Rings of the right (left) almost stable range 1

TL;DR

This paper introduces the concept of rings with right or left almost stable range 1, develops a canonical matrix reduction theory for such rings, and characterizes new classes of noncommutative elementary divisor rings.

Contribution

It defines rings of almost stable range 1, constructs a canonical diagonal reduction for matrices over these rings, and characterizes when certain classes of rings are elementary divisor rings.

Findings

Every D-adequate Bézout domain has almost stable range 1.

Necessary and sufficient conditions for Hermite D-rings to be elementary divisor rings.

Every L-ring of almost stable range 1 is a ring of right almost stable range 1.

Abstract

We introduce a concept of rings of right (left) almost stable range $1$ and we construct a theory of a canonical diagonal reduction of matrices over such rings. A description of new classes of noncommutative elementary divisor rings is done as well. In particular, for B\'ezout $D$-domain we introduced the notions of $D$-adequate element and $D$-adequate ring. We proved that every $D$-adequate B\'ezout domain has almost stable range $1$. For Hermite $D$-ring we proved the necessary and sufficient conditions to be an elementary divisor ring. A ring $R$ is called an $L$-ring if the condition $RaR = R$ for some $a\in R$ implies that $a$ is a unit of $R$. We proved that every $L$-ring of almost stable range $1$ is a ring of right almost stable range $1$.

Equations

$$A\sim D:=\mathrm{diag}(d_1,\dots ,d_k,0,\dots ,0),$$

$$(a+c\lambda )R+(b+c\mu )R=R.$$

$$a=d{a}_1,b=d{b}_1\text{and}{a}_{1}R+b_{1}R=R.$$

$$aR+duR=aR+bR+cR=aR+(b\lambda +c)R=R,$$

$$\overline{a}\overline{R}+\overline{b}\overline{R}+\overline{c}\overline{R}=\overline{R}\text{and}a\notin J(R).$$

$$aR+(b\lambda +c)R=R,\lambda \in R.$$

$$rR+bR=R,sR+cR=R\text{and}rR+sR=R,$$

$$\overline{b}\overline{R}+\overline{c}\overline{R}=\overline{R},\overline{r}\overline{R}+\overline{s}\overline{R}=\overline{R},\overline{b}\overline{R}+\overline{r}\overline{R}=\overline{R}\text{and}\overline{c}\overline{R}+\overline{s}\overline{R}=\overline{R}.$$

$$(\overline{s}\overline{v}{)}^2=\overline{s}\overline{v}=\overline{e}\text{and}(\overline{r}\overline{u}{)}^2=\overline{r}\overline{u}=\overline{1}-\overline{e}.$$

$$\overline{r}\overline{x}+\overline{b}\overline{y}=\overline{1}$$

$$rR+bR=R,sR+cR=R\text{and}rR+sR=R.$$

$$RaR=bR=Rb,$$

$$RaR+RbR+RcR=R\text{and}c\ne 0,$$

$$dR=(\lambda a+b)R+cR\subseteq R(\lambda a+b)R+RcR.$$

$$RaR=a^{\ast }R=R{a}^{\ast }\ne R.$$

$$r^{\ast }R+d^{\ast }R=hR\subset h^{\ast }R=R{h}^{\ast },$$

$$sR\subset s^{\ast }R\subset d^{\ast }R\text{and}{d}^{\ast }R+aR=h^{\ast }R\ne R.$$

$$RaR+RbR+RcR=R.$$

$$\sum _i\sum _{j}R{a}_{ij}R=\sum _k\sum _{s}R{b}_{ks}R,$$

$$\sum _i\sum _{j}R{a}_{ij}R=\sum _i\sum _j{a}_{ij}^{\ast }R=\sum _i\sum _{j}R{a}_{ij}^{\ast }=\alpha R=R\alpha .$$

$$P\left[\begin{array}[]{cc}a&0\\ b&c\\ \end{array}\right]Q=\left[\begin{array}[]{cc}z&0\\ &*\\ \end{array}\right]\qquad\text{and}\qquad RzR=R.$$

$$\left[\begin{array}[]{cc}\lambda&1\\ 1&0\\ \end{array}\right]\left[\begin{array}[]{cc}a&0\\ b&c\\ \end{array}\right]=\left[\begin{array}[]{cc}\lambda a+b&c\\ a&0\end{array}\right].$$

$$\left[\begin{array}[]{cc}\lambda&1\\ 1&0\end{array}\right]\left[\begin{array}[]{cc}a&0\\ b&c\end{array}\right]Q=\left[\begin{array}[]{cc}d&*\\ &*\end{array}\right],$$

$$\left[\begin{array}[]{cc}d&*\\ &*\end{array}\right]S=\left[\begin{array}[]{cc}z&0\\ &*\end{array}\right]$$

$$RaR+RbR+RcR=R.$$

Full text

The research was supported by UAEU UPAR grants G00002160 and G00003658.

Dedicated to $65^{\text{th}}$ birthday of Professor Volodymyr Shchedryk

1. Introduction

The stable range of a ring is one of the most important invariants of the algebraic $K$-theory, which was introduced by H. Bass [2]. Rings of small stable range are of particular interest and are being actively studied. Therefore, at present, the theory of rings of stable range $1$ and $2$ is a rather extensive branch of the theory of rings and modules. However, it should be noted that while studies on stable range $1$ rings are a very popular topic, the same is not true for stable range $2$ rings (see [30, p. 2]).

Commutative rings with restrictions on their elements, resembling the definition of the stable range, were considered in [21]. Commutative Bézout rings that are of stable range $1.5$, i.e. rings in which for any $a\not=0,b,c\in R$ there exists $r$ such that $(a+rb,c)=1$, were introduced in [25] and actively studied in [5, 23, 26].

However, it is worth emphasizing that W. McGovern [20] is credited with the first study of commutative rings with the properties that non-trivial homomorphic images are a ring of stable range $1$, i.e. rings of almost stable range $1$. This ground breaking study has given impetus to further research in this area. The historical overview and current state of the topic are covered in books [24, 28] and the articles [4, 5, 12, 8, 21, 9].

In the present article, we introduce a notion of non-commutative rings with almost stable range $1$. We show that every ring of stable range $1$ is a ring of right (left) almost stable range $1$. In addition, the concept of an element of right (left) almost stable range $1$ is introduced. Based on the above, we introduce a notion of rings of right (left) almost stable range $1$ and study their properties and connections with other ring properties.

Let $R$ be an associative ring (not necessary commutative) with $1\not=0$, let $U(R)$ be the group of units of $R$ and let $R^{n\times m}$ be the vector space of ${n\times m}$ matrices over $R$ with $n,m\geq 1$. Let $\rm GL_{n}(R)$ be the group of units of the matrix ring $R^{n\times n}$.

The matrix $D:={\rm diag}(d_{1},\ldots,d_{s})\in R^{n\times m}$ means a (possibly rectangular) matrix having $d_{1},\ldots,d_{s}$ (in which $s:=\min(n,m)$) on the main diagonal and zeros elsewhere. By the main diagonal we mean the one beginning at the upper left corner.

Two matrices $A$ and $B$ over a ring $R$ are called equivalent if there exist invertible matrices $P$ and $Q$ over $R$ of suitable sizes such that $A=PBQ$ and it is denoted by $A\sim B$.

According to I. Kaplansky (see [15, p. 465]), a ring $R$ is called an elementary divisor ring if for any matrix $A\in R^{n\times m}$ we have

[TABLE]

in which $d_{i}$ is a total divisor of $d_{i+1}$, i.e. $Rd_{i+1}R\subseteq d_{i}R\cap Rd_{i}$ for each $i=1,\ldots,k-1$. In this case, we say that the matrix $A$ has a canonical diagonal reduction over $R$.

The class of elementary divisor rings is contained in the class of Bézout rings (for example, see [15, 24, 28]), i.e. rings with nonzero unit in which every finitely generated one-sided ideal is a principal one-sided ideal. Note that elementary divisor rings are Hermite rings, i.e. rings in which each $1\times 2$ and $2\times 1$ matrix has a diagonal reduction, i.e. $(a,b)P=(c,0)$ and $Q(a,b)^{T}=(d,0)^{T}$, where $a,b,c,d\in R$ and $P,Q\in\rm GL_{2}(R)$ (see [15, 28]). Every right Hermite ring is a right Bézout ring.

2. Rings of almost stable range $1$

Set R^{2}:=R\times R=\big{\{}(a,b):a,b\in R\big{\}} and R^{3}:=\big{\{}(a,b,c):a,b,c\in R\big{\}}.

A ring $R$ has stable range $1$ if for each pair $(a,b)\in R^{2}$ the equality $aR+bR=R$ implies $(a+b\lambda)R=R$ for some $\lambda\in R$. Similarly, a ring $R$ has stable range $2$ if for each triple $(a,b,c)\in R^{3}$ the equality $aR+bR+cR=R$ implies that there exist $\lambda,\mu\in R$ such that

[TABLE]

An element $a\in R\setminus\{0\}$ has right (left) almost stable range $1$, if for each pair $(b,c)\in R^{2}$ there exists $\lambda\in R$ ($\mu\in R$) such that equality $aR+bR+cR=R$ implies $aR+(b+c\lambda)R=R$ ($Ra+Rb+Rc=R$ implies $Ra+R(b+\mu c)=R$, respectively). If each nonzero element of $R$ is an element of right (left) almost stable range $1$, then $R$ is a ring of right (left) almost stable range $1$.

This notion generalizes the notion of rings of almost stable range $1$, which was introduced by W. McGovern for commutative rings. A right (left) Bézout ring is a ring in which every right (left) finitely generated ideal is a principal right (left) ideal. A Bézout ring is a ring that is both right and left Bézout.

Let $(a,b)\in R^{2}$ be such that $aR+bR=R$. The pair $(a,b)\in R^{2}$ is called right diadem, if there exists $\lambda\in R$ such that for each triple $(a+b\lambda,c,d)\in R^{3}$ with the property $(a+b\lambda)R+cR+dR=R$ there exists $\mu\in R$ such that $(a+b\lambda)R+(c+d\mu)R=R$.

The concept of a left diadem is introduced by analogy. A diadem is a pair that is both right and left diadem at the same time. We say that $R$ is a ring of right dyadic range $1$ if for each pair $(a,b)\in R^{2}$ the equality $aR+bR=R$ implies that the pair $(a,b)$ is a right diadem. Similarly we define rings of left dyadic range $1$. A ring of dyadic range $1$ is a ring that is both of right dyadic range $1$ and of left dyadic range $1$ at the same time (see [30, Definition 2.1 and Definition 2.3]). Note that if $R$ has right almost stable range $1$, then each pair $(a,b)\in R\setminus\{0\}\times R$ is a right diadem, because $aR+(c+b\mu)R=R$, $\lambda=0$, $\mu\in R$.

Lemma 1**.**

Let $R$ be a right Bézout ring. If $R$ has stable range $2$, then for each pair $(a,b)\in R^{2}$ there exists a triple $(d,a_{1},b_{1})\in R^{3}$ such that

[TABLE]

Proof.

Since $R$ is a right Bézout ring, $aR+bR=dR$ for some $d\in R$. Clearly, $a=da_{0}$, $b=db_{0}$ and $au+bv=d$ for some $d,a_{0},b_{0},u,v\in R$. Thus $da_{0}u+db_{0}v=d$ and $dc=0$, where $c:=1-a_{0}u-b_{0}v$. Consequently $a_{0}R+b_{0}R+cR=R$. Since $R$ has stable range $2$, we have $(a_{0}+c\lambda)R+(b_{0}+c\mu)R=R$ for some $\lambda,\mu\in R$. Finally, setting $a_{1}:=a_{0}+c\lambda$ and $b_{1}:=b_{0}+c\mu$, we obtain $a_{1}R+b_{1}R=R$, $a=da_{1}$ and $b=db_{1}$. ∎

Theorem 1**.**

The following conditions hold.

• (i)

Each right Bézout ring of stable range $1$ is a ring of right almost stable range $1$.

• (ii)

Each ring of right (left) almost stable range $1$ is a ring of right (left) dyadic range $1$.

• (iii)

Each right (left) Bézout ring of a right (left) almost stable range $1$ is a ring of stable range $2$.

Proof.

Let $aR+bR+cR=R$ and $bR+cR=dR$. Each ring of stable range $1$ is a ring of stable range $2$ (see [2, p. 14]), so $b=db_{1}$, $c=dc_{1}$ and $c_{1}R+b_{1}R=R$ for some $d,c_{1},b_{1}\in R$ by Lemma 1. Since $R$ has stable range $1$, from $c_{1}R+b_{1}R=R$ it follows that $b_{1}\lambda+c_{1}=u\in U(R)$ for some $\lambda\in R$. That yields $db_{1}\lambda+dc_{1}=du$, i.e. $b\lambda+c=du$. Since $aR+bR+cR=R$ and $bR+cR=dR$, we have

[TABLE]

i.e. $R$ is a ring of right almost stable range $1$.

Each Bézout ring of right dyadic range $1$ is a ring of stable range $2$ (see [30, Theorem 2.1]). Two other statements are obvious. ∎

Theorem 2**.**

A ring $R$ has right almost stable range $1$ if and only if $R/J(R)$ has right almost stable range $1$, where $J(R)$ is the Jacobson radical of $R$.

Proof.

Let $\overline{x}:=x+J(R)\in\overline{R}:=R/J(R)$, where $x\in R$. Let $(a,b,c)\in R^{3}$ be such that

[TABLE]

That yields $aR+bR+cR=R$ and $a\neq 0$. Since $R$ has right almost stable range $1$, we get

[TABLE]

Thus $\overline{a\vphantom{b}}\,\overline{R\vphantom{b}}+(\overline{b}\,\overline{\lambda}+\overline{c\vphantom{b}})\overline{R\vphantom{b}}=\overline{R\vphantom{b}}$, so $\overline{R\vphantom{b}}$ has right almost stable range $1$. The rest of the proof is similar. ∎

Theorem 3**.**

Let $R$ be a right Bézout ring. If for each $(a,b)\in R^{2}$ with the property $aR+bR=R$ there exists $\lambda\in R$ such that $a+b\lambda$ is an element of right almost stable range $1$, then $R$ is a ring of stable range $2$.

Proof.

Let $(a,b,c)\in R^{3}$ such that $aR+bR+cR=R$. Since $R$ is a right Bézout ring, we have $bR+cR=dR$ for some $d\in R$, so $aR+dR=R$. By our assumption there exists $\lambda\in R$ such that the element $\nu:=a+d\lambda$ has right almost stable range $1$. From $bR+cR=dR$ we obtain that $a+bx+cy=\nu$ for some $x,y\in R$, so $\nu R+bR+cR=R$, because $aR+bR+cR=R$.

By the definition of an element of right almost stable range $1$ we have $\nu R+(b+c\mu)R=R$ for some $\mu\in R$. Let $\nu s+(b+c\mu)t=1$ for some $s,t\in R$. Since $Rs+Rt=R$, we have $Rs+R(xs+t)R=R$.

Let $us+v(xs+t)=ys+\mu t$ for some $u,v\in R$. Thus $(a+cu)s+(b+cv)(xs+t)=1$, i.e. $(a+cu)R+(b+cv)R=R$, so $R$ is a ring of stable range $2$. ∎

3. Case of commutative rings

Let $R$ be a commutative ring. Then in (1) each $d_{i}$ is a divisor of $d_{i+1}$ for $i=1,\ldots,k-1$. According to W. McGovern [20, p. 393], we say that a commutative ring $R$ has almost stable range $1$, if its every proper homomorphic image has stable range $1$. We have the following assertion.

Theorem 4**.**

For a commutative ring $R$ the following statements are equivalent:

• (i)

$R$* is a ring of almost stable range $1$,*

• (ii)

for every $(a,b,c)\in R^{3}$ with the properties $a\neq 0$ and $aR+bR+cR=R$, there exists $\lambda\in R$ such that $aR+(b+c\lambda)R=R$.

Proof.

The proof of the part $\textsl{(i)}\Rightarrow\textsl{(ii)}$ can be found in [20, Theorem 3.6] putting $a\neq 0$.

Let $(a,b,c)\in R^{3}$ such that $aR+bR+cR=R$ and $a\neq 0$. Assume there exists $\lambda\in R$ such that $aR+(b+c\lambda)R=R$.

Set $\overline{R\vphantom{b}}:=R/aR$ and $\overline{x}:=x+aR$, where $x\in R$. If $(b,c)\in R^{2}$ such that $\overline{b}\,\overline{R\vphantom{b}}+\overline{c\vphantom{b}}\,\overline{R\vphantom{b}}=\overline{R\vphantom{b}}$, then $aR+bR+cR=R$. According to the assumption we have $aR+(b+c\lambda)R=R$ for some $\lambda\in R$. Evidently $(\overline{b}+\overline{c\vphantom{b}}\,\overline{\lambda})\overline{R\vphantom{b}}=\overline{R\vphantom{b}}$, where $\overline{b}=b+aR$, i.e. $\overline{R\vphantom{b}}$ is a ring of stable range $1$. If $I$ is a nonzero ideal of $R$, then for every $0\not=a\in I$ we have the isomorphism R/I\big{/}aR/I\simeq R/aR. If $R/aR$ is a ring of stable range $1$, then $R/I$ has stable range $1$ as well. ∎

A ring is called clean if its every nonzero element is a sum of a unit and an idempotent [22]. A ring in which every proper homomorphic image is clean is called neat (see [19, p. 248]). An element $a\in R$ is called neat if $R/aR$ is a clean ring.

An $R$-module $M$ has finite exchange property if for any $R$-module $N$ with the decomposition $N=M^{\prime}\oplus P=\bigoplus_{i\in I}Q_{i}$ in which $M^{\prime}\cong M$ and $I$ is finite, there exist submodules $Q_{i}^{\prime}\subseteq Q_{i}$ for each $i\in I$, such that $N=M^{\prime}\oplus(\oplus Q_{i}^{\prime})$. A ring $R$ is called exchange if its regular module $R_{R}$ is a finite exchange $R$-module.

A commutative ring $R$ has neat range $1$ if for each pair $(a,b)\in R^{2}$ with the property $aR+bR=R$ there exists a neat element $c\in R$ such that $a+bt=c$ for some $t\in R$. For the properties of the above rings and their relationship with other types of rings, see [1, 16, 17, 19, 31, 32].

Theorem 5**.**

Let $R$ be a commutative Bézout ring. If for each $(a,b,c)\in R^{3}$ with the properties $aR+bR+cR=R$ and $a\in R\backslash\{0\}$ there exists a decomposition $a=rs$ for some $r,s\in R$ such that

[TABLE]

then $R/aR$ is a clean ring.

Proof.

Set $\overline{R}:=R/aR$ and $\overline{x}:=x+aR$, where $x\in R$. Since $aR+bR+cR=R$ and (2) holds, we get

[TABLE]

It follows that $\overline{r}\,\overline{u}+\overline{s}\,\overline{v}=\overline{1}$ and $\overline{r}\,\overline{s}=\overline{0}$ for some $u,v\in R$. This yields $\overline{r}^{2}\overline{u}=\overline{r}$ and $\overline{s}^{2}\overline{v}=\overline{s}$. Obviously,

[TABLE]

Taking into account that $\overline{r\vphantom{b}}\,\overline{R\vphantom{b}}+\overline{b}\,\overline{R\vphantom{b}}=\overline{R\vphantom{b}}$, we get

[TABLE]

for some $\overline{x},\overline{y}\in\overline{R}$. Thus $\overline{s\vphantom{b}}\,\overline{v\vphantom{b}}\,\overline{r\vphantom{b}}\,\overline{x\vphantom{b}}+\overline{s\vphantom{b}}\,\overline{v\vphantom{b}}\,\overline{b}\,\overline{y\vphantom{b}}=\overline{s\vphantom{b}}\,\overline{v\vphantom{b}}$ and $\overline{e\vphantom{b}}\,\overline{b}\,\overline{y\vphantom{b}}=\overline{e\vphantom{b}}$, i.e. $\overline{e\vphantom{b}}\in\overline{b}\,\overline{R\vphantom{b}}$. Using the fact that $\overline{s\vphantom{R}}\,\overline{R}+\overline{c\vphantom{R}}\,\overline{R}=\overline{R}$, we obtain $\overline{1}-\overline{e\vphantom{R}}\in\overline{c\vphantom{R}}\,\overline{R}$, so $\overline{R}$ is an exchange ring by [15, Proposition 1.1 and Theorem 2.1]. Finally, $\overline{R}$ is a clean ring by [19, p. 244]. ∎

In the case when $R$ is a domain, the following result is known.

Theorem 6** (​​[32, Theorem 31]).**

Let $R$ be a commutative Bézout domain and let $a\in R\backslash\{0\}$. If $R/aR$ is a clean ring, then for each $(b,c)\in R^{2}$ with the property $aR+bR+cR=R$ there exists a decomposition $a=rs$ for some $r,s\in R$ such that

[TABLE]

Now, using Theorems 5 and 6 we have the following.

Theorem 7**.**

Let $R$ be a commutative Bézout domain. The following statements hold:

• (i)

$R$* has neat range $1$ if and only if for each $(a,b)\in R^{2}$ with the property $aR+bR=R$ there exists $\lambda\in R$ such that $R/(a+b\lambda)R$ is clean;*

• (ii)

$R$* is an elementary divisor ring if and only if $R$ has neat range $1$ (see [32, Theorem 33]);*

• (iii)

if the stable range of $R/aR$ is not $1$ for each $a\in R\setminus\{0\}$, then $R$ is not an elementary divisor ring (see **[31, Theorem 5]**).

Each commutative clean ring $R$ has stable range $1$ (see [19, p. 244] and [22, Proposition 1.8]). Furthermore, a neat ring is a ring of almost stable range $1$ and a neat element is an element of almost stable range $1$, so if $R$ is an elementary divisor domain that is not a ring of stable range $1$, then at $R\setminus U(R)$ there exists a neat element that has almost stable range $1$ (see [29, Theorem 7]).

4. Rings with Dubrovin property

The coboundary of a one-sided ideal $I$ of a ring $R$ is a two-sided ideal which equals to the intersection of all two-sided ideals which contain $I$. Note that this definition is left-right symmetric.

A ring $R$ in which for every $a\in R\setminus\{0\}$ there exists $b\in R$ such that

[TABLE]

(in other words, the coboundary of $R$ is a principal ideal) is called a ring with Dubrovin property ($D$-property). Simple elementary divisor rings [28, Section 4.2], quasi-duo elementary divisor rings [7, Theorem 1] and semi-local semi-primitive elementary divisor rings [11, Theorem 1], [10] are examples of elementary divisor rings with $D$-property.

In [3], the investigation of elementary divisor rings of stable range $1$ with Dubrovin and Dubrovin-Komarnytsky property was done. In the same article, a theory of a canonical diagonal reduction of matrices over such rings was constructed as well, and new families of non-commutative rings of elementary divisor rings were presented.

A two-sided analogue of one-sided rings of almost stable range $1$ is the following.

A ring $R$ has almost stable range $1$ if for each triple $(a,b,c)\in R^{3}$ with the properties

[TABLE]

there exists $\lambda\in R$ such that $R(\lambda a+b)R+RcR=R$.

Theorem 8**.**

Let $R$ be a Bézout ring. The following statements are equivalent:

• (i)

$R$* is a ring of almost stable range $1$;*

• (ii)

for each $(a,b,c)\in R^{3}$ with the properties $RaR+RbR+RcR=R$ and $c\neq 0$, there exist $\lambda,d\in R$ such that $(\lambda a+b)R+cR=dR$ and $RdR=R$.

Proof.

Let $RaR+RbR+RcR=R$ and $(\lambda a+b)R+cR=dR$ in which $RdR=R$ for some $\lambda\in R$. Since $cR\subset RcR$ and $(\lambda a+b)R\subset R(a\lambda+mb)R$, we have

[TABLE]

Hence, $RdR=R$ and $d\in R(\lambda a+b)R+RcR$, so we have $R(\lambda a+b)R+RcR=R$.

Let the condition $RaR+RbR+RcR=R$ imply that $R(\lambda a+b)R+RcR=R$ with $c\neq 0$ and some $\lambda\in R$. It is easy to see that $(\lambda a+b)R+cR=dR$ for some $d\in R$, because $R$ is a Bézout ring. Using the fact that $(\lambda a+b)R\subseteq dR$, we obtain that $cR\subseteq dR$, so $RcR\subseteq RdR$ and $R(\lambda a+b)R\subseteq RdR$. Consequently from $R(\lambda a+b)R+RcR=R$ we obtain that $RdR=R$, which proves the equivalence of (i) and (ii). ∎

A ring $R$ is called an $L$-ring if the condition $RaR=R$ for some $a\in R$ implies that $a$ is a unit of $R$. This definition arise from conditions of [18, Proposition 7.3(2)].

Theorem 9**.**

Every $L$-ring of almost stable range $1$ is a ring of right almost stable range $1$.

Proof.

Let $aR+bR+cR=R$ and $c\neq 0$. Obviously $RaR+RbR+RcR=R$ and $R$ is a ring of almost stable range $1$ by Theorem  8. Thus, $(\lambda a+b)R+cR=dR$ in which $RdR=R$. Since $R$ is an $L$-ring, $(\lambda a+b)R+cR=R$, i.e. $R$ is a ring of a right almost stable range $1$. ∎

Note that every elementary divisor $L$-ring is a ring with $D$-property [28, Theorem 4.7.1]. According to Theorem 1 and Theorem 9 we have the following.

Theorem 10**.**

Every Bézout $L$-ring of almost stable range $1$ has stable range $2$. Moreover, each Bézout ring $R$ of stable range $1$ has almost stable range $1$.

Proof.

The first part follows from Theorems 1 and 9.

Let $RaR+RbR+RcR=R$ and $Ra+Rb=Rd$. Since a ring of stable range $1$ has stable range $2$ (see [2, p. 14]), we have $a=a_{0}d$, $b=b_{0}d$ and $Ra_{0}+Rb_{0}=R$ by Lemma 1. The ring $R$ has stable range $1$ and $Ra_{0}+Rb_{0}=R$, so $a_{0}+\lambda b_{0}=u\in U(R)$. Then $a+\lambda b=ud$, i.e. $R(a+\lambda b)=Rud=Rd$ and $a=a_{0}d$, $b=b_{0}d$. Let $dR+cR=zR$. Similarly we have $dy+c=z$. Then we have $(a+\lambda b)R+cR=zR$. Since $a=a_{0}d+a_{0}zd_{0}$, we have $b=b_{0}d=b_{0}zd_{0}$ and $c=c_{0}z$ for some $d_{0},c_{0}\in R$. It follows that $RaR\subseteq RzR$, $RbR\subseteq RzR$, and $RcR\subseteq RzR$. Since $RaR+RbR+RcR=R$, we have $R=RzR$ and $R$ has almost stable range $1$. ∎

5. Rings with $D$-adequacy property

The notion of the adequacy of commutative domains was introduced by O. Helmer [14]. A.I. Gatalevych [12, Definition 1, p. 116] was the first who extended this notation to non-commutative rings. Now it is known that the generalized right adequate (in the sense of Gatalevych) duo Bézout domain is an elementary divisor domain (see [12, Theorem 2, p. 117]).

In [13], it was explored the problem when a ring of matrices over either adequate rings or elementary divisor rings inherits the property of adequacy.

Let $A$ and $E$ be adequate and elementary divisor domains, respectively. Let $A^{2\times 2}$ and $E^{2\times 2}$ be rings of ${2\times 2}$ matrices over rings $A$ and $E$, respectively. In [13], it was proved that the set of full nonsingular matrices from $A^{2\times 2}$ is an adequate set in $A^{2\times 2}$ and the set of full singular matrices from $E^{2\times 2}$ is an adequate set in the set of full matrices in $E^{2\times 2}$.

In [6], another definition of adequate rings was proposed, which differs from the one proposed by A.I. Gatalevych [12, Definition 1, p. 116].

We propose a new version of the definition of adequacy, which is very close to the definitions of adequacy that were given in [12, 13, 6].

The introduction of new definitions is related to Dubrovin’s condition for non-commutative rings. For a class of duo rings our definition is equivalent to the definition of A.I. Gatalevich [12, Definition 1, p. 116].

Let $R$ be a Bézout $D$-domain and let $a\in R\setminus\{0\}$ be such that

[TABLE]

The element $a$ is called $D$-adequate if for each $b\in R$ the following conditions hold:

• (i)

$a=rs$ for some $r,s\in R$ such that $RrR=r^{*}R=Rr^{*}$, $RsR=s^{*}R=Rs^{*}$, and $RbR=b^{*}R=Rb^{*}$;

• (ii)

$r^{*}R+b^{*}R=R$;

• (iii)

for each non-trivial divisor $s^{\prime*}$ of $s^{*}$ we have $s^{\prime*}R+b^{*}R\neq R$ in which $Rs^{\prime}R=s^{\prime*}R=Rs^{\prime*}$.

A ring $R$ is called $D$-adequate if each element $0\not=a\in R$ with the property $RaR\neq R$ is $D$-adequate.

Theorem 11**.**

Every $D$-adequate Bézout $D$-domain has almost stable range $1$.

Proof.

Let $(a,b,c)\in R^{3}$ such that $RaR+RbR+RcR=R$ and $c\neq 0$. If $RcR=R$, then the statement of our theorem is obvious. Let $RcR\neq R$. It follows that

• (i)

$c=rs$;

• (ii)

$r^{*}R+a^{*}R=R$;

• (iii)

for each non-trivial divisor $s^{\prime*}$ of $s^{*}$ we have $s^{\prime*}R+b^{*}R=R$.

Let $(a+rb)R+cR=dR$. Using the fact that $RdR=d^{*}R=Rd^{*}$, we obtain $dR\subset RdR=d^{*}R=Rd^{*}$. Hence, we obtain $(a+rb)R\subset dR\subset d^{*}R$. If

[TABLE]

then $aR\subset h^{*}R\neq R$. Clearly $h^{*}R$ is a two-sided ideal, so $a^{*}R\subset h^{*}R$, which contradicts to (ii). Hence $r^{*}R+d^{*}R=R$, so $r^{*}u+d^{*}v=1$ and $r^{*}us^{*}+d^{*}vs^{*}=s^{*}$ for some $u,v\in R$. Since $s^{*}$ is a duo element, then $r^{*}s^{*}u^{\prime}+d^{*}s^{*}v^{\prime}=s^{*}$ for some $u^{\prime},v^{\prime}\in R$. It follows that

[TABLE]

It follows that $aR\subset h^{*}R$, $bR\subset h^{*}R$, $cR\subset h^{*}R$ and $h^{*}R\neq R$. Since $h^{*}R=Rh^{*}$, we get $RaR\subset h^{*}R$, $RbR\subset h^{*}R$, and $RcR\subset h^{*}R$, which contradicts the condition

[TABLE]

Thus, we proved that $(a+rb)R+cR=dR$ and $RdR=R$, i.e. $R$ has almost stable range $1$. ∎

6. Hermite $D$-rings

We denote by $A_{*}$ a two-sided ideal in a ring $R$ generated by all elements of the matrix $A=(a_{ij})\in R^{m\times n}$. Evidently $A_{*}=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}Ra_{ij}R$.

We need the following well-known results.

Lemma 2**.**

If $A\sim B$, then $A_{*}=B_{*}$.

Proof.

Let $A=(a_{ij}),B=(b_{ij})\in R^{m\times n}$ be such that $A\sim B$. It follows that $a_{ij}\in\sum_{k}\sum_{s}Rb_{ks}R$ and $b_{ij}\in\sum_{k}\sum_{s}Ra_{ks}R$ for each $i$, $j$, $k$, $s$. Consequently

[TABLE]

i.e. $A_{*}=B_{*}$. ∎

Lemma 3**.**

Each Hermite ring $R$ with $D$-property is an elementary divisor ring if and only if every $A=(a_{ij})\in R^{n\times n}$ with the property $\sum\limits_{i}\sum\limits_{j}Ra_{ij}R=R$ has a canonical diagonal reduction.

Proof.

Since the proof of the “if” part is obvious, we start with the proof of the “only if” part.

Let $Ra_{ij}R=a_{ij}^{*}R=Ra_{ij}^{*}R$ for each $i$, $j$ (see (3)). For some duo-element $\alpha\in R$ we have

[TABLE]

That yields $a_{ij}=\alpha a_{ij}^{0}$ and $A=\mathrm{diag}(\alpha,\dots,\alpha)A_{0}$, where $A_{0}=(a_{ij}^{0})$. Since $R$ is Hermite, it is a Bézout ring of stable range $2$ [28, p. 30, Collorarly $2$.1.3, Theorem $1$.2.40], so $\sum_{i}\sum_{j}Ra_{ij}^{\prime}R=R$ by Lemma 1. ∎

Theorem 12**.**

Let $R$ be a Hermite ring of almost stable range $1$. If $a,b,c\in R$ with the property $RaR+RbR+RcR=R$, then there exist $P,Q\in\rm GL_{2}(R)$ such that

[TABLE]

Proof.

Since every Hermite ring is Bézout [28, Corolary 2.13], then $R$ has almost stable range $1$. For each $(a,b,c)\in R^{3}$ with the property $RaR+RbR+RcR=R$, there exists $\lambda\in R$ such that $(\lambda a+b)R+cR=dR$ by Theorem  8. This yields that

[TABLE]

Set $d:=(\lambda a+b)x+cy$, $a_{0}d:=\lambda a+b$ and $c:=c_{0}d$ for some $a_{0},c_{0},x,y\in R$. Each Hermite ring is a Bézout ring of stable range $2$ [28, Theorem 1.2.40], so according to Lemma 1 we have that $\lambda a+b=a_{1}d$ and $c=c_{1}d$, where $a_{1}R+c_{1}R=R$, i.e. $a_{1}u+c_{1}v=1$ for some $u,v\in R$. Since $Ru+Rv=R$ and $R$ is a Hermite ring, then the column \left[\begin{array}[]{c}u\\ v\\ \end{array}\right] is completable to a matrix Q=\left[\begin{array}[]{cc}u&*\\ v&*\end{array}\right]\in\rm GL_{2}(R) (see [28, Corolary 2.1.7]). Evidently

[TABLE]

where $RdR=R$. Using the fact that $R$ is a Hermite ring, we obtain that

[TABLE]

for an invertible matrix $S$. Taking into account that $dR\subset zR$ and $RdR=R$, we get that $R=RdR\subseteq RzR$, i.e. $RzR=R$. ∎

For a canonical diagonal reduction of matrices over a Hermite ring $R$ (see [15, Theorem 5.1]) it is enough to consider such reduction for matrices of the form \left[\begin{array}[]{cc}a&0\\ b&c\end{array}\right], where $a,b,c\in R$. Therefore, according to Lemma 3, for a canonical diagonal reduction of matrices over a Hermite ring $R$ with $D$-property it is enough to consider such reduction for matrices \left[\begin{array}[]{cc}a&0\\ b&c\end{array}\right], where

[TABLE]

Using Theorem 12 we have the following assertion.

Theorem 13**.**

Let $R$ be a Hermite ring with $D$-property of almost stable range $1$. The ring $R$ is an elementary divisor ring if and only if for every $a,b,c\in R$ with $RaR=R$ the matrix \left[\begin{array}[]{cc}a&0\\ b&c\end{array}\right]\in R^{2\times 2} has a canonical diagonal reduction over $R$.

—-

This result shows that the set of $a\in R$ for which $RaR=R$ is crucial for a canonical diagonal reduction of matrices over $R$.

Example. *Let $R$ be a domain. An element $a\in R\setminus\{0,U(R)\}$ is called atom if $a$ has no proper factors. An element which is either a unit or a product of atoms is called finite. Let $R$ be a Bézout domain with $D$-condition of almost stable range $1$. If for every $a\in R$, the condition $RaR=R$ implies that $a$ is a finite element, then $R$ is an elementary divisor ring. Indeed, according to Theorem 13, we can consider a matrix $A$ of the form \left[\begin{array}[]{cc}a&0\\ b&c\end{array}\right]\in R^{2\times 2}, where $RaR=R$ and $a$ is a finite element. There exist $P,Q\in\rm GL_{2}(R)$ (see [27, Theorem 6]) such that PAQ=\left[\begin{array}[]{cc}z&0\\ 0&d\end{array}\right] in which $RdR\subseteq zR\cap Rz$. *

Acknowledgements

The current article finishes the joint collaboration, which had started prior to Professor Bohdan Zabavsky’s demise in August 2020.