Diagonal reduction of matrices over Bezout rings of stable range 1 with the Kazimirsky condition

TL;DR

This paper develops a theory for diagonalizing matrices over certain Bezout rings with the Kazimirsky condition, establishing conditions under which these rings are elementary divisor rings.

Contribution

It introduces a diagonalizability theory for matrices over Bezout rings with the Kazimirsky condition and characterizes when such rings are elementary divisor rings.

Findings

A ring of stable range 1 with the Kazimirsky condition is an elementary divisor ring if and only if it is a duo ring.

The paper establishes a link between the Kazimirsky condition and elementary divisor rings.

Provides criteria for diagonalizability over specific classes of rings.

Abstract

We constuct the theory of diagonalizability for matrices over Bezout rings of stable range 1 with the Kazimirsky condition. It is shown that a ring of stable range 1 with the right (left) Kazimirsky condition is an elementary divisor ring if and only if it is a duo ring.

Full text

Obviously, an unit-central ring is a ring with the Kazimirsky condition.

Let $Ra=Rau$. Since $R$ is a ring of stable range 1 than by [8] we have $va=au$ for some invertible $v\in R$. ∎

Recall that a ring $R$ is a right (left) quasi-duo ring if any right (left) maximal ideal is an ideals. A right and left quasi-duo ring is called a quasi-duo ring. By Theorem 1 we have the following result.

Since $m\in M$ than $xm\in M$, i.e. $Rn+Rxm\subseteq M$. This is of course impossible, so $M$ must be a two-sided ideal. By [9] the Proposition is proved. ∎

According to the results of [7, 9] we have the following result.

Note that a duo ring is a ring with the Kazimirsky condition. Really, if $R$ is a duo ring and for any $a\in R$ and any invertible $u\in R$ we have $au=xa$ ($ua=ay$) for some $x,y\in R$, i.e. $Rau\subseteq Ra$ ($uaR\subseteq aR$).

Recall that a ring $R$ is said to have the unit stable range 1 if $aR+bR=R$ implies that there exists an invertible $u\in R$ such that $(a+bu)R=R$.

We note that a regular ring $R$ has the unit stable range 1 if and only if it is unit-regular and for any idempotent $e,f\in R$, $eR+fR=R$ implies that there exist invertible $u,v\in R$ such that $eu+fv=1$. Moreover $M_{2}(\mathbb{Z}_{2})$ has the unit stable range 1, while $\mathbb{Z}_{2}$ does not have the unit stable range 1 [1].

A ring of unit stable range 1 is a ring in which any additively generated by its units.

From these results we have the following consequence.

Example. In the article [6] it was proved that if the skew polynomial ring $R[x;\sigma]$ is left or right duo, then $R[x;\sigma]$ is commutative.

Let $R=\mathbb{Z}[\sqrt{-7}]$ and $K=\mathbb{Q}(\sqrt{-7})$ be the quotient field of the ring $R$. Will consider an automorphism $\sigma\colon K\to K$ such that $\sigma(a+b\sqrt{-7})=a-b\sqrt{-7}$, $a,b\in\mathbb{Q}$. We will consider the subring $S$ of the ring $K[x;\sigma]$ which consists of polynomials the free member of which belongs $R$. The only units in the ring $S$ are 1 and $-1$, therefore consequently the ring $S$ is a unit-central ring but not commutative.