A Bezout ring of stable range 2 which has square stable range 1

TL;DR

This paper introduces a new class of rings with specific stable range properties and characterizes their structure, linking square stable range 1 to elementary divisor rings and Toeplitz rings.

Contribution

It defines rings of stable range 2 with square stable range 1 and establishes their equivalence to known classes like elementary divisor rings and Toeplitz rings under certain conditions.

Findings

Hermitian rings with square stable range 1 are elementary divisor rings if and only if they are duo rings of neat range 1.

Commutative Hermitian rings are Toeplitz rings if and only if they have square stable range 1.

Abstract

In this paper we introduced the concept of a ring of stable range 2 which has square stable range 1. We proved that a Hermitian ring $R$ which has (right) square stable range 1 is an elementary divisor ring if and only if $R$ is a duo ring of neat range 1. And we proved that a commutative Hermitian ring $R$ is a Toeplitz ring if and only if $R$ is a ring of (right) square range 1.

Equations

$$\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$$

$$S=\left(\begin{array}{cc}a& -b\\ t& a\end{array}\right),K=\left(\begin{array}{cc}{u}^{-1}& 0\\ 0& u^{-1}\end{array}\right).$$

$$(a,b)S=(u,0),(u,0)K=(1,0),$$

$$(a,b)SK=(1,0).$$

$$\left(\begin{array}{cc}a& -b\\ t& a\end{array}\right)\left(\begin{array}{cc}{u}^{-1}& 0\\ 0& u^{-1}\end{array}\right)=\left(\begin{array}{cc}a{u}^{-1}& -b{u}^{-1}\\ -t{u}^{-1}& a{u}^{-1}\end{array}\right)=T$$

$$\left(\begin{array}{cc}{a}_0& -b_0\\ t& a_0\end{array}\right)\left(\begin{array}{cc}{u}^{-1}& 0\\ 0& u^{-1}\end{array}\right).$$

$$T=\left(\begin{array}{cc}a& b\\ x& a\end{array}\right),$$

$$PAQ=\left(\begin{array}{ccccc}{e}_1& 0& \dots & 0& 0\\ 0& e_2& \dots & 0& 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & e_s& 0\\ 0& 0& \dots & 0& A_0\end{array}\right)$$

$$PAQ=\left(\begin{array}{cc}{e}_1& 0\\ 0& e_2\end{array}\right),$$

$$A=\left(\begin{array}{cc}a& b\\ 0& c\end{array}\right),$$

$$PAQ=\left(\begin{array}{cc}1& x\\ y& z\end{array}\right)=A_1.$$

$$\left(\begin{array}{cc}1& 0\\ -y& 1\end{array}\right)A_1\left(\begin{array}{cc}1& -x\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& ac\end{array}\right),$$

$$SPAQT=\left(\begin{array}{cc}1& 0\\ 0& ac\end{array}\right).$$

Full text

A Bezout ring of stable range 2 which has square stable range 1

Bohdan Zabavsky,   Oleh Romaniv

Department of Mechanics and Mathematics

Ivan Franko National University of Lviv, Ukraine

zabavskii@gmail.com, oromaniv@franko.lviv.ua

November, 2018

Abstract: In this paper we introduced the concept of a ring of stable range 2 which has square stable range 1. We proved that a Hermitian ring $R$ which has (right) square stable range 1 is an elementary divisor ring if and only if $R$ is a duo ring of neat range 1. And we proved that a commutative Hermitian ring $R$ is a Toeplitz ring if and only if $R$ is a ring of (right) square range 1. We proved that if $R$ be a commutative elementary divisor ring of (right) square stable range 1, then for any matrix $A\in M_{2}(R)$ one can find invertible Toeplitz matrices $P$ and $Q$ such that $PAQ=\left(\begin{smallmatrix}e_{1}&0\\ 0&e_{2}\end{smallmatrix}\right),$ where $e_{i}$ is a divisor of $e_{2}$.

Key words and phrases: Hermitian ring, elementary divisor ring, stable range 1, stable range 2, square stable range 1, Toeplitz matrix, duo ring, quasi-duo ring.

Mathematics Subject Classification: 06F20,13F99.

1 Introduction

The notion of a stable range of a ring was introduced by H. Bass, and became especially popular because of its various applications to the problem of cancellation and substitution of modules. Let us say that a module $A$ satisfies the power-cancellation property if for all modules $B$ and $C$, $A\oplus B\cong A\oplus C$ implies that $B^{n}\cong C^{n}$ for some positive integer $n$ (here $B^{n}$ denotes the direct sum of $n$ copies of $B$). Let us say that a right $R$-module $A$ has the power-substitution property if given any right $R$-module decomposition $M=A_{1}\oplus B_{1}=A_{2}\oplus B_{2}$ which each $A_{i}\cong A$, there exist a positive integer $n$ and a submodule $C\subseteq M^{n}$ such that $M^{n}=C\oplus B_{1}^{n}=C\oplus B_{2}^{n}$.

Prof. K. Goodearl pointed out that a commutative rind $R$ has the power-substitution property if and only if $R$ is of (right) power stable range 1, i.e. if $aR+bR=R$ than $(a^{n}+bx)R=R$ for some $x\in R$ and some integer $n\geq 2$ depending on $a,b\in R$ [1].

Recall that a ring $R$ is said to have 1 in the stable range provided that whenever $ax+b=1$ in $R$, there exists $y\in R$ such that $a+by$ is a unit in $R$. The following Warfield’s theorem shows that 1 in the stable range is equivalent to a substitution property.

Theorem 1**.**

[1]** Let $A$ be a right $R$-module, and set $E=\mathrm{End}_{R}(A)$. Then $E$ has 1 in the stable range if and only if for any right $R$-module decomposition $M=A_{1}\oplus B_{1}=A_{2}\oplus B_{2}$ with each $A_{i}\cong A$, there exists a submodule $C\subseteq M$ such that $M=C\oplus B_{1}=C\oplus B_{2}$.

A ring $R$ is said to have 2 in the stable range if for any $a_{1},\dots,a_{r}\in R$ where $r\geq 3$ such that $a_{1}R+\dots+a_{r}R=R$, there exist elements $b_{1},\dots,b_{r-1}\in R$ such that $(a_{1}+a_{r}b_{1})R+(a_{2}+a_{r}b_{2})R+\dots+(a_{r-1}+a_{r}b_{r-1})R=R$.

K. Goodearl pointed out to us the following result.

Proposition 1**.**

[1]** Let $R$ be a commutative ring which has 2 in the stable range. If $R$ satisfies right power-substitution, then so does $M_{n}(R)$, for all $n$.

Our goal this paper is to study certain algebraic versions of the notion of stable range 1. In this paper we study a Bezout ring which has 2 in the stable range and which is a ring square stable range 1.

A ring $R$ is said to have (right) square stable range 1 (written $ssr(R)=1$) if $aR+bR=R$ for any $a,b\in R$ implies that $a^{2}+bx$ is an invertible element of $R$ for some $x\in R$. Considering the problem of factorizing the matrix $\left(\begin{smallmatrix}a&0\\ b&0\end{smallmatrix}\right)$ into a product of two Toeplitz matrices. D. Khurana, T.Y. Lam and Zhou Wang were led to ask go units of the form $a^{2}+bx$ given that $aR+bR=R$.

Obviously, a commutative ring which has 1 in the stable range is a ring which has (right) square stable range 1, but not vice versa in general. Examples of rings which have (right) square stable range 1 are rings of continuous real-valued functions on topological spaces and real holomorphy rings in formally real fields [2].

Proposition 2**.**

[2]** For any ring $R$ with $ssr(R)=1$, we have that $R$ is right quasi-duo (i.e. $R$ is a ring in which every maximal right ideal is an ideal).

We say that matrices $A$ and $B$ over a ring $R$ are equivalent if there exist invertible matrices $P$ and $Q$ of appropriate sizes such that $B=PAQ$. If for a matrix $A$ there exists a diagonal matrix $D=\mathrm{diag}(\varepsilon_{1},\varepsilon_{2},\dots,\varepsilon_{r},0,\dots,0)$ such that $A$ and $D$ are equivalent and $R\varepsilon_{i+1}R\subseteq\varepsilon_{i}R\cap R\varepsilon_{i}$ for every $i$ then we say that the matrix $A$ has a canonical diagonal reduction. A ring $R$ is called an elementary divisor ring if every matrix over $R$ has a canonical diagonal reduction. If every $(1\times 2)$-matrix ($(2\times 1)$-matrix) over a ring $R$ has a canonical diagonal reduction then $R$ is called a right (left) Hermitian ring. A ring which is both right and left Hermitian is called an Hermitian ring. Obviously, a commutative right (left) Hermitian ring is an Hermitian ring. We note that a right Hermitian ring is a ring in which every finitely generated right ideal is principal.

Theorem 2**.**

[3]** Let $R$ be a right quasi-duo elementary divisor ring. Then for any $a\in R$ there exists an element $b\in R$ such that $RaR=bR=Rb$. If in addition all zero-divisors of $R$ lie in the Jacobson radical, then $R$ is a duo ring.

Recall that a right (left) duo ring is a ring in which every right (left) ideal is two-sided. A duo ring is a ring which is both left and right duo ring.

We have proved the next result.

Theorem 3**.**

Let $R$ be an elementary divisor ring which has (right) square stable range 1 and which all zero-divisors of $R$ lie in Jacobson radical of $R$, then $R$ is a duo ring.

Proof.

By Proposition 2 we have that $R$ is a right quasi-duo ring. By Theorem 2 we have that $R$ is a duo ring. ∎

Proposition 3**.**

Let $R$ be a Hermitian duo ring. For every $a,b,c\in R$ such that $aR+bR+cR=R$ the following conditions are equivalent:

there exist elements $p,q\in R$ such that $paR+(pb+qc)R=R$;

• 2)

there exist elements $\lambda,u,v\in R$ such that $b+\lambda c=vu$, where $uR+aR=R$ and $vR+cR=R$.

Proof.

1)$\Rightarrow$2) Since $paR+(pb+qc)R=R$ we have $pR+qcR=R$ and since $R$ is a duo ring we have $pR+cR=R$. Than $Rp+Rc=R$, i.e. $vp+jc=1$ for some elements $v,j\in R$. Then $vpb+jcb=b$ and $b-vpb=jcb=cj^{\prime}b=ct$ where $t=j^{\prime}b$ and $jc=cj^{\prime}$. Element $j^{\prime}$ exist, since $R$ is a duo ring. Then $v(pb+qc)=vpb+vqc=b+ct+vqc=b+ct+ck$, where $vqc=ck$ for some element $k\in R$. That is, we have $v(pb+qc)-b=c\lambda$ for some element $\lambda\in R$. We have $b+c\lambda=v(pb+qc)$. Let $u=pb+qc$. We have $b+c\lambda=vu$, where $vR+cR=R$, since $vp+cj^{\prime}=1$ and $uR+aR=R$, since $paR+(pb+qc)R=R$.

2)$\Rightarrow$1) Since $vR+cR=R$ then $Rv+Rc=R$. Let $pv+jc=1$ for some elements $p,j\in R$. Then $pR+cR=R$. Since $b+\lambda c=vu$, we have $pb=p(vu-\lambda c)=(pv)u-p\lambda c=(1-jc)u+p\lambda c=u-ju^{\prime}c+p\lambda c=u+qc$ for some element $q=p\lambda-ju^{\prime}$, where $cu=u^{\prime}c$ for some element $u^{\prime}\in R$. Since $u=pb+qc$, therefore $(pb+qc)R+aR=R$. Since $R$ is an Hermitian duo ring then we have $pR+qR=dR$ where $p=dp_{1}$, $q=dq_{1}$ and $p_{1}R+q_{1}R=R$. Then $p_{1}R+(p_{1}b+q_{1}c)R=R$ since $pR\subset p_{1}R$ and $pR+cR=R$, $p_{1}R+q_{1}R=R$, i.e. we have $p_{1}R+(p_{1}b+q_{1}c)R=R$. Hence, $aR+(p_{1}b+q_{1}c)R$ we have $p_{1}aR+(p_{1}b+q_{1}c)R=R$. ∎

Remark 1**.**

In Proposition 3 we can choose the elements $u$ and $v$ such that $uR+vR=R$.

Proposition 4**.**

Let $R$ be an Hermitian duo ring. Then the following conditions are equivalent:

$R$* is an elementary divisor duo ring;*

• 2)

for every $x,y,z,t\in R$ such that $xR+yR=R$ and $zR+tR=R$ there exists an element $\lambda\in R$ such that $x+\lambda y=vu$, where $vR+zR=R$ and $uR+tR=R$.

Proof.

1)$\Rightarrow$2) Let $R$ be an elementary divisor ring. By [4] for any $a$, $b$, $c$ such that $aR+bR+cR=R$ there exist elements $p,q\in R$ such that $paR+(pb+qc)R=R$.

Since $xR+yR=R$, $zR+tR=R$ and the fact that $R$ is a Hermitian duo ring we have $zR+xR+ytR=R$. By Proposition 3 we have $x+\lambda yt=uv$ where $uR+zR=R$, $vR+ytR=R$. Since $x+(\lambda t)y=x+\mu y=uv$ where $\mu=\lambda t$, we have $uR+zR=R$, $vR+yR=R$.

2)$\Rightarrow$1) Let $aR+bR+cR=R$ and $Rb+Rc=Rd$ and $b=b_{1}d$, $c=c_{1}d$, where $Rb_{1}=Rc_{1}=R$. Since $R$ is a duo ring then $b_{1}R+c_{1}R=R$. So now $dR=Rd$ and $aR+bR+cR=R$, $Rb+Rc=Rd$ we have $aR+dR=R$, i.e. $dd_{1}+ax=1$ for some elements $d_{1},x\in R$. Then $1-dd_{1}\in aR$.

Since $b_{1}R+c_{1}R=R$, by Conditions 2 of Proposition 3 there exists an element $\lambda_{1}\in R$ such that $b_{1}+c_{1}\lambda=vu_{1}$ where $u_{1}R+(1-dd_{1})R=R$ and $vR+dd_{1}R=R$. Since $(1-dd_{1})\in aR$ and $u_{1}R+(1-dd_{1})R=R$. We have $uR+aR=R$. Let $u=u_{1}d$. Since $u_{1}R+aR=R$ and $dR+aR=R$ we have $uR+aR=R$. Since $b_{1}+c_{1}\lambda=vu_{1}$, we have $b+c\mu+vu$, where $\lambda d=d\mu$.

Recall that $vR+dd_{1}R=R$ then $vR+dR=R$. Since $vR+cR=vR+c_{1}dR=vR+c_{1}R$. So $b_{1}+c_{1}\lambda=vu_{1}$ and $b_{1}R+c_{1}R=R$ then $vR+c_{1}R=R$.

Therefore, $vR+cR=R$. This means that the Condition 2 of Proposition 3 is true. By Proposition 3 we conclude that for every $a,b,c\in R$ with $aR+bR+cR=R$ there exist elements $p,q\in R$ such that $paR+(pb+qc)R=R$, i.e. according to [4], $R$ is an elementary divisor ring. ∎

Definition 1**.**

Let $R$ be a duo ring. We say that an element $a\in R\backslash\{0\}$ is neat if for any elements $b,c\in R$ such that $bR+cR=R$ there exist elements $r,s\in R$ such that $a=rs$, where $rR+bR=R$, $sR+cR=R$, $rR+sR=R$.

Definition 2**.**

We say that a duo ring $R$ has neat range 1 if for every $a,b\in R$ such that $aR+bR=R$ there exists an element $t\in R$ such that $a+bt$ is a neat element.

According to Propositions 3, 4 and Remark 1 we have the following result.

Theorem 4**.**

A Hermitian duo ring $R$ is an elementary divisor ring if and only if $R$ has neat range 1.

The term "neat range 1" substantiates the following theorem.

Theorem 5**.**

Let $R$ be a Hermitian duo ring. If $c$ is a neat element of $R$ then $R/cR$ is a clean ring.

Proof.

Let $c=rs$, where $rR+aR=R$, $sR+(1-a)R=R$ for any element $a\in R$. Let $\bar{r}=r+cR$, $\bar{s}=s+aR$. From the equality $rR+sR=R$ we have $ru+sv=1$ for some elements $u,v\in R$. Hence $r^{2}u+srv=r$ and $rsu+s^{2}v=s$ we have $\bar{r}^{2}\bar{u}=\bar{r}$, $\bar{s}^{2}\bar{v}=\bar{s}$. Let $\bar{s}\bar{v}=\bar{e}$. It is obvious that $\bar{e}^{2}=\bar{e}$ and $\bar{1}-\bar{e}=\bar{u}\bar{r}$. Since $rR+aR=R$, we have $rx+ay=1$ for elements $x,y\in R$. Hence $rxsv+aysv=sv$ we have $rsx^{\prime}v+aysv=sv$ where $xs=sx^{\prime}$ for some element $x^{\prime}\in R$. Then $\bar{a}\bar{y}\bar{e}=\bar{e}$, i.e. $\bar{e}\in\bar{a}\overline{R}$. Similarly from the equality $sR+(1-a)R=R$, it follows $\bar{1}-\bar{e}\in(\bar{1}-\bar{a})\overline{R}$. According to [5] $R/cR$ is an exchange ring. Since $R$ is a duo ring, $R/cR$ is a clean ring. ∎

Taking into account the Theorem 3 and Theorem 4 we have the following result.

Theorem 6**.**

A Hermitian ring $R$ which has (right) square stable range 1 is an elementary divisor ring if and only if $R$ is a duo ring of neat range 1.

Let $R$ be a commutative Bezout ring. The matrix $A$ of order 2 over $R$ is said to be a Toeplitz matrix if it is of the form

[TABLE]

where $a,b,c\in R$.

Notice that if $A$ is an invertible Toeplitz matrix, then $A^{-1}$ is also an invertible Toeplitz matrix.

Definition 3**.**

A commutative Hermitian ring $R$ is called a Toeplitz ring if for any $a,b\in R$ there exist an invertible Toeplitz matrix $T$ such that $(a,b)T=(d,0)$ for some element $d\in R$.

Theorem 7**.**

A commutative Hermitian ring $R$ is a Toeplitz ring if and only if $R$ is a ring of (right) square range 1.

Proof.

Let $R$ be a commutative Hermitian ring of (right) square stable range 1 and $aR+bR=R$ for some elements $a,b\in R$. Then $a^{2}+bt=u$, where $u$ is an invertible element of $R$.

Let

[TABLE]

Then

[TABLE]

i.e. we have

[TABLE]

Since

[TABLE]

we have that $T=SK$ is a Toeplitz matrix. So $(a,b)T=(1,0)$. If $a,b\in R$ and $aR+bR=dR$ then by $a=da_{0}$, $b=db_{0}$ and $a_{0}R+b_{0}R=R$ [4]. Then there exists an element $t\in R$ such that $a_{0}+b_{0}t=u$, where $a$ is an invertible element of $R$.

Let

[TABLE]

Note that $T$ is an invertible Toeplitz matrix. Then $(a,b)T=(d,0)$, i.e. $R$ is a Toeplitz ring.

Let $R$ be a Toeplitz ring and $aR+bR=R$. The exists an invertible Toeplitz matrix $T$ such that $(a,b)T=(1,0)$. Let $S=T^{-1}=\begin{pmatrix}x&t\\ y&x\end{pmatrix}$, where $x,y,t\in R$. So $\det T^{-1}=z^{2}+ty=u$ is an invertible element of $R$. Since $(a,b)=(1,0)T^{-1}$, we have $a=x$, $b=t$. By equality $x^{2}+ty=u$ we have $a^{2}+by=u$, i.e. $R$ is a ring of (right) square stable range 1. ∎

Theorem 8**.**

Let $R$ be a commutative ring of square stable range 1. Then for any row $(a,b)$, where $aR+bR=R$, there exists an invertible Toeplitz matrix

[TABLE]

where $x\in R$.

Proof.

By Theorem 7 we have $(a,b)=(1,0)T$ for some invertible Toeplitz matrix $T$. Let $T=\begin{pmatrix}x&t\\ y&x\end{pmatrix}$. Then $a=x$, $b=t$ and $T=\begin{pmatrix}a&b\\ y&a\end{pmatrix}$ is an invertible Toeplitz matrix. ∎

Recall that $GE_{n}(R)$ denotes a group of $n\times n$ elementary matrices over ring $R$. The following theorem demonstrated that it is sufficient to consider only the case of matrices of order 2 in Theorem 7.

Theorem 9**.**

[4]** Let $R$ be a commutative elementary divisor ring. Then for any $n\times m$ matrix $A$ ($n>2$, $m>2$) one can find matrices $P\in GE_{n}(R)$ and $Q\in GE_{m}(R)$ such that

[TABLE]

where $e_{i}$ is a divisor of $e_{i+1}$, $1\leq i\leq s-1$, and $A_{0}$ is a $2\times k$ or $k\times 2$ matrix for some $k\in\mathbb{N}$.

Theorem 10**.**

Let $R$ be a commutative elementary divisor ring of (right) square stable range 1. Then for any $2\times 2$ matrix $A$ one can find invertible Toeplitz matrices $P$ and $Q$ such that

[TABLE]

where $e_{i}$ is a divisor of $e_{2}$.

Proof.

Since $R$ is a Toeplitz ring it is enough to consider matrices of the form

[TABLE]

where $aR+bR+cR=R$. Since $R$ is an elementary divisor ring by [4] there exist elements $p,q\in R$ such that $paR+(pb+qc)R=R$, i.e. $par+(pb+qc)s=1$ for some elements $r,s\in R$. Since $pR+qR=R$ and $rR+sR=R$, by Theorem 8 we have the invertible Toeplitz matrices $P=\begin{pmatrix}p&q\\ *&*\end{pmatrix}$, $Q=\begin{pmatrix}r&*\\ s&*\end{pmatrix}$ such that

[TABLE]

Then

[TABLE]

where $S=\begin{pmatrix}1&0\\ -y&1\end{pmatrix}$ and $T=\begin{pmatrix}1&-x\\ 0&1\end{pmatrix}$ are invertible Toeplitz matrices. So

[TABLE]

Theorem is proved. ∎

Open Question. Is it true that every commutative Bezout domain of stable range 2 which has (right) square stable range 1 is an elementary divisor ring?