). For
instance, in the matrix ring
$\,R={\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}$, both
$\,a_{1}=\footnotesize{\begin{pmatrix}1&2\\\ 0&0\end{pmatrix}}$ and
$\,a_{2}=\footnotesize{\begin{pmatrix}0&-1\\\ 1&1\end{pmatrix}}$ are in
$\,{\rm sreg}\,(R)$, but
$\,b\\!:=a_{1}a_{2}=\footnotesize{\begin{pmatrix}2&1\\\
0&0\end{pmatrix}}\in{\rm ureg}\,(R)\,$ is not in $\,{\rm sreg}\,(R)\,$ since
$\,b^{2}=2\,b\,$ clearly implies that $\,b\notin b^{2}R$.

To state a consequence of Theorem 4.5, we recall from [KL2] that a ring
$\,R\,$ is said to be IC (“internally cancellable”) if, whenever
$\,R_{R}=A\oplus B=A^{\prime}\oplus B^{\prime}$, $\,A\cong
A^{\prime}\,\Rightarrow\,B\cong B^{\prime}\,$ in the category of right
$\,R$-modules. From [KL2: (1.5)], it is well known that this property is left-
right symmetric, and is equivalent to the condition that $\,{\rm
reg}\,(R)={\rm ureg}\,(R)$. In view of this “element-wise” characterization of
IC rings, the following result is an immediate consequence of Theorem 4.1 and
Theorem 4.5(1).

Corollary 4.7. A ring $\,R\,$ is IC iff every element $\,a\in{\rm reg}\,(R)\,$
has stable range one, iff every element in $\,{\rm reg}\,(R)\,$ is a product
of elements in $\,{\rm ureg}\,(R)$.

Remark 4.8. If $\,a,b\in{\rm ureg}\,(R)$, there is also a known ideal-
theoretic criterion for $\,ab\,$ to be regular (or unit-regular) in $\,R\/$;
namely, that the right ideal $\,{\rm ann}_{r}(a)+bR\,$ be a direct summand of
$\,R_{R}$. For a proof of this interesting criterion, see [La4: Theorem 6.33].
Having proved Corollary 4.7, we should mention that the rings $\,R\,$ with the
property that $\,{\rm reg}\,(R)\,$ is closed under multiplication are called
reg-closed rings in [La4: §6], where it is shown that these are precisely the
SSP rings introduced earlier by Garcia in [Ga]. (The term “SSP” was in
reference to a certain “summand sum property” characterizing such rings.) For
a fuller discussion on these rings and a more complete listing of the relevant
literature on them, see again [La4: §6].

For any commutative ring $\,S\,$ with only trivial idempotents, it is known
from [KL2: Example 5.9(1)] that the matrix ring $\,{\mathbb{M}}_{2}(S)\,$ is
IC. In fact, this conclusion holds even without the assumption that $\,{\rm
idem}\,(S)=\\{0,1\\}$, since a recent argument of Swan reported in [La4:
Theorem 23.16] showed that $\,{\mathbb{M}}_{2}(S)\,$ is IC as long as $\,S\,$
is a commutative ring. Assuming this nice result, we can deduce the following.

Corollary 4.9. For any commutative ring $\,S$, any (von Neumann) regular
matrix in $\,{\mathbb{M}}_{2}(S)\,$ is unit-regular, and hence has stable
range one.

Example 4.10. Even for a commutative domain $S$, ${\mathbb{M}}_{3}(S)$ may
fail to be IC. For instance, it is pointed out after the proof of [La4:
(23.16)] that, for $\,S={\mathbb{R}}\,[x,y,z]/(x^{2}+y^{2}+z^{2}-1)\,$ (the
coordinate ring of the real $2$-sphere), the matrix ring
$\,R={\mathbb{M}}_{3}(S)\,$ is not IC, with the matrix
$\,A=\footnotesize{\begin{pmatrix}x&y&z\\\ 0&0&0\\\ 0&0&0\end{pmatrix}}$ being
regular (with reflexive inverse $\,A^{T}\,$) but not unit-regular, so that
$\,{\rm sr}_{R}(A)\neq 1$. Finally, noting that $\,A\,$ has a factorization
$\,BC\,$ where $\,B=\footnotesize{\begin{pmatrix}1&y&z\\\ 0&0&0\\\
0&0&0\end{pmatrix}}$ is an idempotent and $\,C={\rm diag}\,(x,1,1)$, we see
that $\,{\rm sr}_{R}(C)\neq 1\,$ by Theorem 2.8. This is in sharp contrast to
our forthcoming result Theorem 7.1, which implies that any diagonal matrix
$\,D={\rm diag}\,(a,b,0)\,$ (over any base ring $\,S\,$) has stable range one
in $\,{\mathbb{M}}_{3}(S)$.

To close this section, we shall prove one more result relating the notion of
IC (“internal cancellation”) to the ring-theoretic notion of stable range one.
In the case of an exchange ring $\,R$, it has been well known since 1995 that
$\,{\rm sr}\,(R)=1\,$ iff $\,R\,$ is IC, iff $\,{\rm reg}\,(R)={\rm
ureg}\,(R)$. This nice result was first proved by Yu in [Yu1: Theorem 9] and
Camillo-Yu in [CY1: Theorem 3], and later recapitulated by Chen in [Ch2, Ch3],
and Khurana-Lam in [KL2]. While this classical result was purely expressed in
terms of the “global” ring-theoretic properties of $\,R$, the element-wise
approach to stable range one taken in our present paper would seem to suggest
that there should exist a corresponding criterion characterizing elements
$\,a\in R\,$ with $\,{\rm sr}\,(a)=1\,$ in the case where, say, all elements
in $\,aR\,$ are suitable (in the sense of Nicholson [Ni1]). Optimally, such an
element-wise criterion should be expressible in terms of “certain” von Neumann
regular elements associated with $\,a\in R\,$ being automatically unit-
regular. Moreover, it should naturally imply the classical Camillo-Yu theorem
in the case where $\,R\,$ itself is an exchange ring. Luckily, such an
element-wise criterion does exist, as we will show in the result below.

Theorem 4.11. Let $\,a\in R\,$ be such that all elements in $\,aR\,$ are
suitable. Then $\,{\rm sr}\,(a)=1\,$ iff, for any $\,f\in{\rm idem}\,(R)$,
$\,fa\in{\rm reg}\,(R)\Rightarrow fa\in{\rm ureg}\,(R)$.

Proof. First assume $\,{\rm sr}\,(a)=1$. If $\,f\in{\rm idem}\,(R)\,$ is such
that $\,fa\in{\rm reg}\,(R)$, then $\,{\rm sr}\,(fa)=1\,$ by Theorem 2.8, so
Theorem 4.1 implies that $\,fa\in{\rm ureg}\,(R)$. Conversely, assume that for
any $\,f\in{\rm idem}\,(R)$, $\,fa\in{\rm reg}\,(R)\Rightarrow fa\in{\rm
ureg}\,(R)$. To show that $\,{\rm sr}\,(a)=1$, we’ll verify the criterion (7)
in Corollary 2.3; that is, given any equation $\,ax+e=1\,$ with $\,e=e^{2}$,
we want to find some $\,b\in R\,$ such that $\,a+e\,b\in{\rm U}(R)$. Letting
$\,f\\!:=1-e=ax$, we have $\,(fa)\,x\,(fa)=f^{3}a=fa$, so $\,fa\in{\rm
reg}\,(R)$. By assumption, this implies that $\,fa\in{\rm ureg}\,(R)$, so
$\,{\rm sr}\,(fa)=1$. From $\,(fa)\,x+e=f^{2}+e=1$, it then follows that
$\,fa+er\in{\rm U}(R)\,$ for some $\,r\in R$. This amounts to
$\,a+e\,(r-a)\in{\rm U}(R)$, as desired.∎

§5. Stable Range One Elements Via Nilpotency

Since units and idempotents in any ring $\,R\,$ have stable range one, it is
natural to wonder which nilpotent elements $\,a\in{\rm nil}\,(R)\,$ might have
the same property. To give some examples of such elements, let us first recall
the following two classical element-wise notions that were closely related to
the notion of nilpotent elements in rings. First, an element $\,a\in R\,$ is
called strongly nilpotent (after J. Levitzki [Lv]) if every sequence
$\,a_{1},a_{2},a_{3},\dots\,$ such that $\,a_{1}=a\,$ and $\,a_{n+1}\in
a_{n}Ra_{n}\,$ for all $\,n\,$ is eventually zero. Secondly, an element
$\,a\in R\,$ is called quasi-nilpotent (after R. Harte [Ha]) if $\,1-as\in{\rm
U}(R)\,$ for every $\,s\in R\,$ commuting with $\,a$. With these two standard
definitions in place, our first main result in this section can be stated as
follows.

Theorem 5.1. Suppose an element $\,a\in R\,$ satisfies one of the following
conditions:

(A) $\,a\,$ is strongly nilpotent ;
(B) $\,a\,$ is central and quasi-nilpotent ;
(C) $\,a\in{\rm nil}\,(R)$, all elements in $\,aR\,$ are suitable, and
$\,ae=ea\,$ for every $\,e\in{\rm idem}\,(R)$.

Then $\,a\in{\rm rad}\,(R)$, and hence $\,{\rm sr}\,(a)=1\,$ by Theorem
2.6(2).

Proof. (A) If $\,a\in R\,$ is strongly nilpotent, Levitzki’s result [Lv:
Theorem 1] guarantees that $\,a\,$ is in the lower nil-radical of $\,R$. (For
an easily accessible proof for this fact, see [La2: Exercise 10.17].) In
particular, we have $\,a\in{\rm rad}\,(R)$.

(B) If $\,a\,$ is central and quasi-nilpotent, then $\,1-as\in{\rm U}(R)\,$
for every $\,s\in R\,$ (since we do have $\,as=sa\,$ here by assumption). By
[La1: Lemma 4.1], this shows that $\,a\in{\rm rad}\,(R)$.

(C) We first show that $\,aR\cap{\rm idem}\,(R)=\\{0\\}$. Indeed, if
$\,e=ax\in{\rm idem}\,(R)\,$ for some $\,x\in R$, then an easy induction shows
that $\,e=a^{n}x^{n}\,$ (noting that $\,e=a^{n}x^{n}\Rightarrow
e=e(ax)=aex=a(a^{n}x^{n})x=a^{n+1}x^{n+1}$). Thus, taking $\,n\,$ to be a
large integer shows that $\,e=0$. To prove that $\,a\in{\rm rad}\,(R)$, it
suffices to show that $\,(1-ay)\,R=R\,$ for every $\,y\in R\,$ (again by [La1:
Lemma 1.4]). Since $\,ay\,$ is a suitable element (by assumption), there
exists an idempotent $\,f\in ayR\,$ such that $\,1-f\in(1-ay)R$. By the first
statement in this paragraph, we must have $\,f=0$, and so $\,1\in(1-ay)R$, as
desired. ∎

While the elements $\,a\in R\,$ studied in Theorem 5.1 all turned out to be in
$\,{\rm rad}\,(R)$, we will next study the case of regular nilpotent elements
; that is, elements $\,a\in{\rm reg}\,(R)\cap{\rm nil}\,(R)$. This time,
$\,a\,$ is no longer in $\,{\rm rad}\,(R)\,$ unless $\,a=0\,$ (according to
[La4: Theorem 2.20]). For such elements, we have the following two known
results (1) and (2) from the literature.

Theorem 5.2. Let $\,a\in{\rm reg}\,(R)$, with $\,a^{n}=0\,$ for a given
integer $\,n\geq 1$.

(1) If $\,a^{i}\in{\rm reg}\,(R)\,$ for all $\,i\geq 1$, then $\,a\,$ is a
product of $\,n\,$ idempotents in $\,R\,$. In this case, $\,a^{i}\in{\rm
ureg}\,(R)\,$ and $\,{\rm sr}\,(a^{i})=1\,$ for all $\,i\geq 1$.
(2) If $\,R\,$ is an exchange ring, then $\,a\in{\rm ureg}\,(R)$, and $\,{\rm
sr}\,(a^{i})=1\,$ for all $\,i\geq 1$.

Proof. (1) is essentially a result of Beidar, Hannah, O’Meara and Raphael; see
[HO: (1.2)] and [BOR: (3.7)]. For an exposition on this result, see [La4:
(2.46), (6.40)]. Here, we will offer a quick and completely self-contained
proof for (1) in the special case where $\,n=2$. Writing $\,a=axa\,$ for some
$\,x\in R$, let $\,e=ax\in{\rm idem}\,(R)$, and $\,f=1-e$. Then
$\,ae=a\,(ax)=0$, so $\,(e+a)\,f=a\,(1-e)=a$. Here, $\,(e+a)^{2}=e+ea=e+a$, so
$\,a\,$ is a product of two idempotents. Also,
$\,a\,(1+fxe)\,a=a\,(1-e)\,xa=a\,$ shows that $\,a\in{\rm ureg}\,(R)\,$ since
$\,1+fxe\,$ is a unipotent unit. Thus, $\,{\rm sr}\,(a)=1\,$ by Theorem 4.1.
Finally, in the case (2) where $\,R\,$ is assumed to be an exchange ring, the
fact that $\,a\in{\rm reg}\,(R)\cap{\rm nil}\,(R)\Rightarrow a\in{\rm
ureg}\,(R)\,$ is a significant result of Ara from [Ar: Theorem 2]. In this
case, it follows from Theorem 4.1 and Theorem 2.8 that $\,{\rm
sr}\,(a^{i})=1\,$ for all $\,i\geq 1$.∎

In practice, there are many concrete examples of non-central nilpotent
elements $\,a\in R\,$ with the property that $\,{\rm sr}_{R}(a)=1$. In Example
5.3 below, we will work in a $\,2\times 2\,$ matrix ring to produce some such
elements that are even unit-regular.

Example 5.3. For any square-zero element $\,s\in S$, consider the square-zero
matrix $\,A\\!:=\footnotesize{\begin{pmatrix}s&1\\\ 0&-s\end{pmatrix}}\in
R={\mathbb{M}}_{2}(S)$. This matrix has a unit inner inverse
$\footnotesize{\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}}$. Thus, $\,A\in{\rm
ureg}\,(R)$, and so $\,{\rm sr}_{R}(A)=1\,$ by Theorem 4.1. More generally,
for any $\,t\in S$, the “equivalent” matrix
$\,B\\!:=A\,\footnotesize{\begin{pmatrix}1&t\\\
0&1\end{pmatrix}}=\footnotesize{\begin{pmatrix}s&1+st\\\ 0&-s\end{pmatrix}}$
continues to be unit-regular. So again $\,{\rm sr}_{R}(B)=1$, and hence
$\,{\rm sr}_{R}(B^{2})=1\,$ too. Here, $\,B^{2}=-sts\,E_{12}\,$ may not be
zero, although we do have $\,B^{3}=0$.

In general, if $\,a\,$ is just a nilpotent element in a ring $\,R$, we may not
have $\,{\rm sr}_{R}(a)=1\,$ even in the case where $\,a\in{\rm reg}\,(R)$.
Indeed, in [NS: Theorem 3.19] and [ArO: Theorem 2.1], an example of a ring
element $\,a\in{\rm reg}\,(R)\setminus{\rm ureg}\,(R)\,$ (in some ring
$\,R\,$) was given such that $\,a^{3}=0$, so in particular $\,{\rm
sr}_{R}(a)\neq 1\,$ by Theorem 4.1. Without assuming regularity, on the other
hand, we will offer below an example of a square-zero element $\,a\,$ in an
abelian ring $\,R\,$ such that $\,{\rm sr}_{R}(a)\neq 1$.

Example 5.4. Let $\,R\,$ be the ring $\,k\,\langle a,x\rangle\,$ generated
over the $2$-element field $\,k={\mathbb{F}}_{2}\,$ with a single relation
$\,a^{2}=0$. A free $\,k$-basis of $\,R\,$ is given by the set of
(noncommutative) monomials in the “variables” $\,a\,$ and $\,x\,$ which do not
involve a factor $\,a^{2}$. Thus, $R\,$ has an obvious
$\,{\mathbb{Z}}^{+}$-graded ring structure. From this, we see easily that
$\,{\rm idem}\,(R)=\\{0,1\\}$; in particular, $R\,$ is an abelian ring.
Assuming a result of Bergman reported in [DL: (2.5)], the unit group $\,{\rm
U}(R)\,$ is given by $\,1+k\,a+aR\,a\,$. (In particular, all units in $\,R\,$
are unipotents, so $\,R\,$ is a “UU ring” in the sense of Cǎlugǎreanu [Ca].)
We claim that $\,{\rm sr}\,(a)\neq 1$. Indeed, if otherwise, Theorem 2.2 would
imply the existence of an element $\,b\in R\,$ such that $\,a+b+axb\in{\rm
U}(R)$. Writing out $\,b\,$ as a “noncommutative polynomial” in $\,a\,$ and
$\,x\,$ in which $\,a^{2}\,$ does not appear, we see that $\,b\,$ must have
constant term $\,1$. If $\,n\geq 0\,$ is chosen largest such that $\,b\,$ has
a monomial term $\,(ax)^{n}$, then $\,a+b+axb\,$ would have a monomial term
$\,(ax)^{n+1}$. This would contradict the fact that $\,{\rm
U}(R)=1+k\,a+aR\,a$, so we have proved that $\,{\rm sr}\,(a)\neq 1$.

Going back to our earlier work (in the proof of Theorem 5.2(1)) on a square-
zero regular element $\,a=axa\,$ in any ring $\,R$, recall that, for the
idempotent $\,e=ax\,$ and its complementary idempotent $\,f=1-e\,$ there, we
have $\,ae=0\,$ and so $\,a=af=eaf\in eRf$. The conclusion that $\,{\rm
sr}\,(a)=1\,$ in that situation prompted us to the discovery of the following
more general result.

Theorem 5.5. Let $\,e,f\in{\rm idem}\,(R)\,$ be such that $\,ef=fe=0$. Then
$\,{\rm sr}\,(erf)=1\,$ for all $\,r\in R$, and $\,{\rm sr}\,(erfse)=1\,$ for
all $\,r,s\in R$.

Proof. It is easy to check that $\,f+erf\in{\rm idem}\,(R)$. This shows that
$\,erf=e\,(f+erf)\,$ is a product of two idempotents in $\,R$. Since
idempotents have stable range one, the Product Theorem 2.8 implies that
$\,{\rm sr}\,(erf)=1$. We have similarly $\,{\rm sr}\,(fse)=1$, so another
application of the Product Theorem gives $\,{\rm sr}\,(erfse)=1$.∎

Corollary 5.6. Let $\,p,q\in R\,$ be such that $\,pq=qp=0\,$ and
$\,u\\!:=p+q\in{\rm U}(R)$. Then for $\,a\\!:=prq\,$ where $\,r\,$ is any
element of $\,R$, we have $\,{\rm sr}\,(a)=1$.

Proof. Note that the elements $\,p,q,u\,$ pairwise commute. If we define
$\,e=u^{-1}p\,$ and $\,f=u^{-1}q$, then $\,ef=fe=0\,$ and $\,e+f=1$, so
$\,e,f\,$ are complementary idempotents in $\,R$. Since $\,p=ue\,$ and
$\,q=uf$, we have $\,u^{-1}a=u^{-1}prq=eruf$. Thus, Theorem 5.5 gives $\,{\rm
sr}\,(u^{-1}a)=1$, and hence $\,{\rm sr}\,(a)=1\,$ by Theorem 2.4(B).∎

In some other special situations, we may also exploit the conclusion in
Theorem 5.5 to get useful information on $\,{\rm sr}_{R}(r)\,$ for an element
$\,r\,$ in a corner ring $\,eR\,e$. To do this, let us first recall the
following basic notions in general ring theory. First, two elements $\,x,y\in
R\,$ are said to be equivalent if $\,y=uxv\,$ for some $\,u,v\in{\rm U}(R)$.
Second, two idempotents $\,e,g\in R\,$ are said to be isomorphic if
$\,R\,e\cong Rg\,$ as left $\,R$-modules; or equivalently, if $\,R\,e\cong
Rg\,$ as right $\,R$-modules. By standard ring theory (see, e.g. [La1: Prop.
21.20]), this amounts to the existence of two elements $\,a\in eRg\,$ and
$\,b\in gRe\,$ such that $\,e=ab\,$ and $\,g=b\,a$. In terms of the two basic
notions above, our next result gives a sufficient condition for every element
in a corner ring $\,eR\,e\,$ to have stable range one in a given ring $\,R$.

Theorem 5.7. Let $\,e,\,f\,$ be a pair of complementary idempotents in $\,R\,$
such that $\,e\,$ is isomorphic to some idempotent $\,g\in fRf$. Then any
element $\,r\in eR\,e\,$ is equivalent to some element in $\,eRf$. In
particular, $\,{\rm sr}_{R}(r)=1$.

Proof. Given that $\,e\,$ and $\,g\,$ are isomorphic idempotents, we may
choose the two elements $\,a,b\in R\,$ as in the paragraph preceding the
statement of this theorem. Let $\,h=f-g\in{\rm idem}\,(fRf)$. Upon expanding
$\,(a+b+h)^{2}\,$ into nine terms, six of these terms are zero, leaving us
with

$(a+b+h)^{2}=ab+ba+h^{2}=e+g+h=1.$

In particular, $\,a+b+h\in{\rm U}(R)$. Thus, $\,r\in eR\,e\,$ is equivalent to

$r\,(a+b+h)=ra\in(eR\,e)\,(eRg)\subseteq eRf,$

as claimed. Given this information, the last conclusion that $\,{\rm
sr}_{R}(r)=1\,$ follows from Theorem 5.5 and Theorem 2.4(B).∎

Note that the theorem above may no longer hold in general if we try to remove
the hypothesis on $\,e\,$ being isomorphic to some idempotent $\,g\in fRf$.
For instance, let $\,R\,$ be the commutative ring
$\,{\mathbb{Z}}\times{\mathbb{Z}}$, and let $\,e=(1,0)$, $\,f=(0,1)=1-e$.
Here, $\,e\,$ is certainly not isomorphic to any idempotent in $\,fRf$. For
any integer $\,n\geq 2$, $\,(n,0)\in eR\,e\cong{\mathbb{Z}}\,$ does not have
stable range one in $\,R$, for otherwise the projection map from $\,R\,$ to
$\,eR\,e\,$ would have given $\,{\rm sr}_{\mathbb{Z}}(n)=1$, which is not the
case according to Example 2.5B.

Applying Theorem 5.5 to the case of a full matrix ring
$\,R={\mathbb{M}}_{n}(S)$, we will next prove the following very useful result
which produces many matrices of stable range one in such a matrix ring. Here,
as usual, the $\,E_{ij}$’s denote the matrix units in $\,R$.

Theorem 5.8. Let $\,A=(a_{ij})\in R={\mathbb{M}}_{n}(S)\,$ where $\,n\geq 2$,
and assume that, for some integer $\,k\in[1,n]$, $\,a_{ij}=0\,$ for all
$\,i\neq k\,$ and all $\,j$, and that $\,a_{kj}=0\,$ for some $\,j$. Then
$\,{\rm sr}\,(A)=1$. In particular, $\,{\rm sr}\,\bigl{(}aE_{kj}\bigr{)}=1\,$
for all $\,a\in S\,$ and all $\,k,j$.

Proof. After a row permutation, we may assume in view of Theorem 2.4(B) that
$\,k=1$. Following this by a column permutation, we may likewise assume that
$\,a_{11}=0$. For $\,e\\!:=E_{11}$ and $\,f=I_{n}-e$, we have then $\,A\in
eRf$. Therefore, Theorem 5.5 implies that $\,{\rm sr}\,(A)=1$.444In 2020,
after an initial announcement of our results, a somewhat different proof for
$\,{\rm sr}\,\bigl{(}aE_{kj}\bigr{)}=1\,$ was given by Cǎlugǎreanu and Pop in
[CP1: Proposition 7]. ∎

Remark 5.9. Of course, the last statement in Theorem 5.8 implies that, as long
as $\,n\geq 2$, every matrix in $\,{\mathbb{M}}_{n}(S)\,$ is a sum of
$\,n^{2}\,$ matrices of stable range one. We did not state this as a part of
Theorem 5.8, since there is already a much better result in the literature.
According to Henriksen [He2], every matrix in $\,{\mathbb{M}}_{n}(S)\,$ is in
fact a sum of three invertible matrices, each of which has stable range one by
Example 2.1. The foregoing discussion brings into focus the following
question: which rings $\,R\,$ could have the property that $\,R\,$ is
additively spanned by the set of stable range one elements in $\,R\,$?
Examples of such rings include: (1) all rings of stable range one (for
instance those mentioned earlier in Example 2.5A); (2) the ring
$\,{\mathbb{Z}}\,$; and (3) the matrix rings $\,{\mathbb{M}}_{n}(S)\,$ for
$\,n\geq 2$). Other examples include (4) all clean rings in the sense of
Nicholson [Ni1], and more generally, (5) any ring in which every element is a
sum of units and idempotents (e.g. ${\mathbb{Z}}$,
$\,{\mathbb{Z}}\,[\sqrt{2}\,]$, $\,{\mathbb{Z}}\,[\sqrt{3}\,]$, or any group
ring $\,k\,[G]\,$ where $\,k\,$ is a ring in which every element is a sum of
units). On the other hand, any polynomial ring
$\,R={\mathbb{Z}}\,[x_{1},\dots,x_{n}]\;(n\geq 1)\,$ is easily seen to be a
“non-example”, since a quick application of Theorem 2.4(D) shows that the set
of stable range one elements in $\,R\,$ is $\,\\{0,\,{\pm 1}\\}$, whose
additive span is $\,{\mathbb{Z}}\subsetneq R$.

Example 5.10. Using Theorem 5.8, it is easy to come up with examples of ring
elements of stable range one that are not unit-regular (or even regular). For
instance, in the ring $\,R={\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}$,
$\,A=2E_{11}\,$ has stable range one by Theorem 5.8, but $\,A\notin{\rm
reg}\,(R)\,$ since any equation $\,A=ABA\in R\,$ for $\,B=(b_{ij})\in R\,$
would lead to a “bad” equation $\,2=4\,b_{11}\in{\mathbb{Z}}$.

Example 5.11. This example is a generalization of a result of Cǎlugǎreanu and
Pop [CP1: Theorem 9] on $\,2\times 2\,$ matrices over Bézout rings. Let
$\,R={\mathbb{M}}_{2}(S)\,$ where $\,S\,$ is a commutative ring, and let
$\,p,q\in S\,$ be such that $\,pS+qS=aS\,$ for some $\,a\in S\,$ that is not a
$\,0$-divisor in $\,S$. We claim that the matrix
$\,C=\footnotesize{\begin{pmatrix}p&q\\\ 0&0\end{pmatrix}}\in R\,$ has stable
range one. To see this, we write $\,p=as,\;q=at\,$ for some $\,s,t\in S$, and
$\,a=px-qy\,$ for some $\,x,y\in S$. Since $\,a\,$ is not a $\,0$-divisor, we
have $\,sx-ty=1$, so $\,U\\!:=\footnotesize{\begin{pmatrix}s&t\\\
y&x\end{pmatrix}}\in{\rm U}(R)$. From the factorization $\,C=(aE_{11})\,U$, we
see that $\,{\rm sr}_{R}(C)=1\,$ by using Theorem 2.4(B) and Theorem 5.8.

Example 5.12. The conclusion that $\,{\rm sr}\,\bigl{(}aE_{ij}\bigr{)}=1\,$ in
Theorem 5.8 can be more useful than we might have initially thought. For
instance, applying it to a square-zero matrix

$M=\footnotesize{\begin{pmatrix}0&0&a&b\\\ 0&0&c&d\\\ 0&0&0&0\\\
0&0&0&0\end{pmatrix}}$, viewed as a block matrix of the form
$\footnotesize{\begin{pmatrix}{\bf 0}_{2}&N\\\ {\bf 0}_{2}&{\bf
0}_{2}\end{pmatrix}}$

where $\,{\bf 0}_{2}\,$ denotes the $\,2\times 2\,$ zero matrix and
$\,N\in{\mathbb{M}}_{2}(S)$, we may conclude from Theorem 5.8 that $\,{\rm
sr}\,(M)=1\,$ in any matrix ring
$\,{\mathbb{M}}_{4}(S)={\mathbb{M}}_{2}\bigl{(}{\mathbb{M}}_{2}(S)\bigr{)}$.
Similarly, we see that any block diagonal matrix $\,{\rm diag}\,(N,\,{\bf
0}_{2})\,$ has stable range one in $\,{\mathbb{M}}_{4}(S)$. However, Theorem
5.8 does not give any information about the stable range of, say, $\,{\rm
diag}\,(x,z,0)\,$ in $\,{\mathbb{M}}_{3}(S)$. To handle this case, we will
have to use another technique; see Example 7.1 in §7.

While in this section we have mainly focused our attention on the stable range
one problem for nilpotent elements, we should make the following supplemental
remark about some related results in the literature. Using an element-wise
version of an argument developed by Goodearl and Menal for proving their
“Theorem 5.8” in [GM], one can show that, if $\,a\in R\,$ is strongly
$\,\pi$-regular (in the sense that $\,a^{n}\in Ra^{n+1}\cap a^{n+1}R\,$ for
some $\,n\geq 1$), and if additionally $\,a^{n}\in{\rm reg}\,(R)\,$ for all
$\,n\geq 1$, then $\,a\in{\rm ureg}\,(R)\,$ (and hence $\,{\rm sr}\,(a)=1\,$);
see [CY1: Theorem 5], [BOR: Corollary 3.7], [Khu: Theorem 4], or [La4: Theorem
5.24]. However, the condition $\,a^{2}\in{\rm reg}\,(R)\,$ may already fail
(even in the case where $\,a\in{\rm ureg}\,(R)\,$ with $\,a^{3}=0\,$), as was
shown by Camillo-Yu in [CY1: Example 9], Yu in [Yu2], and Patrício-Hartwig in
[PaH]. For a self-contained exposition on this, see [La4: Example 1.12C].

§6. Stable Range One Elements Via Peirce Decompositions

We offer in this section some more constructive results on stable range one
elements in the context of Peirce decompositions with respect to corner rings.
To prepare ourselves for the Suspension Theorem 6.2, we start by first proving
the following simple lemma.

Lemma 6.1. Let $\,e,f\in{\rm idem}\,(R)\,$ be such that $\,e+f=1$, and let
$\,x\in{\rm U}(eRe)$, $\,y\in{\rm U}(fRf)$. Then $\,x+p+y\in{\rm U}(R)\,$ for
every $\,p\in fR\,e$.

Proof. Let $\,x^{\prime}\,$ be the inverse of $\,x\,$ in $\,eRe$, and let
$\,y^{\prime}\,$ be the inverse of $\,y\,$ in $\,fRf$. Letting
$\,p^{\prime}=-y^{\prime}px^{\prime}\in fR\,e$, we have
$\,yp^{\prime}=-yy^{\prime}px^{\prime}=-fpx^{\prime}=-px^{\prime}$, so

$(x+p+y)\,(x^{\prime}+p^{\prime}+y^{\prime})=x\,x^{\prime}+p\,x^{\prime}+y\,p^{\prime}+y\,y^{\prime}=x\,x^{\prime}+y\,y^{\prime}=e+f=1,$

and similarly, $\,(x^{\prime}+p^{\prime}+y^{\prime})\,(x+p+y)=1$. This shows
that $\,x^{\prime}+p^{\prime}+y^{\prime}\,$ is an inverse of $\,x+p+y\,$ in
$\,R$. (In the more suggestive notation of Peirce decompositions, the “Peirce
matrix” $\,\footnotesize{\begin{pmatrix}x&0\\\ p&y\end{pmatrix}}$ is
invertible, with inverse given by the Peirce matrix
$\,\footnotesize{\begin{pmatrix}x^{\prime}&0\\\
p^{\prime}&y^{\prime}\end{pmatrix}}$ since
$\,px^{\prime}+yp^{\prime}=0=p^{\prime}x+y^{\prime}p$.) ∎

The first main result of this section is the “Suspension Theorem” below for
ring elements of stable range one in a corner ring context. In the “iff”
statement of this theorem, the “if” part was essentially first discovered (in
the special case $\,p=0\,$) by Vaserstein in [Va2: Theorem 2.8]. Although
Vaserstein did not stress (or even talk about) the stable range one notion for
elements, a careful examination of the proof for his “Theorem 2.8” will reveal
that he had actually proved the “if” part of the result below in the special
case where $\,p=0$. Here, we offer a new proof for the “if” part (for all
$\,p\in fR\,e$) using rather different ideas. Also, we will complete this “if”
part by proving its converse, which then makes for a rather satisfactory “if
and only if” statement in the theorem below. The name of this theorem came
about since we may think of the process of adding the idempotent $\,p+f\,$ to
$\,a\in eR\,e\,$ as a kind of “suspension” of the element $\,a$.

Suspension Theorem 6.2. Let $\,e,f\in{\rm idem}\,(R)\,$ be such that
$\,e+f=1$, and let $\,a\in eR\,e$, $\,p\in fR\,e$. Then $\,{\rm
sr}_{eR\,e}(a)=1\,$ iff $\,{\rm sr}_{R}(a+p+f)=1$. In this case, we have
necessarily $\,{\rm sr}_{R}(a)=1$.

Proof. It suffices to prove the “iff” statement since, with $\,p=0$, $\,{\rm
sr}_{R}(a+f)=1\,$ and $\,{\rm sr}_{R}(e)=1\,$ imply that $\,1={\rm
sr}_{R}(e\,(a+f))={\rm sr}_{R}(a)\,$ by the Product Theorem 2.8.

First, assume that $\,{\rm sr}_{eR\,e}(a)=1\,$ and $\,p\in fR\,e$. To show
that $\,{\rm sr}_{R}(a+p+f)=1$, we take any $\,s\in R$, and want to produce an
element $\,r\in R\,$ satisfying

$a+p+f+(1-(a+p+f)\,s)\,r\in{\rm U}(R).$ $None$

Take $\,r\in eR\,e\,$ such that $\,a+(e-a\,(ese))\,r\in{\rm U}(eRe)$. We have

$\displaystyle a+p+f+(1-(a+p+f)\,s)\,r$ $\displaystyle=$ $\displaystyle
a+p+f+r-asr-psr-fsr$ $\displaystyle=$ $\displaystyle(a+er-aeser)+(p-psr-
fsr)+f.$

Here, $\,a+er-aeser\in{\rm U}(eRe)$, $p-psr-fsr\in fR\,e$, and $\,f\in{\rm
U}(fRf)$. Applying Lemma 6.1, we see that

$a+p+f+(1-(a+p+f)\,s)\,r\in{\rm U}(R).$

This shows that $\,{\rm sr}_{R}(a+p+f)=1$.

Conversely, assume that $\,{\rm sr}_{R}(a+p+f)=1$. Since $\,1-p\in{\rm U}(R)$,
we have $\,1={\rm sr}_{R}((a+p+f)\,(1-p))={\rm sr}_{R}(a+f)$. For any $\,s\in
eR\,e$, there exists $\,t\in R\,$ such that
$\,\alpha\\!:=a+f+(1-(a+f)\,(s+f))\,t\in{\rm U}(R)$. By direct computation, we
have

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle
a+f+(e-as)\,t=a+f+(e-as)\,(ete+etf+fte+ftf)$ $\displaystyle=$
$\displaystyle[\,a+(e-as)\,ete\,]+(e-as)(etf)+f\in eR\,e+eRf+fRf,$

which has a Peirce decomposition matrix of the form
$\footnotesize{\begin{pmatrix}k&q\\\ 0&f\end{pmatrix}}$. Writing down the
Peirce matrix of $\,\alpha^{-1}\,$ in the form
$\footnotesize{\begin{pmatrix}w&x\\\ y&z\end{pmatrix}}$, we see that
$\,f=fz=z$, so $\,yq+zf=f\Rightarrow yq=0$. Since $\,yk=0\,$ and $\,yf=0\,$
too, we have $\,y\alpha=0$, and so $\,y=0$. This implies that $\,kw=e$. As
$\,wk=e\,$ too, we conclude that $\,k=a+(e-as)\,ete\in{\rm U}(eR\,e)$. This
serves to show that $\,{\rm sr}_{eRe}(a)=1$, as desired.∎

Remark 6.4. In the setting of Theorem 6.2 above, the condition $\,{\rm
sr}_{R}(a)=1\,$ itself is in general not sufficient to imply that $\,{\rm
sr}_{eR\,e}(a)=1$. For instance, if we take
$\,R={\mathbb{M}}_{2}\bigl{(}{\mathbb{Z}}\bigr{)}\,$ with $\,e=E_{11}\,$ and
$\,f=E_{22}$, then $\,a\\!:=2E_{11}\in eR\,e\,$ has the property that $\,{\rm
sr}_{R}(a)=1\,$ by Theorem 5.8, but $\,{\rm sr}_{eR\,e}(a)={\rm
sr}_{\mathbb{Z}}(2)\neq 1\,$ as we have seen in Example 2.5B.

Using the the Suspension Theorem 6.2 in the special case where $\,p=0$, we can
easily retrieve the following well-known classical corner ring result of
Vaserstein [Va2: Theorem 2.8] for rings of stable range one.

Corollary 6.5. If $\,R\,$ is a ring of stable range one, so is every corner
ring $\,eR\,e\,$ of $\,R$.

Proof. Let $\,f=1-e\in{\rm idem}\,(R)$, and consider any $\,a\in eR\,e$. Since
$\,R\,$ has stable range one, we have $\,{\rm sr}_{R}(a+f)=1$. By the
Suspension Theorem, we will automatically have $\,{\rm sr}_{eR\,e}(a)=1$. This
shows that the corner ring $\,eR\,e\,$ also has stable range one.∎

In some sense, Corollary 6.5 is a “companion” to the result of Ara and
Goodearl in [AG], which implies that, in the case where $\,e\,$ is a full
idempotent of $\,R\,$ (defined by the properties that $\,e^{2}=e\,$ and
$\,R\,eR=R\,$), the corner ring $\,eRe\,$ having stable range one would
guarantee that $\,R\,$ itself has stable range one. Combining this with
Corollary 6.5, we see, for instance, that a matrix ring
$\,R={\mathbb{M}}_{n}(S)\,$ (for a fixed integer $\,n$) has stable range one
iff the base ring $\,S\,$ has stable range one. This is another well known
result of Vaserstein in [Va1], which was also reiterated later in [Va2:
Theorem 2.4].

By suitably applying the Suspension Theorem 6.2 and the Product Theorem 2.8,
we can now easily prove the following “self-extensions” of Theorem 6.2.

Theorem 6.6. Let $\,e,f\in{\rm idem}\,(R)\,$ with $\,e+f=1$.

(1) If $\,a\in eR\,e$, $\,p\in fR\,e\,$ and $\,b\in fRf\,$ are such that
$\,{\rm sr}_{eR\,e}(a)=1\,$ and $\,{\rm sr}_{fRf}(b)=1$, then $\,{\rm
sr}_{R}(a+p+b)=1$.

(2) If $\,a\in eR\,e$, $\,p\in fR\,e$ and $\,u\in{\rm U}(fRf)$, then $\,{\rm
sr}_{eR\,e}(a)=1\,$ iff $\,{\rm sr}_{R}(a+p+u)=1$.

Proof. (1) Since $\,{\rm sr}_{eRe}(a)=1$, the Suspension Theorem 6.2 shows
that $\,{\rm sr}_{R}(a+p+f)=1$. Similarly, $\,{\rm sr}_{fRf}(b)=1\,$ shows
that $\,{\rm sr}_{R}(e+b)=1$. Since

$(a+p+f)\,(e+b)=ae+p\,e+fb=a+p+b,$

the Product Theorem 2.8 yields the desired conclusion that $\,{\rm
sr}_{R}(a+p+b)=1$.

(2) The “only if” part follows from (1) and the trivial fact that $\,{\rm
sr}_{fRf}(u)=1$. For the “if” part, assume that $\,{\rm sr}_{R}(a+p+u)=1$.
Letting $\,v\,$ be the inverse of $\,u\,$ in $\,fRf$, we have $\,{\rm
sr}_{R}(e+v)=1\,$ since $\,e+v\in{\rm U}(R)$. By the Product Theorem 2.8, we
have

$1={\rm sr}_{R}\bigl{(}(a+p+u)\,(e+v)\bigr{)}={\rm sr}_{R}(a+p+f),$

so the Suspension Theorem 6.2 gives $\,{\rm sr}_{eR\,e}(a)=1$.∎

Upon applying Theorem 6.6 to unit-regular elements (instead of just stable
range one elements), we obtain quickly the following result on $\,{\rm
ureg}\,(R)\,$ (for any ring $\,R\,$) which recovers a part of the main theorem
in the paper of Lam and Murray [LM: §2].

Theorem 6.7. Let $\,e,f\in{\rm idem}\,(R)\,$ with $\,e+f=1$, and let $\,a\in
eR\,e$, $\,u\in{\rm U}(fRf)$. Then $\,a+u\in{\rm ureg}\,(R)\,$ iff $\,a\in{\rm
ureg}\,(eR\,e)$.

Proof. The “if” part is trivial, so we need only prove the “only if” part.
Assuming that $\,a+u\in{\rm ureg}\,(R)$, we have $\,{\rm sr}_{R}(a+u)=1\,$ by
Theorem 4.1. Therefore, Theorem 6.6(2) (applied with $\,p=0\,$) shows that
$\,{\rm sr}_{eR\,e}(a)=1$. On the other hand, if $\,r\,$ is any inner inverse
of $\,a+u\,$ in $\,R$, we see easily (by using Peirce’s decomposition theorem)
that $\,ere\,$ is an inner inverse of $\,a\,$ in $\,eR\,e$, so $\,a\in{\rm
reg}\,(eR\,e)$. Coupling this with the information that $\,{\rm
sr}_{eR\,e}(a)=1$, another application of Theorem 4.1 shows that $\,a\in{\rm
ureg}\,(eR\,e)$.∎

Next, we will prove the theorem below on the stable range of certain lower
triangular matrices in matrix rings. In the special case where such matrices
have a zero diagonal, for instance, this result leads to a large supply of
nilpotent matrices with stable range one.

Theorem 6.8. Let $\,R={\mathbb{M}}_{n}(S)$, where $\,n\geq 1\,$ and $\,S\,$ is
any ring. If $\,A=(a_{ij})\in R\,$ is a lower triangular matrix with $\,{\rm
sr}_{S}(a_{ii})=1\,$ for all $\,i\in[1,\,n]$, then $\,{\rm sr}_{R}(A)=1$.

Proof. The proof of this theorem will be carried out by induction on $\,n\geq
1$. For $\,n=1$, the result is trivial. For the inductive step, write $\,A\,$
in the block form $\footnotesize{\begin{pmatrix}A^{\prime}&0\\\
\alpha&a_{nn}\end{pmatrix}}$, where $\,A^{\prime}\in{\mathbb{M}}_{n-1}(S)\,$
satisfies the same hypothesis as $\,A$, and $\,\alpha\,$ is a row vector of
length $\,n-1$. By the inductive hypothesis, $\,A^{\prime}\,$ has stable range
one in $\,{\mathbb{M}}_{n-1}(S)$. Applying Theorem 6.6(1) (with $\,b=a_{nn}\in
fRf\,$ having stable range one) for the two complementary idempotents
$\,e=E_{11}+\cdots+E_{n-1,n-1}\,$ and $\,f=E_{n,n}$, we see that $\,{\rm
sr}_{R}(A)=1$. ∎

Remark 6.9. Needless to say, the theorem above also holds for upper triangular
matrices $\,A=(a_{ij})\,$ satisfying the same hypothesis that $\,{\rm
sr}_{S}(a_{ii})=1\,$ for all $\,i$. After taking such a matrix $\,A\,$ and
applying some row and column permutations to it, we can come up with some
examples of possibly non-nilpotent matrices $\,A\in R={\mathbb{M}}_{n}(S)\,$
with $\,{\rm sr}_{R}(A)=1$. For instance, for $\,n=3$, we may start with
$\,A=\footnotesize{\begin{pmatrix}a&x&y\\\ 0&b&z\\\
0&0&c\end{pmatrix}}\in{\mathbb{M}}_{3}(S)$, where $\,x,y,z\in S\,$ are
arbitrary and (say) $\,\\{a,b,c\\}\subseteq{\rm idem}\,(S)\cup{\rm U}(S)$.
Interchanging the first and third rows of $\,A\,$ and then interchanging the
first and second columns of the resulting matrix, we see that any matrix of
the form $\footnotesize{\begin{pmatrix}0&0&c\\\ b&0&z\\\ x&a&y\end{pmatrix}}$
has stable range one in $\,{\mathbb{M}}_{3}(S)\,$ for any base ring $\,S$.

For our next result, we will prove Theorem 6.10 below which shows that what we
have called “Super Jacobson’s Lemma 3.2” in §3 holds not only for units,
regular elements and unit-regular elements, but also for the set of stable
range one elements in an arbitrary ring $\,S$. The reason for presenting this
result here instead of in §3 is that the proof of this new result for stable
range one elements in any ring $\,S\,$ makes substantial use of the Suspension
Theorem 6.2 in this section.

Theorem 6.10. For any three elements $\,a,b,x\in S$, $\,{\rm
sr}_{S}(a+b-axb)=1\,$ if and only if $\,{\rm sr}_{S}(a+b-bxa)=1$. (In
particular, after replacing $\,a,\,b\,$ by $\,1-a,\;1-b\,$ and letting
$\,x=1$, it follows that $\,{\rm sr}_{S}(1-ab)=1\,$ iff $\,{\rm
sr}_{S}(1-ba)=1$.)

Proof. Working in the matrix ring $\,R={\mathbb{M}}_{2}(S)$, we first observe
that, by an elementary column operation followed by an elementary row
operation, we can bring the matrix
$\,M\\!:=\footnotesize{\begin{pmatrix}1&a\\\ b&c\end{pmatrix}}$ first to
$\,\footnotesize{\begin{pmatrix}1&0\\\ b&c-ba\end{pmatrix}}$, and then to
$\,\footnotesize{\begin{pmatrix}1&0\\\ 0&c-ba\end{pmatrix}}$. Thus, the
Suspension Theorem 6.2 shows that $\,{\rm sr}_{R}(M)=1\,$ iff $\,{\rm
sr}_{S}(c-ba)=1$. To prove Theorem 6.10, we note further that, by an
elementary column operation followed by a switch of rows, we can bring a
matrix $\,P=\footnotesize{\begin{pmatrix}1&-b\\\ 1-ax&a\end{pmatrix}}$ to
$\,Q=\footnotesize{\begin{pmatrix}1&a\\\ 1-bx&-b\end{pmatrix}}$. Thus, $\,{\rm
sr}_{R}(P)=1\,$ iff $\,{\rm sr}_{R}(Q)=1$. Applying now the observation in the
first part of our proof, we see that $\,{\rm sr}_{S}(a-(1-ax)(-b))=1\,$ iff
$\,{\rm sr}_{S}(-b-(1-bx)\,a)=1$. After a sign change, we conclude that
$\,{\rm sr}_{S}(a+b-axb)=1\,$ iff $\,{\rm sr}_{S}(a+b-bxa)=1$. ∎

Example 6.11. In contrast to the last statement in Theorem 6.10, it is not
difficult to produce a ring $\,S\,$ with three elements $\,a,b,c\,$ such that
$\,{\rm sr}_{S}(c-ab)=1\,$ but $\,{\rm sr}_{S}(c-ba)\neq 1$. To get such an
example, we take $\,S={\mathbb{M}}_{2}({\mathbb{Z}})$, $\,a=E_{12}$,
$\,b=E_{11}$ and $\,c=2E_{21}\,$ in $\,S$. With these choices,
$\,c-ab=c=2E_{21}\,$ has stable range one by Theorem 5.8. On the other hand,
$\,c-ba=\footnotesize{\begin{pmatrix}0&-1\\\ 2&0\end{pmatrix}}$ is matrix-
equivalent to $\,{\rm diag}\,(2,1)$, which according to Theorem 6.6(2) does
not have stable range one in $\,S={\mathbb{M}}_{2}({\mathbb{Z}})\,$ since
$\,{\rm sr}_{\mathbb{Z}}(2)\neq 1$.

To give an application of Example 6.11, we trace our path back to Corollary
3.10, which proved the invariance of the stable range one property with
respect to the transposition of $\,n\times n\,$ matrices over a commutative
ring $\,S$. In the following, we will offer an example to show that the
commutativity assumption on $\,S\,$ cannot be dropped, even in the case
$\,n=2$. This should not be a huge surprise if we recall, for instance from
[Ja1: p. 24] (c. 1953), that the transpose of a $\,2\times 2\,$ invertible
matrix over a noncommutative ring need not be invertible in general. Such a
classical fact suggests that we should not expect a whole lot about the
preservation of matrix properties under transposition when we work with
matrices over noncommutative rings, as the following concrete example shows.

Example 6.12. Let $\,S\,$ be a ring with three elements $\,a,b,c\,$ such that
$\,{\rm sr}_{S}(c-ab)=1$, but $\,{\rm sr}_{S}(c-ba)\neq 1$. (For instance, we
can take the ring $\,S\,$ that was constructed in Example 6.11.) In the matrix
ring $\,R\\!:={\mathbb{M}}_{2}(S)$, let
$\,M\\!:=\footnotesize{\begin{pmatrix}1&a\\\ b&c\end{pmatrix}}$, with
transpose $\,M^{T}=\footnotesize{\begin{pmatrix}1&b\\\ a&c\end{pmatrix}}$.
Using the criterion for $\,{\rm sr}_{R}(M)=1\,$ obtained in the proof of
Theorem 6.10 (and its analogue for $\,M^{T}\,$), we see that $\,{\rm
sr}_{R}(M)=1$, but $\,{\rm sr}_{R}(M^{T})\neq 1$.

Example 6.13. Let $\,R\\!:={\mathbb{M}}_{4}(\mathbb{Z})$ and
$\,M\\!:=\footnotesize{\begin{pmatrix}1&a\\\ b&c\end{pmatrix}}\in R$ with
$a,\,b,\,c$ as in Example 6.11. We saw in Example 6.12 that $\,{\rm
sr}_{R}(M)=1\,$. Note that $\,{\rm det}\,(M)=2$ and $\,{\rm
sr}_{\mathbb{Z}}(2)\neq 1\,$. This shows that the determinant of a matrix with
stable range one over a commutative ring may not have stable range one, in
answer to a question raised by a referee of this paper.

§7. Integral Matrices of Stable Range One

By “integral matrices” in the above caption, we mean the matrices in the rings
$\,{\mathbb{M}}_{n}\bigl{(}{\mathbb{Z}}\bigr{)}$. In this section, we shall
determine precisely the set of all such matrices with stable range one, for
any $\,n\geq 1$. To begin with, we first study the case of diagonal matrices
$\,A\,$ over any ring $\,S$. In the case where some diagonal entry of $\,A\,$
is zero, we have the following slightly surprising result.

Theorem 7.1. Let $\,A={\rm diag}\,(a_{1},\dots,a_{n})\in
R={\mathbb{M}}_{n}(S)$, with some $\,a_{i}=0$. Then $\,{\rm sr}_{R}(A)=1$.

Proof. We may assume, of course, that $\,n\geq 2$. Without loss of generality,
we may also assume that $\,a_{n}=0$. After switching the first and the last
rows, we obtain a matrix $\,B\,$ with last row $\,(a_{1},0,\dots,0)$, and with
northwest $\,(n-1)\times(n-1)\,$ block given by $\,D={\rm
diag}\,(0,a_{2},\dots,a_{n-1})$. By invoking an inductive hypothesis here, we
may assume that $\,{\rm sr}\,(D)=1\,$ in $\,{\mathbb{M}}_{n-1}(S)$. Applying
now Theorem 6.6(1) (with respect to the idempotent
$\,e=E_{11}+\cdots+E_{n-1,n-1}$), we see that $\,{\rm sr}_{R}(A)={\rm
sr}_{R}(D)=1$, as desired. ∎

In the paper [CP1] of Cǎlugǎreanu and Pop, $2\times 2$ matrices over
$\,{\mathbb{Z}}\,$ of stable range one were completely determined; namely,
these were shown to be precisely the $\,2\times 2\,$ integral matrices $\,A\,$
with $\,{\rm det}\,(A)\in\\{0\\}\cup{\rm U}({\mathbb{Z}})=\\{0,\,{\pm 1}\\}$.
With the aid of Theorem 7.1, we can now extend this theorem to all $\,n\times
n\,$ matrices over $\,{\mathbb{Z}}$. To do this, recall that according to
Henriksen [He1], a ring $\,S\,$ is called an elementary divisor ring if every
square matrix over $\,S\,$ is equivalent to a diagonal matrix; that is, for
any $\,A\in{\mathbb{M}}_{n}(S)$, there exist $\,U,V\in{\rm GL}_{n}(S)\,$ such
that $\,UAV\,$ is a diagonal matrix. For instance, it is well known that every
commutative PID (principal ideal domain) is an elementary divisor ring.

Theorem 7.2. Let $\,A\in R={\mathbb{M}}_{n}(S)$, where $\,n\geq 1\,$ and
$\,S\,$ is a commutative elementary divisor domain. If $\,{\rm
det}\,(A)\in\\{0\\}\cup{\rm U}(S)$, then $\,{\rm sr}_{R}(A)=1$.555This
implies, in particular, that any nilpotent matrix in $\,R\,$ has stable range
one. In the special case where $\,S={\mathbb{Z}}$, the converse of this
statement also holds.

Proof. To prove the first conclusion, we may assume (in view of the Product
Theorem 2.8) that $\,A={\rm diag}\,(a_{1},\dots,a_{n})$. If $\,{\rm
det}\,(A)\in\\{0\\}\cup{\rm U}(S)$, then either some $\,a_{i}=0\,$ or
$\,A\in{\rm U}(R)$. In the latter case, we have trivially $\,{\rm sr}\,(A)=1$.
In the former case, the same conclusion holds also in view of Theorem 7.1. To
prove the last statement in the theorem, we work now in the case
$\,S={\mathbb{Z}}$, and assume that $\,{\rm sr}\,(A)=1$. Replacing $\,a_{i}\,$
by $\,-a_{i}\,$ if necessary, we may assume that $\,a_{i}\geq 0\,$ for all
$\,i$. If $\,d\\!:={\rm det}\,(A)=a_{1}\cdots a_{n}\notin\\{0\\}\cup{\rm
U}(S)=\\{0,\,{\pm 1}\\}$, then $\,d\geq 2$. Letting

$D={\rm diag}\,(a_{1},\dots,a_{n})\cdot{\rm
diag}\,(a_{2},\dots,a_{n},a_{1})\cdots{\rm
diag}\,(a_{n},a_{1},\,\dots\,a_{n-1})=d\cdot I_{n}\,,$ $None$

we have $\,{\rm sr}\,(D)=1\,$ by Theorem 2.8, since each factor of $\,D\,$
above has stable range one. Applying the stable range one criterion in Theorem
2.2(3), we see that there exists a matrix
$\,B\in{\mathbb{M}}_{n}\bigl{(}{\mathbb{Z}}\bigr{)}\,$ such that

$M\\!:=D+\bigl{(}I_{n}+D\cdot d^{n}I_{n}\bigr{)}\cdot
B=D+(1+d^{n+1})\,B\in{\rm GL}_{n}\bigl{(}{\mathbb{Z}}\bigr{)}.$ $None$

Computing $\,{\rm det}\,(M)\,$ over the factor ring
$\,{\mathbb{Z}}/(1+d^{n+1})\,{\mathbb{Z}}\,$ and noting that $\,{\rm
det}\,(M)={\pm 1}$, we get a congruence relation $\,{\rm
det}\,(D)=d^{n}\equiv{\pm 1}\;\bigl{(}{\rm mod}\;(1+d^{n+1})\bigr{)}$. Since
$\,d\geq 2$, this is clearly impossible, so we have proved that $\,{\rm
det}\,(A)\in\\{0,\,{\pm 1}\\}$, as desired. ∎

By using the theorem above in the special case $\,S={\mathbb{Z}}$, any issue
concerning integral matrices of stable range one can be easily settled through
a simple computation of determinants. The following consequence of Theorem 7.2
gives some good instances of this.

Corollary 7.5. For the matrix ring
$\,R={\mathbb{M}}_{n}\bigl{(}{\mathbb{Z}}\bigr{)}\,$ where $\,n\geq 1$, the
following holds.

(1) A diagonal matrix $\,{\rm diag}\,(a_{1},\dots,a_{n})\in R\,$ has stable
range one iff $\,a_{i}=0\,$ for some $\,i$, or $\,a_{i}\in\\{{\pm 1}\\}\,$ for
all $\,i$.
(2) For any two matrices $\,A,B\in R$, $\,{\rm sr}_{R}(AB)=1\,$ iff $\,{\rm
sr}_{R}(BA)=1\,$.

Proof. (1) follows directly from Theorem 7.2, while (2) also follows from
Theorem 7.2 since $\,{\rm det}\,(AB)={\rm det}\,(A)\cdot{\rm det}\,(B)={\rm
det}\,(BA)$. ∎

Coming back once more to the last “converse” statement in Theorem 7.2, we
should note that, while that statement holds in the special case
$\,S={\mathbb{Z}}$, it may not hold if $\,S\,$ is just a commutative PID. For
instance, if $\,S\,$ is a commutative semilocal PID (e.g. a discrete valuation
domain), it is well known that both $\,S\,$ and $\,R={\mathbb{M}}_{n}(S)\,$
are rings of stable range one. In this case, $\,{\rm sr}_{R}(A)=1\,$ holds for
all $\,A\in R$, but we certainly may not have any conclusion such as $\,{\rm
det}\,(A)\in\\{0\\}\cup{\rm U}(S)$. Finally, we should also point out that, if
$\,S\,$ is commutative domain and $\,A\in R={\mathbb{M}}_{n}(S)\,$ has $\,{\rm
det}\,(A)=0$, we may not have $\,{\rm sr}_{R}(A)=1\,$ either. For instance, if
$\,S\,$ is a polynomial ring $\,k\,[x,y]\,$ over a field $\,k$, the matrix
$\,A=\footnotesize{\begin{pmatrix}x&y\\\ 0&0\end{pmatrix}}$ with $\,{\rm
det}\,(A)=0\,$ will not have stable stable range one in
$\,{\mathbb{M}}_{2}(S)$. The theoretical justification for this statement will
be more fully discussed in our later work.

§8. Open Questions

In spite of the work done in this paper, there are still many significant
unsolved problems in the study of stable range one elements in rings. To
stimulate further work in this area, we collect and formulate three such open
problems below.

The first open problem is prompted by the fact that we have not been able to
fully understand the relationship between nilpotent elements, strongly
$\pi$-regular elements, and elements of stable range one in an arbitrary ring
$\,R$, in spite of some of our partial results in §5 and §6.

Question 8.1. For any given nilpotent or strongly $\,\pi$-regular element
$\,a\in R$, what is a necessary and sufficient condition for $\,{\rm
sr}_{R}(a)=1\,$?

From Theorem 5.1, Theorem 5.2 and Theorem 6.8, we do have some partial answers
to the question above if we impose some further assumptions on the element
$\,a\in R\,$ or on the ring $\,R$. A general answer for Question 8.1, even
just for an exchange ring $\,R$, should lead to an interesting overview for
some of these earlier results in §5 and §6. In the meantime, any such
characterization results in the case of strongly $\pi$-regular elements should
be helpful toward a further understanding of Ara’s highly significant result
[Ar: Theorem 4] that all strongly $\,\pi$-regular rings have stable range one.

A second question we want to raise concerns the behavior of stable range one
elements in a ring $\,R\,$ with respect to the passage from $\,R\,$ to its
various factor rings. In their 1974 paper [FS], Fisher and Snider proved
essentially that a ring element $\,a\in R\,$ is strongly $\,\pi$-regular iff
its images $\,\overline{a}=a+P\,$ are strongly $\,\pi$-regular in $\,R/P\,$
for every prime ideal $\,P\subset R$. In view of this classical result, it
would be tempting to ask the following.

Question 8.2. If a ring element $\,a\in R\,$ is such that its images
$\,\overline{a}=a+P\,$ have stable range one in $\,R/P\,$ for every prime
ideal $\,P\subset R$, does it follow that $\,{\rm sr}_{R}(a)=1\,$?

Our third question concerns the general behavior of the element-wise stable
range one notion under ring extensions.

Question 8.3. Let $\,a\,$ be a given element in a ring $\,R$.

(A) When does there exist a ring $\,K\supseteq R\,$ (with $\,1_{K}=1_{R}\,$)
such that $\,{\rm sr}_{K}(a)=1\,$?

(B) When is it true that $\,{\rm sr}_{K}(a)=1\,$ for every ring $\,K\supseteq
R\,$ (with $\,1_{K}=1_{R}\,$)?

Classically, if $\,R\,$ is a commutative integral domain, the answer to
Question (A) is trivially “always”, as we can take $\,K\,$ to be the quotient
field of $\,R$. More generally, if $\,R\,$ is a commutative ring, the answer
to Question (A) is still “always” since a result of Goodearl and Menal [GM:
Prop. 7.6] guarantees that $\,R\,$ can be unitally embedded into a commutative
ring $\,K\,$ of stable range one. If $\,R\,$ is a noncommutative ring,
however, the situation would become more complicated. In the case where
$\,a\,$ is a product of units and idempotents in $\,R\,$ (in any order), we
will surely have $\,{\rm sr}_{K}(a)=1\,$ in every ring $\,K\supseteq R\,$
(with $\,1_{K}=1_{R}\,$). On the other hand, suppose $\,a,\,b\in R\,$ are such
that $\,ab=1\neq ba$. Then for every ring $\,K\supseteq R\,$ (with
$\,1_{K}=1_{R}\,$), we cannot have $\,{\rm sr}_{K}(a)=1$. Indeed, if $\,{\rm
sr}_{K}(a)=1$, then $\,a=aba\in{\rm reg}\,(K)\Rightarrow a\in{\rm
ureg}\,(K)\,$ according to Theorem 4.1. This would imply that $\,a\in{\rm
U}(K)$, and hence $\,ab=1\in K\Rightarrow ba=1\in K$, a contradiction.

Acknowledgements. It is the authors’ great pleasure to thank Professors Pere
Ara, Ebrahim Ghashghaei, Kenneth Goodearl, Pace Nielsen, and two anonymous
referees, for a number of insightful comments and suggestions on this paper.

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Department of Mathematics
Panjab University
Chandigarh-160014, India

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Department of Mathematics
University of California
Berkeley, CA 94720, USA

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