On Fusible Rings

TL;DR

This paper provides new characterizations of semiprime left Goldie rings, showing they are equivalent to being regular left fusible with finite Goldie dimension, and addresses previously posed questions in ring theory.

Contribution

It introduces a novel characterization of semiprime left Goldie rings using the concept of regular left fusibility and Goldie dimension, answering open questions.

Findings

Semiprime left Goldie rings are characterized by regular left fusibility and finite Goldie dimension.

Two previously posed questions in the literature are answered negatively.

The paper establishes a new equivalence involving fusible rings and Goldie rings.

Abstract

We answer in negative two of questions posed in [4]. We also establish a new characterization of semiprime left Goldie rings by showing that a semiprime ring R is left Goldie iff it is regular left fusible and has finite left Goldie dimension.

Full text

1. Introduction

Throughout this paper, $R$ denotes an associative ring with unity. For a nonempty subset $S\subseteq R$, $\mbox{\rm lann}_{R}({A})$ stands for the left annihilator of $A$ in $R$, i.e. $\mbox{\rm lann}_{R}({A})=\{r\in R\mid aA=0\}$.

An element $a\in R$ is a left zero-divisor if $\mbox{\rm lann}_{R}({a})\neq 0$. Elements which are not left zero-divisor are called left regular.

Ghashghaei and McGovern in [4] introduced and investigated fusible rings. A nonzero element $r$ of a ring $R$ is called left fusible if $r$ can be presented in a form $r=c+w$, where $c$ is a left zero-divisor and $w$ is left regular. In such situation we will say that $r=c+w$ is a left fusible decomposition of $r$. Our ring $R$ is called left fusible if every nonzero element of $R$ is left fusible. Let us point out that our notation differs from the one introduced in [4] ”left zero divisors” are right zero divisors in the meaning of [4]. Thus our left fusible rings are right fusible in the language of [4]. Clearly both domains and clean rings are examples of fusible rings (i.e., left and right fusible). It was proved in [4] that polynomial rings and matrix rings over left fusible rings are left fusible.

The aim of the paper is to continue a study of fusible and related rings. In particular, we answer in negative questions (1) and (3) posed in [4]. We also introduce regular left fusible rings, a class which is slightly wider than that of left fusible rings and discuss relations between those two classes. As a result a new description of semiprime left Goldie rings is given in Theorem 2.14. We also prove (cf. Theorem 2.17) that the regular left fusible ring property lifts to matrix rings for rings having left quotient rings. Some questions are formulated.

2. Results

We begin with the following definition.

Definition 2.1**.**

A ring $R$ is said to be regular left fusible if for any nonzero element $r\in R$ there exists a regular (i.e., left and right regular) element $s\in R$ such that the element $sr$ is left fusible, i.e. $sr=c+w$, where $c$ is a left zero divisor and $w$ is left regular.

Since every left regular element is left fusible, our definition reduces to the requirement that for every nonzero left zero divisor $a$ there exists a regular element $s$ in $R$ such that the element $sa$ is left fusible. It is also clear that every left fusible ring is regular left fusible (it is enough to take $s=1$). The following example shows that the class of left fusible rings is strictly smaller than the one of regular left fusible rings.

Example 2.2**.**

Let $F$ be a field and set $R=F<x,y\mid x^{2}=0>$. By [3, Theorem 1], the set of all left zero divisors of $R$ is equal to $xR$. In particular, a difference of any two left zero divisors is a left zero divisor as well. Thus $R$ is not left fusible. Notice that $y$ is a regular element of $R$ and $yr=0+yr$ is a left fusible decomposition of $yr$, for any nonzero element $r\in R$. Thus $R$ is a regular left fusible ring. Similarly, one can see that $R$ is not right fusible but it is regular right fusible.

We will use the above example to answer Question (1) of [4]. Before doing so let us recall some notions. A ring $R$ is a left p.p. ring if any principal left ideal of $R$ is projective and $R$ is said to be left p.q.-Baer (i.e., left principally quasi-Baer) if the left annihilators of principal left ideals are generated as left ideals by an idempotent. Birkenmeier et al. in [1, Corollary 1.15] proved that a ring $R$ is an abelian (i.e., with all idempotents central) left p.p. ring if and only if $R$ is a reduced p.q.-Baer ring. It was proved in [4, Corollary 2.15] that every abelian left (right) p.p. ring is fusible and that, in general, p.q.-Baer rings do not have to be left fusible (see [4, Example 2.17]). Question (1) of [4] asks whether every abelian p.q.-Baer ring is fusible. The following example shows that the answer to this question is negative.

Example 2.3**.**

Let $R=F<x,y\mid x^{2}=0>$ be the ring defined in Example 2.2. Since elements from $yR$ are left regular, we get $ayb\neq 0$, for any nonzero $a,b\in R$. Thus $R$ is a prime ring, so it is also a p.q.-Baer ring. The only idempotents of $R$ are 0 and 1 (so it is abelian). This can be checked directly or one can apply a classical result of Herstein (cf. [5]). By [2, Example 6.], the set of all nilpotent elements of $R$ is equal to $xRx$. The subring $F[x]$ is invariant under all inner automorphisms adjoint to elements of the form $(1+n)$, $n\in xRx$ but $F[x]$ is neither contained in the center nor contains a nonzero ideal of $R$. Therefore, by [5, Theorem], the ring $R$ does not contain nontrivial idempotents. Thus $R$ is an abelian p.q.-Baer ring. We have already seen in Example 2.2 that $R$ is neither left nor right fusible.

Example 2.2 also suggests that although abelian p.q.-Baer rings need not be fusible but maybe they have to be regular fusible. We are unable to answer the following question:

Question 2.4**.**

Is every abelian left p.q.-Baer ring regular left fusible?

Now we will move to some basic properties of regular left fusible rings. Recall that a left ideal $I$ of a ring $R$ is essential (denoted by $I<_{e}{{}_{R}}R$) if $A\cap I\neq 0$ for every nonzero left ideal $A$ of $R$.

Lemma 2.5**.**

Let $a,b\in R$ be such that $\mbox{\rm lann}_{R}({a})<_{e}{{}_{R}}R$ and $\mbox{\rm lann}_{R}({b})\neq 0$. Then $\mbox{\rm lann}_{R}({a-b})\neq 0$. In particular, every element element $a$ such that $\mbox{\rm lann}_{R}({a})<_{e}{{}_{R}}R$ can not be left fusible.

Proof.

The imposed assumptions imply that $\mbox{\rm lann}_{R}({a})\cap\mbox{\rm lann}_{R}({b})\neq 0$, so the thesis is clear. ∎

The left singular ideal of a ring $R$ is $\mbox{\rm Sing}_{l}({R})=\{x\in R\mid\mbox{\rm lann}_{R}({x})<_{e}{{}_{R}}R\}$. It is well known that $\mbox{\rm Sing}_{l}({R})$ is a two-sided ideal of $R$. A ring $R$ is called left nonsingular if $\mbox{\rm Sing}_{l}({R})=0$.

The following lemma generalizes [4, Proposition 2.11].

Lemma 2.6**.**

Every regular left fusible ring $R$ is left nonsingular.

Proof.

Assume $0\neq a\in\mbox{\rm Sing}_{l}({R})$. Since $R$ is regular left fusible, we can pick a regular element $s\in R$ such that $sa$ is left fusible. Since $\mbox{\rm Sing}_{l}({R})$ is a two-sided ideal, $sa\in\mbox{\rm Sing}_{l}({R})$. Thus, by Lemma 2.5, $sa$ is not left fusible, a contradiction. ∎

Let $S$ denote the left Ore set consisting of regular element of the ring $R$ and let $S^{-1}R$ be the left Ore localization of $R$. Writing $Q_{l}(R)$ we will mean that $R$ possesses a classical left quotient ring which is equal to $Q_{l}(R)$.

Lemma 2.7**.**

Let $s\in S$ and $a\in R$. Then $\mbox{\rm lann}_{S^{-1}R}({s^{-1}a})\neq 0$ if and only if $\mbox{\rm lann}_{R}({a})\neq 0$.

Proof.

Suppose $bs^{-1}a=0$, for some nonzero $b\in S^{-1}R$. Eventually multiplying on the left by a suitable element from $S$, we may assume $0\neq b\in R$. Since $S$ is the left Ore set, there exists $t\in S$ and $0\neq c\in R$ such that $bs^{-1}=t^{-1}c$. Then $0\neq c\in\mbox{\rm lann}_{R}({a})$.

Suppose $ca=0$, for some $0\neq c\in R$. Then $(cs)s^{-1}a=0$, so $s^{-1}a$ is a left zero divisor in $S^{-1}R$. ∎

Proposition 2.8**.**

Let $R$ be a ring.

1. (1)

If $R$ is left fusible, then so is $S^{-1}R$. 2. (2)

If $R$ is regular left fusible, then $Q_{l}(R)$ is left fusible. 3. (3)

If $S^{-1}R$ is regular left fusible, then $R$ is regular left fusible. 4. (4)

Suppose every regular element of $R$ is invertible. Then $R$ is regular left fusible iff $R$ is left fusible.

Proof.

(1) Let $s^{-1}r\in S^{-1}R$. As $R$ is left fusible, $r$ has a left fusible decomposition, say $r=a+w$. Lemma 2.7 shows that $s^{-1}r=s^{-1}a+s^{-1}w$ is a left fusible decomposition of $s^{-1}r$.

(2) Suppose $R$ is regular left fusible. Let $0\neq s^{-1}r\in Q_{l}(R)$. Then there exists a regular element $t\in R$ such that $tr$ has a fusible decomposition $tr=a+w$. By Lemma 2.7, $s^{-1}r=(ts)^{-1}a+(ts)^{-1}w$ is a fusible decomposition of $s^{-1}r$ in $Q_{l}(R)$.

(3) Let $r\in R$. Since $S^{-1}R$ is regular left fusible, there exists a regular element $s_{1}^{-1}t_{1}\in S^{-1}R$ such that the element $z={s_{1}}^{-1}t_{1}r$ has a left fusible decomposition, say ${s_{1}}^{-1}t_{1}r={s_{2}}^{-1}a_{2}+{s_{3}}^{-1}w_{3}$. Taking the left common denominator, we may write $z=s^{-1}tr=s^{-1}a+s^{-1}w$ for suitable $s\in S,\;t,a,w\in R$. By Lemma 2.7, $t$ is regular, $w$ is left regular and $a$ is a left zero divisor of $R$. This implies that the element $tr$ is left fusible in $R$ and hence $R$ is regular left fusible.

(4) By assumption, $R=Q_{l}(R)$. Let $r\in R$. Suppose $R$ is regular left fusible. There exists a regular element $t\in R$ such that $tr$ has a fusible decomposition $tr=a+w$. Then, by Lemma 2.7, $r=t^{-1}a+t^{-1}w$ is a fusible decomposition of $r$. This yields the proof of (4). ∎

The above proposition gives immediately the following:

Corollary 2.9**.**

Suppose $R$ has the left quotient ring $Q_{l}(R)$. The following conditions are equivalent:

1. (1)

$R$* is regular left fusible.* 2. (2)

$Q_{l}(R)$* is left fusible.* 3. (3)

$Q_{l}(R)$* is regular left fusible.*

In [4, Propsition 3.14], the authors observed that the statement $(1)$ of Proposition 2.8 holds when $R$ is a commutative ring and asked whether the reverse implication holds. In the context of Corollary 2.9, their question can be restated as:

Question 2.10**.**

Let $R$ be a regular left fusible ring having the classical left quotient ring. Is $R$ left fusible?

Notice that, although the ring $R$ in Example 2.2 is not left fusible, the ring $R$ is regular left fusible. Thus the above question has a negative answer without the assumption that $R$ possesses the classical left quotient ring.

Using the same arguments as in [4, Lemma 2.19.] and making use of the fact that in a commutative ring $R$ an element $r$ is nilpotent if and only if the ideal $rR$ is nilpotent we get the following modification of [4, Lemma 2.19.].

Lemma 2.11**.**

Every commutative regular fusible ring is reduced.

In [4, Propsition 3.7], the authors showed that a commutative ring with only finitely many minimal prime ideals is fusible if and only if it is reduced. Using this characterization we obtain a positive, partial answer to Question 2.10:

Proposition 2.12**.**

Let $R$ be a commutative ring with only finitely many minimal prime ideals. The following conditions are equivalent:

1. (1)

$R$* is fusible.* 2. (2)

$R$* is regular fusible.* 3. (3)

$R$* is reduced.*

Proof.

The implication $(1)\Rightarrow(2)$ is a tautology. The implications $(2)\Rightarrow(3)$ and $(3)\Rightarrow(1)$ are given by Lemma 2.11 and [4, Propsition 3.7], respectively. ∎

[4, Example 3.13] offers a commutative ring which is reduced but is not fusible, i.e. the equivalence $(1)\iff(3)$ does not hold for arbitrary commutative rings. The ring from this example is also not regular fusible. We do not know of any example of a commutative ring which is regular fusible but not fusible.

A ring $R$ is called unit-regular if for each $a\in R$ there exists a unit $u\in U(R)$ such that $aua=a$. In [4, Proposition 5.3], the following proposition was proved. We offer a short direct argument.

Proposition 2.13**.**

Every unit-regular ring is left fusible.

Proof.

Let $R$ be a unit-regular ring and $r\in R$, i.e., $0\neq r=rur$ where $u\in R$. Then $e=ur$ is an idempotent and $e=(-1+e)+(1-2e)$ is its fusible decomposition. This shows that $R$ is regular left fusible. Now, since every regular element in $R$ is invertible, the thesis is a direct consequence of Proposition 2.8(3). ∎

Proposition 2.8 and Corollary 2.9 enable us to give the following new characterization of semiprime left Goldie rings:

Theorem 2.14**.**

For a semiprime ring $R$, the following conditions are equivalent:

1. (1)

$R$* is left Goldie.* 2. (2)

$R$* is regular left fusible and has finite left Goldie dimension*

Proof.

Suppose $R$ is left Goldie. Then, by Goldie’s Theorem, $Q_{l}(R)$ is a semisimple artinian ring. In particular it has finite left Goldie dimension and it is a unit-regular ring. Thus, by Proposition 2.8, $Q_{l}(R)$ is left fusible. Now Corollary 2.9 yields that $R$ is regular left fusible. This completes the proof of $(1)\Rightarrow(2)$.

Let $R$ be as in $(2)$. Then, by Lemma 2.6, $R$ is left nonsingular. Now the implication $(2)\Rightarrow(1)$ is a consequence of Goldie’s Theorem. ∎

Recall that a ring $R$ a right complemented, if for each $a\in R$, there is a $b\in R$ such that $ab=0$ and $a+b$ is regular (see [4]). By [4, Proposition 3.4], every right complemented ring is left fusible. [4, Question (3)] asks whether a right complemented ring have always possesses a right classical quotient ring? Since any domain is complemented on both sides and there are domains having no a classical right quotient ring, the answer to this question is no in general. However the above theorem gives immediately the following:

Corollary 2.15**.**

Every right (left) complemented ring of finite left Goldie dimension has a classical left quotient ring.

Proof.

As observed in [4, Proposition 3.4], every right (left) complemented ring is reduced, so it is semiprime. Now the thesis is a consequence of the left version of Theorem 2.14. ∎

One can easily adopt the proof of [4, Theorem 2.18] to get the following:

Proposition 2.16**.**

Suppose that for any finite set $r_{1},\ldots,r_{n}\in R\setminus\{0\}$ there exists a regular element $s\in R$ such that the elements $sr_{1},\ldots,sr_{n}$ are left fusible. Then the matrix ring $M_{n}(R)$ is regular left fusible.

With the help of the above proposition we get:

Theorem 2.17**.**

Suppose $R$ is a regular left fusible ring having the classical left quotient ring. Then the matrix ring $M_{n}(R)$ is regular left fusible.

Proof.

By Corollary 2.9, $Q_{l}(R)$ is left fusible and Proposition 2.16 shows that $M_{n}(Q_{l}(R))$ is regular left fusible. Let us notice that $M_{n}(Q_{l}(R))=S^{-1}M_{n}(R)$, where $S$ denotes the set of all diagonal matrices $diag(s,\ldots,s)$, where $s$ ranges over all regular elements of $R$. Now the thesis is a direct consequence of Proposition 2.8(3). ∎

Let $R$ be the ring from Example 2.2. Then $y\in R$ is regular and, for any $0\neq r\in R$, $yr$ is left regular, so $R$ satisfies the assumptions of Proposition 2.16. Therefore we have:

Example 2.18**.**

Let $R$ be the ring from Example 2.2. Then the matrix ring $M_{n}(R)$ is regular fusible, for any $n\geq 1$.

We were unable to answer the following:

Question 2.19**.**

Is the matrix ring $M_{n}(R)$ over a regular left fusible ring $R$ itself regular left fusible?

We close the paper with an observation that regular left fusible property behaves well under polynomial extensions. Essentially the same proof as in [4, Proposition 2.9] gives the following

Proposition 2.20**.**

If $R$ is regular left fusible, then so are the rings $R[x;\sigma],R[x,x^{-1};\sigma]$ and $R[[x;\sigma]]$, where $\sigma$ is an automorphism of $R$.