J-Noetherian Bezout domain which is not a ring of stable range 1

TL;DR

This paper investigates a specific class of algebraic structures called J-Noetherian Bezout domains, demonstrating the existence of a nonunit adequate element in such domains that are not of stable range 1.

Contribution

It provides the first example of a J-Noetherian Bezout domain lacking stable range 1 with a nonunit adequate element.

Findings

Existence of nonunit adequate element in certain Bezout domains

J-Noetherian Bezout domains can lack stable range 1

New insights into the structure of Bezout domains

Abstract

We proved that in J-Noetherian Bezout domain which is not a ring of stable range 1 exists a nonunit adequate element (element of almost stable range 1

Equations

$$R=\{z_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n+\cdots \mid z_0\in \mathbb{Z},a_i\in \mathbb{Q},i=1,2,\dots \}.$$

$$D(I)=D(I)\cap mspecR=\{M\in mspecR\mid J⊈M\}$$

$$V(I)=V(I)\cap mspecR=\{M\in mspecR\mid J\subseteq M\},$$

$$\overline{R}/\mathrm{rad}\overline{R}\cong \overline{R}/{\overline{P}}_1\oplus \overline{R}/{\overline{P}}_2\oplus \dots \oplus \overline{R}/{\overline{P}}_n.$$

$$\overline{R}={\overline{e}}_1\overline{R}\oplus {\overline{e}}_2\overline{R}\oplus \cdots \oplus {\overline{e}}_n\overline{R}$$

$$\overline{R}={\overline{e}}_1\overline{R}\oplus {\overline{e}}_2\overline{R}\oplus \dots \oplus {\overline{e}}_n\overline{R}.$$

Full text

Therefore, the subject of research of this paper is a Bezout domain which is a comaximal factorization domain.

Let us start with the following Henriksen example [3]

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This domain is a two-dimensional Bezout domain having a unique prime ideal $J(R)$ (Jacobson radical) of height one and having infinitely many maximal ideals corresponding to the maximal ideals of $\mathbb{Z}$. The elements of $J(R)$ are contained in infinitely many maximal ideals of $R$ while the elements of $R\setminus J(R)$ are contained in only finitely many prime ideals. This ring is a ring of stable range 2 and $R$ is not a ring of stable range 1.

The Henriksen example is an example of a ring in which every nonzero nonunit element has only finitely many prime ideals minimal over it. By [7] in Henriksen’s example every nonzero nonunit element has a complete comaximal factorization. The element of $J(R)$ has a complete comaximal factorization $cx^{n}$, where $c\in\mathbb{Q}$ ($cx^{n}$ is a pseudo-irreducible element). Moreover, every nonzero element of $J(R)$ is pseudo-irreducible. The elements of $R\setminus J(R)$ are contained in only finitely many prime ideals and have a complete comaximal factorization corresponding their factorization in $\mathbb{Z}$.

It is shown [7] that an integral domain $R$ has a complete comaximal factorization if either 1) each nonzero nonunit of $R$ has only finitely many minimal primes or 2) each nonzero nonunit of $R$ is contained in only finitely many maximal ideals.

For a commutative ring $R$, let $spec\,R$ and $mspec\,R$ denote the collection of all prime ideals and all maximal ideals of $R$ respectively. The Zariski topology on $spec\,R$ is the topology obtained by taking the collection of sets of the form $D(I)=\{P\in spec\,R\mid J\nsubseteq P\}$ (respectively $V(I)=\{P\in spec\,R\mid J\subseteq P\}$), for every ideal $I$ of $R$ as the open (respectively closed) sets.

A topological $space\,X$ is called Noetherian if every nonempty set of closed subsets of $X$ ordered by inclusion has a minimal element. An ideal $J$ of $R$ is called a $J$-radical ideal if it is the intersection of all maximal ideals containing it. Clearly, $J$-radical ideals of $R$ correspond to closed subset of $mspec\,R$.

When considered as a subspace of $space\,R$, $mspace\,R$ is called a max-spectrum of $R$. So its open and closed subsets are

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and

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respectively. Clearly max-spectrum of $R$ is Noetherian if and only if $R$ satisfies the ascending chain condition for $J$-radical ideals, i.e. $R$ is a $J$-Noetherian ring [9].

For a commutative Bezout domain $R$ this condition is equivalent to a condition that every nonzero nonunit element of $R$ has only finitely many prime ideals minimal over it [9].

Let $R$ be a domain and $a\in R$. Denote by $minspec\,a$ the set prime ideals minimal over $a$.

Let $\overline{R}$ be a connected ring. Suppose that $a$ is not a pseudo-irreducible element, i.e. $a=bc$, where $b\notin U(R)$, $c\notin U(R)$, and $bR+cR=R$. Than $bu+cv=1$ for some elements $u,v\in R$. Then $\bar{b}^{2}\bar{u}=\bar{b}$, where $\bar{b}=b+aR$, $\bar{u}=u+aR$. Since $\bar{b}\bar{u}$ is an idempotent and $\overline{R}$ is connected, we have $\bar{b}\bar{u}=\bar{0}$ or $\bar{b}\bar{u}=\bar{1}$. If $\bar{b}\bar{u}=\bar{0}$ then $bu=at$ for some element $t\in R$. Then $bu=bct$, i.e. $u=ct$. It is impossible since $bu+cv=1$ and $c\notin U(R)$. If $\bar{b}\bar{u}=\bar{1}$ then $aR+bR=R$, i.e. $b\in U(R)$. Thus, we proved that $a$ is pseudo-irreducible element. ∎

We will notice that in a local ring, every nonzero nonunit element is pseudo-irreducible. If $a$ is a element of domain $R$ where $|minspec\,a|=1$ then $a$ is also pseudo-irreducible. In Henriksen’s example $R=\{z_{0}+a_{1}x+a_{2}x^{2}+\dots+a_{n}x^{n}+\dots\mid z_{0}\in\mathbb{Z},\;a_{i}\in\mathbb{Q},\;i=1,2,\dots\}$, $x$ is pseudo-irreducible. Note that 6 is not pseudo-irreducible but $x=6\cdot\frac{1}{6}x$, i.e. a nonunit divisor of a pseudo-irreducible element is not pseudo-irreducible. In the case of a neat element it is not true.

(1)

$PAQ=D$* is diagonal matrix, $D=(d_{ii})$;*

(2)

$d_{i+1,i+1}R\subset d_{ii}R$.

By [11], we have the following result.

By [13], for a Bezout domain we have the following result.

(1)

$R$* is an elementary divisor ring;*

(2)

for any elements $x,y,z,t\in R$ such $xR+yR=R$ and $zR+tR=R$ there exists an element $\lambda\in R$ such that $x+\lambda y=r\cdot s$, where $rR+zR=R$, $sR+tR=R$ and $rR+sR=R$.

According to Theorem 3 we will obtain the following result.

By [12], any adequate element of a commutative Bezout ring is a neat element.

(1)

$a=rs$;

(2)

$rR+bR=R$;

(3)

$s^{\prime}R+bR\neq R$* for any $s^{\prime}\in R$ such that $sR\subset s^{\prime}R\neq R$.*

A domain $R$ is called adequate if every nonzero element of $R$ is adequate.

The most trivial examples of adequate elements are units, atoms in a ring, and also square-free elements [10].

Henriksen observed that in an adequate domain every nonzero prime ideal is contained in an unique maximal ideal [3].

If $b\in M$ we have $a=1\cdot a$ and for each nonunit divisor $s$ of $a$ we have $sR+bR\in M$, i.e. $sR+bR\neq R$.

If $b\notin M$, we have $aR+bR=R$. ∎

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Since any prime ideal of $\mathrm{spec}\,aR$ is contained in a unique maximal ideal, then $\overline{R}/\overline{P}_{i}$ are valuation rings. Moreover, there exist pairwise orthogonal idempotents $\overline{\overline{e}}_{1}$, …, $\overline{\overline{e}}_{n}$, where $\overline{\overline{e}}_{i}\in\overline{R}/\overline{P}_{i}$ such that $\overline{\overline{e}}_{1}+\ldots+\overline{\overline{e}}_{n}=\overline{1}$. Then, by lifting the idempotent $\overline{\overline{e}}_{i}$ modulo $\mathrm{rad}\,\overline{R}$ to pairwise orthogonal idempotents $\overline{e}_{1},\ldots,\overline{e}_{n}\in\overline{R}$ we find that $1-(e_{1}\ldots+e_{n})$ is an idempotent and $1-(e_{1}+\ldots+e_{n})\in\mathrm{rad}\,\overline{R}$, which is possible only if it is zero. Therefore,

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and each $\overline{e}_{i}\overline{R}$ is a homomorphic image of $\overline{R}$, i.e. a commutative Bezout ring. Since any prime ideal of $\overline{R}$ is contained in a unique maximal ideal, then $\overline{e}_{i}\overline{R}$ is a valuation ring. ∎

A minor modification of the proof of Theorem 9 gives us the following result.

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Since any prime ideal of the ring $\overline{R}$ is contained in a finite set of maximal ideals, then $\overline{e}_{i}\overline{R}$ is a semilocal ring. ∎

Obviously, if a commutative ring $R$ is a direct sum of valuation rings $R_{i}$, then $R$ is a commutative Bezout ring. Let $a=(a_{1},\ldots,a_{n})$, $b=(b_{1},\ldots,b_{n})$ be any elements of $R$, where $a_{i},b_{i}\in R_{i}$, $i=1,2,\ldots,n$. Since $R_{i}$ is a valuation ring, then $a_{i}=r_{i}s_{i}$, where $r_{i}R+b_{i}R=R$ and $s_{i}^{\prime}R_{i}+b_{i}R_{i}\neq R_{i}$ for any non-invertible divisor $s_{i}^{\prime}$ of the element $s_{i}$. If $r=(r_{1},\ldots,r_{n})$, $s=(s_{1},\ldots,s_{n})$ then obviously $a=rs$, $rR+bR=R$. For each $i$ such that $s_{i}^{\prime}$ is a non-invertible divisor of $s_{i}\in R_{i}$, we have $s_{i}R_{i}+b_{i}R_{i}\neq R_{i}$. Hence $s^{\prime}R+bR\neq R$, i.e. $a$ is an adequate element.

A ring $R$ is said to be everywhere adequate if any element of $R$ is adequate.

Note that, as shown above, in the case of a commutative ring, which is a direct sum of valuation rings, any element of the ring (in particular zero) is adequate, i.e. this ring is everywhere adequate.

Any ring of stable range 1 is a ring of almost stable 1 [8]. But not every element of stable range 1 is an element of almost stable range 1. For example, let $e$ be a nonzero idempotent of a commutative ring $R$ and $eR+aR=R$. Then $ex+ay=1$ for some elements $x,y\in R$ and $(1-e)ex+(1-e)ay=1-e$, so $e+a(1-e)y=1$. And we have that $e$ is an element of stable range 1 for any commutative ring. However if you consider the ring $R=\mathbb{Z}\times\mathbb{Z}$ and the element $e=(1,0)\in R$ then, as shown above, $e$ is an element of stable range 1, by $R/eR\cong\mathbb{Z}$, and $e$ is not an element of almost stable range 1. Moreover, if $R$ is a commutative principal ideal domain (e.g. ring of integers), which is not of stable range 1, then every nonzero element of $R$ is an element of almost stable range 1.

The first example of a ring of almost stable range 1 is a ring of stable range 1. Also, every commutative principal ideal ring which is not a ring of stable range 1 (for example, the ring of integers) is a ring of almost stable range 1 which is not a ring of stable range 1.

We note that a semilocal ring is an example of a ring of stable range 1. Moreover, a direct sum of rings of stable range 1 is a ring of stable range 1. As a consequence, we obtain the result from the previous theorems.

Open Question. Is it true that every commutative Bezout domain in which any non-zero prime ideal is contained in a finite set of maximal ideals is an elementary divisor ring?