A Generalization of Reflexive Rings

TL;DR

This paper introduces the concept of J-reflexive rings, a broad generalization of reflexive and J-reversible rings, providing characterizations, properties, and invariance results for this new class.

Contribution

It defines J-reflexive rings, extends known results from reflexive rings to this new class, and shows the property is Morita invariant.

Findings

J-reflexive rings generalize reflexive rings.

Results of reflexive rings extend to J-reflexive rings.

J-reflexive property is Morita invariant.

Abstract

In this paper, we introduce a class of rings which is a generalization of reflexive rings and $J$-reversible rings. Let $R$ be a ring with identity and $J(R)$ denote the Jacobson radical of $R$. A ring $R$ is called {\it $J$-reflexive} if for any $a$, $b \in R$, $aRb = 0$ implies $bRa \subseteq J(R)$. We give some characterizations of a $J$-reflexive ring. We prove that some results of reflexive rings can be extended to $J$-reflexive rings for this general setting. We conclude some relations between $J$-reflexive rings and some related rings. We investigate some extensions of a ring which satisfies the $J$-reflexive property and we show that the $J$-reflexive property is Morita invariant.

Equations

$$S=\left\{\left(\begin{array}[]{ccc}a&b&c\\ 0&a&d\\ 0&0&a\\ \end{array}\right)~{}:~{}a,b,c,d\in R\right\}.$$

$$R/I\cap K\cong (R/IK)/(I\cap K/IK)$$

Full text

2010 MSC: 13C99, 16D80, 16U80

Key words: Reflexive ring, reversible ring, $J$-reflexive ring, $J$-reversible ring, ring extension.

1. Introduction

Throughout this paper all rings are associative with identity unless otherwise stated. We write $M_{n}(R)$ for the ring of all $n\times n$ matrices and $T_{n}(R)$ for the ring of all $n\times n$ upper triangular matrices over a ring $R$. Also we write $R[x]$, $R[[x]]$, $N(R)$, $U(R)$ and $J(R)$ for the polynomial ring, the power series ring over a ring $R$, the set of all nilpotent elements, the set of all invertible elements and the Jacobson radical of a ring $R$, respectively. Also $\mathbb{Z}$ denote the ring of integers.

In [6], Mason introduced the reflexive property for ideals. Let $R$ be a ring (without identity) and $I$ an ideal of $R$. Then $I$ is called reflexive, if $aRb\subseteq I$ for $a,b\in R$ implies $bRa\subseteq I$. It is clear that every semiprime ideal is reflexive. Also, the ring $R$ is called reflexive, if [math] is a reflexive ideal (i.e $aRb=0$ implies $bRa=0$ for $a,b\in R$). In [4], Kwak and Lee studied reflexive rings. They investigate reflexive property of rings related to matrix rings and polynomial rings. According to Cohn [2], a ring $R$ is said to be reversible if for any $a$, $b\in R$, $ab=0$ implies $ba=0$. It is clear that every reversible ring is reflexive. Recently, as a generalization of a reversible ring, so-called $J$-reversible ring has been studied in [1]. A ring $R$ is called $J$-reversible, $ab=0$ implies that $ba\in J(R)$ for $a,b\in R$. As an application it is shown that every $J$-clean ring is directly finite. Motivated by these studies, we introduce a class of rings which generalize $J$-reversible rings and reflexive rings. A ring $R$ is called $J$-reflexive, $bRa\subseteq J(R)$ whenever $aRb=0$ for $a,b\in R$.

We summarize the contents of this paper. In Section 2, we study main properties of $J$-reflexive rings. We give some characterizations of $J$-reflexive rings. We prove that every $J$-reversible ring is $J$-reflexive and we supply an example (Example 2.4) to show that the converse is not true in general. Moreover, we see that if $R$ is a Baer ring, then $J$-reversible rings are $J$-reflexive. It is clear that reflexive rings are $J$-reflexive. Example 2.6 shows that $J$-reflexive rings need not be reflexive. We give a necessary and sufficient condition for a quotient ring to be $J$-reflexive. Also we conclude some results which investigate relations between $J$-reflexive rings and some class of rings. With our finding, we prove that every uniquely clean ring is $J$-reflexive and quasi-duo rings are $J$-reflexive. Moreover, we shows that the converse is not true in general. Being Morita invariant propety is very important for class of rings. A ring-theoretic property $\mathcal{P}$ is Morita invariant if and only if whenever a ring $R$ satisfies $\mathcal{P}$ so does $eRe$, for any full idempotent $e$ and $M_{n}(R)$ for any $n>1$. There are a lot of studies on Morita invarivant property of rings. In Section 3, we prove that the $J$-reflexive property is Morita invariant. Furthermore, we study the $J$-reflexive property in several kinds of ring extensions (Dorroh extension, upper triangular matrix ring, Laurent polynomial ring, trivial extension etc.).

2. $J$-Reflexive Rings

In this section we define the $J$-reflexive property of a ring. We investigate some properties of $J$-reflexive rings and exert relations between $J$-reflexive rings and some related rings.

Definition 2.1**.**

A ring $R$ is called $J$-reflexive, $aRb=0$ implies that $bRa\subseteq J(R)$, for $a,b\in R$.

For a nonempty subset $X$ of a ring $R$, the set $r_{R}(X)=\{a\in R:Xa=0\}$ is called the right annihilator of $X$ in $R$ and the set $l_{R}(X)=\{b\in R:bX=0\}$ is called the left annihilator of $X$ in $R$.

Now we give our main characterization for $J$-reflexive rings.

Theorem 2.2**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is $J$-reflexive.* 2. (2)

For all $a\in R$, $r_{R}(aR)Ra\subseteq J(R)$ and $aRl_{R}(Ra)\subseteq J(R)$. 3. (3)

$IRK=0$* implies $KRI\subseteq J(R)$ for every nonempty subsets $I,K$ of $R$.* 4. (4)

$<a><b>=0$* implies $<b><a>\subseteq J(R)$ for any $a,b\in R$.* 5. (5)

$IK=0$* implies $KI\subseteq J(R)$ for every right (left) ideals $I,K$ of $R$.* 6. (6)

$IK=0$* implies $KI\subseteq J(R)$ for every ideals $I,K$ of $R$.*

Proof.

$(1)\Rightarrow(2)$ Let $b\in r_{R}(aR)$. Then $aRb=0$ for $a,b\in R$. Since $R$ is $J$-reflexive, $bRa\subseteq J(R)$. So we have $r_{R}(Ra)Ra\subseteq J(R)$. Similarly, one can show that $aRl_{R}(Ra)\subseteq J(R)$.

$(2)\Rightarrow(1)$ Assume that $aRb=0$ for $a,b\in R$. Then, $b\in r_{R}(aR)$. By (2) we have $bRa\subseteq J(R)$. So $R$ is a $J$-reflexive ring.

$(3)\Rightarrow(4)\Rightarrow(5)\Rightarrow(6)$ It is clear.

$(6)\Rightarrow(1)$ Let $aRb=0$ for $a,b\in R$. Then $RaRRbR=0$ . By hypothesis, $RbRRaR\subseteq J(R)$. As $bRa\subseteq RbRRaR$, we have $bRa\subseteq J(R)$.

$(1)\Rightarrow(3)$ Assume that $IRK=0$ for nonempty subsets $I,K$ of $R$. Then for any $a\in I$ and $b\in K$, $aRb=0$. As $R$ is $J$-reflexive, $bRa\subseteq J(R)$. This implies that $KRI\subseteq J(R)$. ∎

Examples of $J$-reflexive rings are abundant. All reduced rings, symmetric rings, reversible rings and reflexive rings are $J$-reflexive. In the sequel, we show that every $J$-reversible ring, uniquely clean ring and every right (left) quasi-duo ring is $J$-reflexive.

Proposition 2.3**.**

Every $J$-reversible ring is $J$-reflexive.

Proof.

Let $R$ be a $J$-reversible ring and $aRb=0$ for some $a,b\in R$. Then $ab=0$ and $abr=0$ for any $r\in R$. As $R$ is $J$-reversible, $bra\in J(R)$. Hence, $bRa\subseteq J(R)$. ∎

The converse statement of Proposition 2.3 is not true in general as the following example shows.

Example 2.4**.**

Consider the ring $R=M_{2}(\mathbb{Z})$. It can be easily shown that $R$ is a $J$-reflexive ring. Let A=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right), B=\left(\begin{array}[]{cc}1&1\\ 0&0\\ \end{array}\right)\in R. Although $AB=0$, BA=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right)\notin J(R). So $R$ is not $J$-reversible.

Recall that a ring $R$ is called Baer if the right (left) annihilator of every nonempty subset of $R$ is generated by an idempotent (see for details [3]). We show that the converse statement of Proposition 2.3 is true for Baer rings.

Theorem 2.5**.**

Let $R$ be a Baer ring. Then the following are equivalent.

1. (1)

$R$* is a $J$-reversible ring.* 2. (2)

$R$* is a $J$-reflexive ring.*

Proof.

$(1)\Rightarrow(2)$ It is clear by Proposition 2.3.

$(2)\Rightarrow(1)$ Let $ab=0$ for $a,b\in R$. Then $abR=0$ and so $a\in l_{R}(bR)$. As $R$ is a Baer ring, there exists an idempotent $e\in R$ such that $l_{R}(bR)=eR$. Then we have $eRbR=0$. Since $R$ is $J$-reflexive, $bReR\subseteq J(R)$ and so $ba\in J(R)$, as desired. ∎

Though reflexive rings are $J$-reflexive, $J$-reflexive rings are not reflexive as the following example shows.

Example 2.6**.**

Let $R$ be a commutative ring. Consider the ring

[TABLE]

By [1, Propoosition 3.7], $S$ is $J$-reversible and by Proposition 2.3, it is $J$-reflexive. For A=\left(\begin{array}[]{ccc}0&0&0\\ 0&0&1\\ 0&0&0\\ \end{array}\right), B=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&0\\ 0&0&0\\ \end{array}\right)\in S, $ASB=0$ but $BSA\neq 0$. Thus, $S$ is not a reflexive ring.

The following result can be easily obtain by the definition of $J$-reflexive rings.

Corollary 2.7**.**

The following are equivalent for a ring $R$.

1. (1)

If $R/J(R)$ is reflexive, then $R$ is $J$-reflexive. 2. (2)

If $R/J(R)$ is commutative, then $R$ is $J$-reflexive.

An element $a$ in a ring $R$ is called uniquely clean if $a=e+u$ where $e^{2}=e\in R$ and $u\in U(R)$ and this representation is unique. A ring $R$ is called a uniquely clean ring if every element of $R$ is uniquely clean (see [7]).

Corollary 2.8**.**

Every uniquely clean ring is $J$-reflexive.

Proof.

Assume that $R$ is uniquely clean. Then $R/J(R)$ is Boolean by [7, Theorem 20]. Hence, $R$ is $J$-reflexive by Corollary 2.7. ∎

The converse statement of Corollary 2.8 is not true in general as the following example shows.

Example 2.9**.**

For a commutative ring $R$, consider the ring $M_{2}(R)$. Since $M_{2}(R)$ is not an abelian ring, $M_{2}(R)$ is not a uniquely clean ring. Also, it can be easily shown that $M_{2}(R)$ is a $J$-reflexive ring by Theorem 3.1.

Proposition 2.10**.**

Let $R$ be a ring. If $N(R)\subseteq J(R)$, then $R$ is $J$-reflexive.

Proof.

Assume that $aRb=0$ for $a,b\in R$. Then for any $r\in R$, $arb=0$ and so $ab=0$. Hence, $(bra)^{2}=brabra=0$ for all $r\in R$. So $bra\in N(R)$. By hypothesis we have $bra\in J(R)$, as asserted. ∎

A ring $R$ is called right (left) quasi-duo if every right (left) maximal ideal of $R$ is an ideal [5].

Corollary 2.11**.**

Every right (left) quasi-duo ring is $J$-reflexive.

Proof.

It is clear by [8, Lemma 2.3]. ∎

We now give a necessary and sufficient condition for a quotient ring to be $J$-reflexive.

Theorem 2.12**.**

Let $R$ be a ring and $I$ a nilpotent ideal of $R$. Then $R$ is $J$-reflexive if and only if $R/I$ is $J$-reflexive.

Proof.

Let $R/I=\overline{R}$, $a+I=\overline{a}\in\overline{R}$ and $\overline{aRb}=\overline{0}$ for $\overline{a},\overline{b}\in\overline{R}$. So $aRb\subseteq I$. As $I$ is nilpotent there exists $k\in\mathbb{Z}^{+}$ such that $(RaRbR)^{k}=0$. $(RbRaR)^{k}\subseteq J(R)$, since $R$ is $J$-reflexive. Thus $RbRaR\subseteq J(R)$ as Jacobson radical is semiprime. Hence $\overline{RbRaR}\subseteq J(R)/I=J(\overline{R})$. So $\overline{bRa}\subseteq J(\overline{R})$.

Conversely, assume that $aRb=0$ for $a,b\in R$. Then $\overline{aRb}=\overline{0}$. So $aRb\subseteq I$ and $RaRbR\subseteq I$. Therefore there exists $k\in\mathbb{Z}^{+}$ such that $I^{k}=0$, and so $(RaRbR)^{k}=RaRbRaRbR\cdots RaRbR=0$. Hence, $\overline{(RaRbR)^{k}}=\overline{0}$. Since $R/I$ is $J$-reflexive, $\overline{(RbRaRbRaR\cdots RbRaR)}=\overline{(RbRaR)}^{k}=\subseteq J(\overline{R})$. As Jacobson radical is semiprime ideal, we have $RbRaR\subseteq J(\overline{R})$. Thus, $\overline{bRa}\subseteq J(\overline{R})$. Hence, for all $\overline{r}\in\overline{R}$, we have $\overline{1}-(\overline{bra})\overline{x}\in U(\overline{R})$ for some $\overline{x}\in J(\overline{R})$. Then, there exists $\overline{s}\in\overline{R}$ such that $(\overline{1}-(\overline{bra})\overline{x})\overline{s}=\overline{1}$. Hence, $1-(1-bra)xs\in I$. As every nilpotent ideal is nil, $1-(brax)s\in U(R)$. This implies that $bra\in J(R)$ and so $bRa\subseteq J(R)$, as desired. ∎

Corollary 2.13**.**

Let $R$ be a ring. Then the following are satisfied.

1. (1)

If $J(R)$ is a nilpotent ideal, then $R$ is $J$-reflexive if and only if $R/J(R)$ is $J$-reflexive. 2. (2)

If $R$ is an Artinian ring, then $R$ is $J$-reflexive if and only if $R/J(R)$ is $J$-reflexive.

Proof.

$(1)$ It is clear by Theorem 2.12.

$(2)$ Since Jacobson radical of Artinian ring is nilpotent, it is clear by (1). ∎

Proposition 2.14**.**

Let $R$ be a ring and $I$ an ideal of $R$ with $I\subseteq J(R)$. If $R/I$ is $J$-reflexive, then $R$ is $J$-reflexive.

Proof.

Let $\overline{R}=R/I$ and for $\overline{a}=a+I\in R/I$. Assume that $aRb=0$ for $a,b\in R$. So $\overline{aRb}=\overline{0}$. Since $\overline{R}$ is $J$-reflexive, $\overline{bRa}\subseteq J(\overline{R})$ and $\overline{bra}\in J(\overline{R})$ for any $\overline{r}\in\overline{R}$. Thus, for all $\overline{x}\in\overline{R}$ we have $\overline{1}-(\overline{bra})\overline{x}\in U(\overline{R})$. Then, there exists $\overline{s}\in\overline{R}$ such that $(\overline{1}-(\overline{bra})\overline{x})\overline{s}=\overline{1}$. Hence, $1-(1-brax)s\in I$. As $I$ is contained $J(R)$, $(1-brax)s\in U(R)$. This implies that $bra\in J(R)$ and so $bRa\subseteq J(R)$, as desired. ∎

Proposition 2.15**.**

Let $R$ be a ring and $I$ a reflexive ideal of $R$. Then $R/I$ is $J$-reflexive.

Proof.

Let $\overline{R}=R/I$ and for $\overline{a}=a+I\in R/I$. Suppose that $\overline{aRb}=\overline{0}$ for $\overline{a},\overline{b}\in\overline{R}$. Then $aRb\subseteq I$. Since $I$ is a reflexive ideal, we have $bRa\subseteq I$. Hence, $\overline{bRa}=\overline{0}\in J(\overline{R})$. ∎

Theorem 2.16**.**

Every subdirect product of $J$-reflexive ring is $J$-reflexive.

Proof.

Let $R$ be a ring, $I,K$ ideals of $R$ and $R$ a subdirect product of $R/I$ and $R/K$. Assume that $R/I$ and $R/K$ are $J$-reflexive. Let $aRb=0$ for $a,b\in R$. Then $\overline{aRb}=\overline{0}$ in $R/I$ and $R/K$. Since $R/I$ and $R/K$ are $J$-reflexive, $\overline{bRa}\subseteq J(R/I)$ and $\overline{bRa}\subseteq J(R/K)$. Then for each $x\in R$ we have $\overline{1-brax}\in U(R/I)$ and $\overline{1-brax}\in U(R/K)$. Hence, there exist $y\in R/I$ and $z\in R/K$ such that $\overline{(1-brax)y}=\overline{1}\in R/I$ and $\overline{(1-brax)z}=\overline{1}\in R/K$. So $1-(1-brax)y\in I$ and $1-(1-brax)z\in K$. If we multiply the last two elements, we have $(1-(1-brax)y)(1-(1-brax)z)\in IK\subseteq I\cap K=0$. Thus, $1-(1-brax)t=0$ and $(1-brax)t=1$. This implies that $bRa\subseteq J(R)$. ∎

Corollary 2.17**.**

Let $I$ and $K$ be ideals of a ring $R$. If $R/I$ and $R/K$ are $J$-reflexive, then $R/I\cap K$ is $J$-reflexive.

Proof.

Let $\alpha:R/(I\cap K)\rightarrow R/I$ and $\beta:R/(I\cap K)\rightarrow R/K$ where $\alpha(r+(I\cap K))=r+I$ and $\beta(r+(I\cap K))=r+K$. It can be shown that $\alpha$ and $\beta$ surjective ring homomorphisms and ker$\alpha\cap$ker$\beta=0$. Hence $R/(I\cap K)$ is subdirect product of $R/I$ and $R/K$. Therefore, $R/(I\cap K)$ by Theorem 2.16. ∎

Corollary 2.18**.**

Let $R$ be a ring and $I,K$ of ideals of $R$. If $R/I$ and $R/K$ are $J$-reflexive, then $R/IK$ is $J$-reflexive.

Proof.

Assume that $R/I$ and $R/K$ are $J$-reflexive. Since

[TABLE]

and $(I\cap K/IK)^{2}=0$, we complete the proof by Theorem 2.12. ∎

3. Extensions of $J$-reflexive rings

In this section we show that several extensions (Dorroh extension, upper triangular matrix ring, Laurent polynomial ring, trivial extension etc.) of a $J$-reflexive ring are $J$-reflexive. In particular, it is proved that the $J$-reflexive condition is Morita invariant.

Two rings $R$ and $S$ are said to be Morita equivalent if the categories of all right $R$-modules and all right $S$-modules are equivalent. Properties shared between equivalent rings are called Morita invariant properties. $\mathcal{P}$ is Morita invariant if and only if whenever a ring $R$ satisfies $\mathcal{P}$, then so does $eRe$ for every full idempotent $e$ and so does every matrix ring $M_{n}(R)$ for every positive integer $n$.

Next result shows that the property of $J$-reflexivitiy is Morita invariant.

Theorem 3.1**.**

Let $R$ be a ring. Then we have the following.

1. (1)

If $R$ is $J$-reflexive, then $eRe$ is $J$-reflexive for all idempotent $e\in R$. 2. (2)

$R$* is a $J$-reflexive ring if and only if $M_{n}(R)$ is $J$-reflexive for any positive integer $n$.*

Proof.

$(1)$ Assume that $R$ is a $J$-reflexive ring. Let $a,b\in eRe$ with $aeReb=0$. As $R$ is $J$-reflexive, $ebRae=ebRea\subseteq J(eRe)=eJ(R)e$. This implies that $eRe$ is a $J$-reflexive ring.

$(2)$ Assume that $M_{n}(R)$ is $J$-reflexive ring. It is clear that $R$ is $J$-reflexive by $(1)$. Conversely, suppose that $R$ is $J$-reflexive and $I,K$ are an ideals of $M_{n}(R)$ such that $IK=0$. Then, there exist $I_{1},K_{1}$ ideals of $R$ such that $I=M_{n}(I_{1})$ and $K=M_{n}(K_{1})$. So $0=IK=M_{n}(I_{1})M_{n}(K_{1})=M_{n}(I_{1}K_{1})$. Thus, $I_{1}K_{1}=0$. Since $R$ is $J$-reflexive, $K_{1}I_{1}\subseteq J(R)$. This implies that $KI=M_{n}(K_{1})M_{n}(I_{1})=M_{n}(K_{1}I_{1})\subseteq J(M_{n}(R))=M_{n}(J(R))$. This completes the proof. ∎

Corollary 3.2**.**

Let $M$ be a finitely generated projective modules over a $J$-reflexive ring $R$. Then End${}_{R}(M)$ is $J$-reflexive.

Proof.

It is obvious by Theorem 3.1. ∎

Proposition 3.3**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is $J$-reflexive.* 2. (2)

M=\left\{\left(\begin{array}[]{cccc}r&x_{12}&\cdots&x_{1n}\\ 0&r&\cdots&x_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ 0&\cdots&\cdots&r\\ \end{array}\right):r\in R,~{}x_{ij}\in R\right\}* is $J$-reflexive.*

Proof.

$(1)\Leftrightarrow(2)$ Take I=\left(\begin{array}[]{cccc}0&x_{12}&\cdots&x_{1n}\\ 0&0&\cdots&x_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ 0&\cdots&\cdots&0\\ \end{array}\right). The proof is clear by Theorem 2.12. ∎

Recall that the trivial extension of $R$ by an $R$-module $M$ is the ring denoted by $R\propto M$ whose underlying additive group is $R\oplus M$ with multiplication given by $(r,m)(r^{\prime},m^{\prime})=(rr^{\prime},rm^{\prime}+mr^{\prime})$. The ring $R\propto M$ is isomorphic to S=\left\{\left(\begin{array}[]{cc}x&y\\ 0&x\\ \end{array}\right):x\in R,~{}y\in M\right\} under the usual matrix operations.

Proposition 3.4**.**

The following are equivalent for a ring $R$.

1. (1)

The trivial extension $R\propto R$ of the ring $R$ is $J$-reflexive. 2. (2)

$R$* is a $J$-reflexive ring.*

Proof.

$(1)\Rightarrow(2)$ Assume that $R\propto R$ is $J$-reflexive. Let $aRb=0$ for $a,b\in R$. Then, for A=\left(\begin{array}[]{cc}a&0\\ 0&a\\ \end{array}\right),~{}B=\left(\begin{array}[]{cc}b&0\\ 0&b\\ \end{array}\right)\in R\propto R, we have A(R\propto R)B=\left(\begin{array}[]{cc}aRb&aRb\\ 0&aRb\\ \end{array}\right)=\left(\begin{array}[]{cc}0&0\\ 0&0\\ \end{array}\right). As $R\propto R$ is $J$-reflexive, $B(R\propto R)A\subseteq J(R\propto R)$. Hence, $bRa\subseteq J(R)$.

$(2)\Rightarrow(1)$ Suppose that $R$ is $J$-reflexive. Let $A(R\propto R)B=0$ for A=\left(\begin{array}[]{cc}a&x\\ 0&a\\ \end{array}\right),~{}B=\left(\begin{array}[]{cc}b&y\\ 0&b\\ \end{array}\right)\in R\propto R. Then for any M=\left(\begin{array}[]{cc}s&t\\ 0&s\\ \end{array}\right)\in R\propto R, we have AMB=\left(\begin{array}[]{cc}asb&asy+atb+xsb\\ 0&asb\\ \end{array}\right)=\left(\begin{array}[]{cc}0&0\\ 0&0\\ \end{array}\right). Since $R$ is $J$-reflexive and $aRb=0$, we conclude that $bsa\in J(R)$ for any $s\in R$. Note that J(R\propto R)=\left(\begin{array}[]{cc}J(R)&R\\ 0&J(R)\\ \end{array}\right). Hence, $B(R\propto R)A\subseteq J(R\propto R)$, as asserted. ∎

Proposition 3.5**.**

Let $\{R_{i}\}_{i\in\mathcal{I}}$ indexed set of the ring $R_{i}$. Then $R_{i}$ is $J$-reflexive for all $i\in\mathcal{I}$ if and only if $\prod_{i\in\mathcal{I}}R_{i}$ is $J$-reflexive.

Proof.

$\Rightarrow$ Let $\prod_{i\in\mathcal{I}}M_{i}K_{i}=0$ for ideals $\prod_{i\in\mathcal{I}}M_{i},\prod_{i\in\mathcal{I}}K_{i}$ of $\prod_{i\in\mathcal{I}}R_{i}$. Then $\prod_{i\in\mathcal{I}}M_{i}K_{i}=0$. Therefore, $M_{i}K_{i}=0$ for all $i\in\mathcal{I}$. Since $R_{i}$ is $J$-reflexive, $K_{i}M_{i}\subseteq J(R_{i})$ for all $i\in\mathcal{I}$. So $\prod_{i\in\mathcal{I}}K_{i}\prod_{i\in\mathcal{I}}M_{i}=\prod_{i\in\mathcal{I}}K_{i}M_{i}\subseteq J(\prod_{i\in\mathcal{I}}R_{i})=\prod_{i\in\mathcal{I}}J(R_{i})$.

$\Leftarrow$ Assume that $M_{\phi}K_{\phi}=0$ for ideals $M_{\phi},K_{\phi}$ of $R_{\phi}$. Choose $M=(M_{\phi})_{\phi\in\mathcal{I}}$ and $K=(K_{\phi})_{\phi\in\mathcal{I}}$ as only $\phi$ components are nonzero ideal. So $M$ and $K$ are ideals of $\prod_{i\in\mathcal{I}}R_{i}$. Also we have $MK=0$. As $\prod_{i\in\mathcal{I}}R_{i}$ is $J$-reflexive, $KM\subseteq J(\prod_{i\in\mathcal{I}}R_{i})$. Thus, $K_{\phi}M_{\phi}\subseteq J(R_{\phi})$. ∎

Proposition 3.6**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is a $J$-reflexive ring.* 2. (2)

$T_{n}(R)$* is $J$-reflexive for any $n\in\mathbb{Z}^{+}$.*

Proof.

$(1)\Rightarrow(2)$ For $n=1$ it is clear. Consider the ring $T_{2}(R)$. Choose the ideal I=\left(\begin{array}[]{cc}0&R\\ 0&0\\ \end{array}\right). It is clear that $I^{2}=0$. So $T_{2}(R)/I\cong R\times R$. By Proposition 3.5, $T_{2}(R)/I$ is $J$-reflexive. Hence $T_{2}(R)$ is $J$-reflexive, by Theorem 2.12. By induction, $T_{n}(R)$ is $J$-reflexive for any $n\in\mathbb{Z}^{+}$.

$(2)\Rightarrow(1)$ It is evident from Theorem 3.1(1). ∎

Proposition 3.7**.**

Let $R$ be a ring and $e^{2}=e\in R$ is central. Then, $R$ is a $J$-reflexive ring if and only if $eR$ and $(1-e)R$ are $J$-reflexive.

Proof.

The necessity is obvious by Theorem 3.1. For the sufficiency suppose that $eR$ and $(1-e)R$ are $J$-reflexive for a central idempotent $e\in R$. It is well-known that $R\cong eR\times(1-e)R$. By Proposition 3.5, $R$ is $J$-reflexive. ∎

For an algebra $R$ over a commutative ring $S$, the * Dorroh extension* $I(R;S)$ of $R$ by $S$ is the additive abelian group $I(R;S)=R\oplus S$ with multiplication $(r,v)(s,w)=(rs,rw+vs+vw)$.

Proposition 3.8**.**

Let $R$ be a ring and $M=I(R;S)$ Dorroh extension of $R$ by commutative ring $S$. Assume that for all $s\in S$ there exists $s^{\prime}\in S$ such that $s+s^{\prime}+ss^{\prime}=0$. Then the following are equivalent.

1. (1)

$R$* is $J$-reflexive.* 2. (2)

$M$* is $J$-reflexive.*

Proof.

$(1)\Rightarrow(2)$ Let $(a_{1},b_{1})M(a_{2},b_{2})=(0,0)$ for $(a_{1},b_{1}),(a_{2},b_{2})\in M$. So for any $(x,y)\in M$, we have $(a_{1},b_{1})(x,y)(a_{2},b_{2})=(0,0)$. Then $(a_{1}xa_{2},a_{1}xb_{2}+a_{1}ya_{2}+b_{1}xa_{2}+b_{1}ya_{2}+a_{1}yb_{2}+b_{1}xb_{2}+b_{1}yb_{2})=(0,0)$. Hence, $a_{1}xa_{2}=0$ and $a_{1}xb_{2}+a_{1}ya_{2}+b_{1}xa_{2}+b_{1}ya_{2}+a_{1}yb_{2}+b_{1}xb_{2}+b_{1}yb_{2}=0$. As $R$ is $J$-reflexive, $a_{2}xa_{1}\in J(R)$ for any $x\in R$. Thus, $(a_{2},b_{2})(x,y)(a_{1},b_{1})=(a_{2}xa_{1},*)$. By hypothesis, $(0,S)\subseteq J(M)$. It can be easy to show that $(a_{2}xa_{1},0)\in J(M)$ for each $x\in R$. Therefore, $(a_{2},b_{2})S(a_{1},b_{1})\subseteq J(M)$.

$(2)\Rightarrow(1)$ Let $aRb=0$ for $a,b\in R$. Then $(a,0)M(b,0)=(0,0)$. Since $M$ is $J$-reflexive, $(b,0)M(a,0)\subseteq J(M)$. By hypothesis, $(0,S)\subseteq J(M)$. This implies that $(bRa,0)\subseteq J(S)$. Hence, $bRa\subseteq J(R)$. ∎

If $R$ is a ring and $f:R\rightarrow R$ is a ring homomorphism, let $R[[x,f]]$ denote the ring of skew formal power series over $R$; that is all formal power series in $x$ with coefficients from $R$ with multiplication defined by $xr=f(r)x$ for all $r\in R$. Note that $J(R[[x,f]])=J(R)+<x>$. Since $R[[x,f]]\cong I(R;<x>)$ where $<x>$ is the ideal generated by $x$, we have the following result.

Corollary 3.9**.**

Let R be a ring and $f:R\rightarrow R$ a ring homomorphism. Then the following are equivalent.

1. (1)

$R$* is a $J$-reflexive ring.* 2. (2)

$R[[x,f]]$* is $J$-reflexive.*

If $f$ is taken $f=1_{R}:R\rightarrow R$ $(i.e.~{}1_{R}(r)=r~{}\mbox{for all}~{}r\in R)$, we have $R[[x]]=R[[x,1_{R}]]$ is the ring of formal power series over $R$.

Corollary 3.10**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is a $J$-reflexive ring.* 2. (2)

$R[[x]]$* is $J$-reflexive.*

Let $R$ be a ring and $u\in R$. Recall that $u$ is right regular if $ur=0$ implies $r=0$ for $r\in R$. Similarly, left regular element can be defined. An element is regular if it is both left and right regular.

Proposition 3.11**.**

Let $R$ be a ring and $M$ multiplicatively closed subset of $R$ consisting of central regular elements. Then the following are equivalent.

1. (1)

$R$* is $J$-reflexive.* 2. (2)

$S=M^{-1}R=\{\frac{a}{b}:a\in R,b\in M\}$* is $J$-reflexive.*

Proof.

$(1)\Rightarrow(2)$ Let $aSb=0$ for $a,b\in S$. So there exist $a_{1},b_{1}\in R$ and $u^{-1},v^{-1}\in M$ such that $a=a_{1}u^{-1}$ and $b=b_{1}v^{-1}$. Then $0=aSb=a_{1}u^{-1}Sb_{1}v^{-1}=a_{1}Sbv^{-1}$. Hence for any $rs^{-1}\in S$ we have $a_{1}rs^{-1}bv^{-1}$. Thus, $a_{1}rb_{1}=0$ for each $r\in R$. As $R$ is $J$-reflexive, $b_{1}ra_{1}\in J(R)$. This implies that $b_{1}v^{-1}rs^{-1}a_{1}u^{-1}\in J(R)$. As $J(R)\subseteq J(S)$, $aSb\subseteq J(S)$.

$(2)\Rightarrow(1)$ Let $aRb=0$ for $a,b\in R$ and $u,v\in M$. So we have $auRbv=0$. Then for any $m\in M$ and $r\in R$ $aurmbv=0$. Since $S$ is $J$-reflexive, $bvrmau\in J(S)$. If we multiply $bvrmau$ with inverses of $u,m,v$, then we have $bra\in J(R)$ for any $r\in R$. This completes the proof. ∎

The following result is a direct consequence of Proposition 3.11.

Corollary 3.12**.**

Let $R$ be a ring. Then the following are equivalent.

1. (1)

$R[x]$* is $J$-reflexive.* 2. (2)

$R[x,x^{-1}]$* is $J$-reflexive.*