Some Notes on $CN$ Rings

Abstract

A ring $R$ is $CN$ if and only if for any $x\in N(R)$ and $y\in R$, $((1+x)y)^{n+k}=(1+x)^{n+k}y^{n+k}$, where $n$ is a fixed positive integer and $k=0, 1,2$; (2) Let $R$ be a $CN$ ring and $n\ge 1$. If for any $x, y\in R\backslash N(R)$, $(xy)^{n+k}=x^{n+k}y^{n+k}$, where $k=0, 1,2$, then $R$ is commutative; (3) Let $R$ be a ring and $n\ge 1$. If for any $x\in R\backslash N(R)$ and $y\in R$, $(xy)^k=x^ky^k$, $k=n, n+1, n+2$, then $R$ is commutative; (4) $NLI$ exchange rings are clean.

1 Introduction

Throughout this paper, all rings are associative with identity. Let $R$ be a ring, we use $N(R)$, $J(R)$, $E(R)$, $Z(R)$, and $U(R)$ to denote the set of all nilpotent elements, the Jacobson radical, the set of all idempotent elements, the center, and the set of all invertible elements of $R$, respectively. For any nonempty subset $X$ of a ring $R$, $r(X)=r_R(X)$ and $l(X)=l_R(X)$ denote the right annihilator of $X$ and the left annihilator of $X$, respectively.

Following [6], a ring $R$ is called CN if $N(R)\subseteq Z(R)$. Clearly, commutative rings and reduced rings (thai is, a ring $R$ with $N(R)=0$) are $CN$.

A theorem of Herstein [10] stated that a ring $R$ which satisfies the identity $(xy)^n=x^ny^n$, where $n$ is a fixed positive integer greater than 1, must have nil commutator ideal. In [3], Bell proved that if $R$ is an $n$-torsion-free ring with identity 1 and satisfies the two identities $(xy)^n=x^ny^n$ and $(xy)^{n+1}=x^{n+1}y^{n+1}$, then $R$ is commutative. In [1], Khuzam proved that if $R$ is $n(n-1)$-torsion-free ring with 1 and satisfies the identity $(xy)^n=x^ny^n$, then $R$ is commutative. In [2], Khuzam proved that if $R$ is a semiprime ring in which for each $x$ in $R$ there exists a positive integer $n=n(x)> 1$ such that $(xy)^n=x^ny^n$ for all $y\in R$, then $R$ is commutative. In [11], Ligh and Richou proved that if $R$ is a ring with 1 which satisfies the identities: $(xy)^k=x^ky^k$, $k=n, n+1, n+2$, where $n$ is a positive integer, then $R$ is commutative. The purpose of this note is to generalize these results.

2 Main Results

We begin with the following theorem which generalizes [6, Theorem 5].

Theorem 2.1

The following conditions are equivalent for a ring $R$:

• (1) $R$ is a $CN$ ring;

• (2) For any $a\in N(R)$, there exists $n=n(a)\ge 2$ such that $a-a^n\in Z(R)$;

• (3) For any $a\in N(R)$ and $b\in R$, there exists $c=c(a, b)\in R$ such that $[ a-a^2c, b]=0$.

Proof

$(1)\Longrightarrow (i), i=2,3$ are trivial.

$(2)\Longrightarrow (1)$ Assume that $a\in N(R)$ with $a^m=0$ for some $m\ge 2$. By (2), there exists $n_1=n_1(a)\ge 2$ such that $a-a^{n_1}\in Z(R)$. Since $a^{n_1}\in N(R)$, by (2), there exists $n_2=n_2(a^{n_1})\ge 2$ such that $a^{n_1}-a^{n_1n_2}\in Z(R)$. Continuing this process, there exists $n_s=n_s(a^{n_1n_2\cdots n_{s-1}})\ge 2$ such that $a^{n_1n_2\cdots n_{s-1}}-a^{n_1n_2\cdots n_{s-1}n_s}\in Z(R)$ and $n_1n_2\cdots n_{s-1}n_s\ge m$. Hence $a^{n_1n_2\cdots n_{s-1}n_s}=0$ and $a=a-a^{n_1n_2\cdots n_{s-1}n_s}=(a-a^{n_1})+(a^{n_1}-a^{n_1n_2})+\cdots +(a^{n_1n_2\cdots n_{s-1}}-a^{n_1n_2\cdots n_{s-1}n_s})\in Z(R)$.

$(3)\Longrightarrow (1)$ Assume that $a\in N(R)$ with $a^n=0$ for some $n\ge 2$. By induction on $n$, we claim that $a\in Z(R)$. For each $x\in R$, by (3), there exists $c=c(a, x)\in R$ such that $[a-a^2c, x]=0$. If $n=2$, then $a\in Z(R)$, we are done. Now we assume that $n> 2$ and assume that for each $y\in N(R)$ with the index of nilpotence at most $n-1$, we have $y\in Z(R)$. Since $(a^2)^{n-1}=0$, by the induction hypothesis, $a^2\in Z(R)$. For any $z\in (a^2)=a^2R$, we have $z^{n-1}\in a^{2(n-1)}R=0$, so $z\in Z(R)$ by the induction hypothesis. This implies $a^2R\subseteq Z(R)$. Hence $0=[a-a^2c, x]=[a, x]$ for any $x\in R$, so $a\in Z(R)$, this shows that $R$ is $CN$. $\square$

A ring $R$ is called NLI if $N(R)$ is a $Lie-$ideal of $R$ (that is, for any $a\in N(R)$ and $b\in R$, $ab-ba\in N(R)$ and $N(R)$ is an additive subgroup of $R$). Clearly, $NI$ rings (that is, $N(R)$ forms an ideal of $R$ ) are $NLI$. A ring $R$ is called QCN if for any $a\in N(R)$ and $b\in R$, there exist $n=n(a, b)> 1$ and $c\in R$ such that $ab-ba=(ab-ba)^nc$. Clearly, $CN$ rings are $QCN$.

Theorem 2.2

The following conditions are equivalent for a ring $R$:

• (1) $R$ is a $CN$ ring;

• (2) $R$ is a $QCN$ $NI$ ring;

• (3) $R$ is a $QCN$ $NLI$ ring.

Proof

$(1)\Longrightarrow (2)\Longrightarrow (3)$ is trivial. $(3)\Longrightarrow (1)$ Assume that $a\in N(R)$ and $b\in R$. Since $R$ is a $QCN$ ring, $ab-ba=(ab-ba)^nc$ for some $n=n(a, b)$ and $c\in R$. Since $R$ is an $NLI$ ring, $ab-ba\in N(R)$. Let $m\ge 1$ such that $(ab-ba)^m=0$. Clearly, $(n-1)m+1\ge m$. Since $ab-ba=(ab-ba)^{(n-1)m+1}c^m$, $ab-ba=0$. Hence $a\in Z(R)$ and $R$ is a $CN$ ring. $\square$

In preparation for the proof of our next theorem, we first state the following known lemma ([12, Lemma 2]).

Lemma 2.1

Let $x, y\in R$. Suppose that for some positive integer $n$, $xy^n=0=x(1+y)^n$. Then $x=0$.

Theorem 2.3

The following conditions are equivalent for a ring $R$:

• (1) $R$ is a $CN$ ring;

• (2) For any $x\in N(R)$ and $y\in R$, $((1+x)y)^{n+k}=(1+x)^{n+k}y^{n+k}$, where $n$ is a fixed positive integer and $k=0, 1,2$;

• (3) For any $x\in N(R)$ and $y\in R$, $((1+x)y)^{n+k}=y^{n+k}(1+x)^{n+k}$, where $n$ is a fixed positive integer and $k=0, 1, 2$.

Proof

$(1)\Longrightarrow (i), i=2, 3$ are trivial. $(2)\Longrightarrow (1)$ Assume that $x\in N(R)$ and $y\in R$. Then by the hypothesis,

$$\begin{aligned} (1+x)^{n+1}y^{n+1}&= (1+x)^ny^n(1+x)y. \end{aligned}$$

(2.1)

$$\begin{aligned} (1+x)^{n+2}y^{n+2}&= (1+x)^{n+1}y^{n+1}(1+x)y. \end{aligned}$$

(2.2)

Since $1+x$ is invertible in $R$, (2.1) gives

$$\begin{aligned} (xy^n-y^nx)y=0, \end{aligned}$$

(2.3)

(2.2) gives

$$\begin{aligned} (xy^{n+1}-y^{n+1}x)y=0. \end{aligned}$$

(2.4)

Multiply (2.3) on the left by $y$, one gets

$$\begin{aligned} (yxy^n-y^{n+1}x)y=0. \end{aligned}$$

(2.5)

From (2.4) and (2.5) we have

$$\begin{aligned} (xy-yx)y^{n+1}=0. \end{aligned}$$

(2.6)

Since (2.6) holds for all $y\in R$, substitute $y+1$ for $y$, to get

$$\begin{aligned} (xy-yx)(1+y)^{n+1}=0. \end{aligned}$$

(2.7)

From (2.6), (2.7) and Lemma 2.1, we have

$$\begin{aligned} xy=yx. \end{aligned}$$

(2.8)

Hence $R$ is $CN$.

$(3)\Longrightarrow (1)$ Suppose that $x\in N(R)$ and $y\in R$. Since $((1+x)y)^{n+1}=(1+x)y((1+x)y)^n$, by the hypothesis, we have

$$\begin{aligned} y^{n+1}(1+x)^{n+1}&= (1+x)y^{n+1}(1+x)^n, \end{aligned}$$

(2.9)

$$\begin{aligned} y^{n+2}(1+x)^{n+2}&= (1+x)y^{n+2}(1+x)^{n+1}. \end{aligned}$$

(2.10)

Since $1+x$ is invertible in $R$, (2.9) gives

$$\begin{aligned} xy^{n+1}=y^{n+1}x, \end{aligned}$$

(2.11)

(2.10) gives

$$\begin{aligned} xy^{n+2}=y^{n+2}x. \end{aligned}$$

(2.12)

Multiply (2.11) on the left by $y$, from (2.12), one gets

$$\begin{aligned} (xy-yx)y^{n+1}=0. \end{aligned}$$

(2.13)

Similar to the proof of $(2)\Longrightarrow (1)$, we have

$$\begin{aligned} xy=yx. \end{aligned}$$

(2.14)

Hence $R$ is $CN$. $\square$

Let $R$ be a $CN$ ring. Then for any $n\ge 2$ and any $a\in N(R)$ and $b\in R$, we have $(ab)^n=a^nb^n=b^na^n$. But the converse is not true in general.

Example 2.1

Let $D$ be a division ring and $R=\left( \begin{array}{cc}D&{}D\\ 0&{}D\end{array}\right)$. Then $N(R)=\left( \begin{array}{cc}0&{}D\\ 0&{}0\end{array}\right)$ with $N(R)^2=0$. Since $N(R)$ is an ideal of $R$, for any $n\ge 2$, any $A\in N(R)$ and $B\in R$, we have $(AB)^n=A^nB^n=B^nA^n=0$. But $R$ is not $CN$.

Theorem 2.4

Let $R$ be a $CN$ ring and $n\ge 1$. If for any $x, y\in R\backslash N(R)$, $(xy)^{n+k}=x^{n+k}y^{n+k}$, where $k=0, 1,2$, then $R$ is commutative.

Proof

It follows immediately from the result in [11].$\square$

Lemma 2.2

Let $R$ be a semiprime ring and $n\ge 2$. If for any $a\in N(R)$ and $b\in R$, $(ab)^n=a^nb^n$, then $R$ is reduced.

Proof

Let $a\in R$ with $a^n=0$. Then $(ax)^n=0$ for each $x\in R$. If $a\ne 0$, then $aR$ is a nonzero nil right ideal of $R$ satisfying the identity $z^n=0$ for all $z\in aR$. Now by [8, Lemma 1.1], $R$ has a nonzero nilpotent ideal which is a contradiction since $R$ is semiprime. Thus $a=0$, this implies $R$ is reduced. $\square$

Theorem 2.5

Let $R$ be a semiprime ring and $n\ge 1$. If for any $x\in R\backslash J(R)$ and $y\in R$, $(xy)^{n+k}=x^{n+k}y^{n+k}$, where $k=0, 1$, then $R$ is commutative.

Proof

If $N(R)\cap J(R)=0$, then by Lemma 2.2, $R$ is reduced. If $N(R)\cap J(R)\ne 0$, then there exists $0\ne a\in N(R)\cap J(R)$ with $a^2=0$. By the hypothesis, for any $y\in R$, we have

$$\begin{aligned} (1+a)^{n+k}(ya)^{n+k}=((1+a)ya)^{n+k}. \end{aligned}$$

(2.15)

Clearly, for any $i\ge 1$, one gets

$$\begin{aligned} ((1+a)ya)^i=(1+a)(ya)^i. \end{aligned}$$

(2.16)

Hence

$$\begin{aligned} (1+a)^{n+k}(ya)^{n+k}=(1+a)(ya)^{n+k}. \end{aligned}$$

(2.17)

Since $1+a$ is invertible in $R$ and $(1+a)^i=1+ia$ for each $i\ge 1$, we have

$$\begin{aligned} (n+k-1)a(ya)^{n+k}=0, k=0, 1. \end{aligned}$$

(2.18)

This implies

$$\begin{aligned} a(ya)^{n+1}=0. \end{aligned}$$

(2.19)

Hence

$$\begin{aligned} (ay)^{n+2}=0. \end{aligned}$$

(2.20)

This leads to $aR$ is a nonzero nil right ideal of $R$ satisfying the identity $z^{n+2}=0$ for all $z\in aR$. Now by [8, Lemma 1.1], $R$ has a nonzero nilpotent ideal which is a contradiction since $R$ is semiprime. Thus $N(R)\cap J(R)=0$ and so $R$ is reduced.

Now suppose $x, y\in R$. If $x, 1+x\notin J(R)$, then by the hypothesis, we have

$$\begin{aligned}&\displaystyle x^{n+1}y^{n+1}=x^ny^nxy, \end{aligned}$$

(2.21)

$$\begin{aligned}&\displaystyle (x+1)^{n+1}y^{n+1}=(x+1)^ny^n(x+1)y. \end{aligned}$$

(2.22)

They give

$$\begin{aligned} x^n(xy^n-y^nx)y&= 0. \end{aligned}$$

(2.23)

$$\begin{aligned} (x+1)^n(xy^n-y^nx)y&= 0. \end{aligned}$$

(2.24)

From Lemma 2.1, (2.23) and (2.24), one gets

$$\begin{aligned} (xy^n-y^nx)y=0. \end{aligned}$$

(2.25)

If $x\in J(R)$, then $x+1$ is invertible in $R$, so

$$\begin{aligned} (x+1)^{n+1}y^{n+1}=(x+1)^ny^n(x+1)y. \end{aligned}$$

(2.26)

This gives

$$\begin{aligned} (xy^n-y^nx)y=0. \end{aligned}$$

(2.27)

If $x\notin J(R)$ and $1+x\in J(R)$, then $x$ is invertible in $R$, so (2.21) implies

$$\begin{aligned} (xy^n-y^nx)y=0. \end{aligned}$$

(2.28)

Hence, in any case, we have

$$\begin{aligned} (xy^n-y^nx)y&= 0, \end{aligned}$$

(2.29)

$$\begin{aligned} (y(xy^n-y^nx))^2&= 0. \end{aligned}$$

(2.30)

Since $R$ is reduced, one gets

$$\begin{aligned} y(xy^n-y^nx)=0. \end{aligned}$$

(2.31)

Clearly, for any $r\in R$, we have

$$\begin{aligned} ((xy^n-y^nx)ry)^2=0. \end{aligned}$$

(2.32)

Hence, for any $r\in R$, we have

$$\begin{aligned} (xy^n-y^nx)ry=0. \end{aligned}$$

(2.33)

that is,

$$\begin{aligned} (xy^n-y^nx)Ry=0. \end{aligned}$$

(2.34)

Thus

$$\begin{aligned} (xy^n-y^nx)R(xy^n-y^nx)=0. \end{aligned}$$

(2.35)

Since $R$ is semiprime, one gets

$$\begin{aligned} xy^n=y^nx. \end{aligned}$$

(2.36)

Since $R$ has no nonzero nil ideals, by [9, Theorem 2], $R$ is commutative.$\square$

Corollary 2.1

Let $R$ be a primitive ring and $n\ge 1$. If for any $x, y\in R$, $(xy)^k=x^ky^k$, where $k=n, n+1$, then $R$ is a field.

Proof

By Theorem 2.5, $R$ is commutative. We claim that $R$ is a division ring. If not, there exists a subring $S$ of $R$ such that $S$ is isomorphic to $2\times 2$ full matrix ring $M_2(D)$ over a division ring $D$. Clearly, for any $x, y\in S$, $(xy)^k=x^ky^k$, where $k=n, n+1$, hence $M_2(D)$ satisfies the same conditions. Now let $A=\left( \begin{array}{cc}0&{}1\\ 0&{}0\end{array}\right) , B=\left( \begin{array}{cc}0&{}0\\ 1&{}0\end{array}\right)$, $(AB)^{n+1}\ne A^{n+1}B^{n+1}$, which is a contradiction. Hence $R$ is a division ring, and so $R$ is a field.$\square$

Theorem 2.6

$R$ is a $CN$ ring if and only if for some positive integer $n\ge 1, m> 1$ and any $x\in R$ and $y\in N(R)$, $[xy-x^ny^m, x]=0$.

Proof

One direction is clear.

Now assume that $n\ge 1, m> 1$ such that for any $x\in R$ and $y\in N(R)$, we have $[xy-x^ny^m, x]=0$. Since $y\in N(R)$, there exists $p\ge 1$ such that $y^{m^p}=0$. The equation $[xy-x^ny^m, x]=0$ gives

$$\begin{aligned} x[x, y]=x^n[x, y^m]. \end{aligned}$$

(2.37)

Since $y^m\in N(R)$, substitute $y^m$ for $y$ in (2.37), one gets

$$\begin{aligned} x^2[x, y]=x^{n+1}[x, y^m]=x^n(x[x, y^m])=x^{2n}[x, y^{m^2}]. \end{aligned}$$

(2.38)

Hence

$$\begin{aligned} x^p[x, y]=x^{np}[x, y^{m^p}]. \end{aligned}$$

(2.39)

This implies

$$\begin{aligned} x^p[x,y]=0. \end{aligned}$$

(2.40)

Since (2.40) holds for all $x\in R$, substitute $x+1$ for $x$ and use Lemma 2.1, we have $[x, y]=0$. Hence $R$ is $CN$. $\square$

Theorem 2.7

Let $R$ be a ring and $n\ge 1$. If for any $x\in R\backslash N(R)$ and $y\in R$, $(xy)^k=x^ky^k$, $k=n, n+1, n+2$, then $R$ is commutative.

Proof

Suppose that $x, y\in R$. If $x\in N(R)$, then $1+x$ is invertible. By the hypothesis,

$$\begin{aligned} (1+x)^ky^k=((1+x)y)^k, k=n, n+1, n+2. \end{aligned}$$

(2.41)

Hence, one gets

$$\begin{aligned} (1+x)^{n-1}y^n&= y((1+x)y)^{n-1}, \end{aligned}$$

(2.42)

$$\begin{aligned} (1+x)^ny^{n+1}&= y((1+x)y)^n, \end{aligned}$$

(2.43)

$$\begin{aligned} (1+x)^{n+1}y^{n+2}&= y((1+x)y)^{n+1}. \end{aligned}$$

(2.44)

Multiply (2.42) on the right by $(1+x)y$, from (2.43), one gets

$$\begin{aligned} y^nxy=xy^{n+1}. \end{aligned}$$

(2.45)

Multiply (2.43) on the right by $(1+x)y$, from (2.44), one gets

$$\begin{aligned} y^{n+1}xy=xy^{n+2}.\end{aligned}$$

(2.46)

Multiply (2.45) on the left by $y$, from (2.46), one gets

$$\begin{aligned} (xy-yx)y^{n+1}=0.\end{aligned}$$

(2.47)

If $x\notin N(R)$, then by the hypothesis, one gets

$$\begin{aligned} (xy)^k=x^ky^k, k=n, n+1, n+2.\end{aligned}$$

(2.48)

If $1+x\in N(R)$, then $x$ is invertible in $R$. Similar to the proof of above, (2.48) implies

$$\begin{aligned} (xy-yx)y^{n+1}=0.\end{aligned}$$

(2.49)

If $1+x\notin N(R)$, then one has

$$\begin{aligned} ((1+x)y)^k=(1+x)^ky^k, k=n, n+1, n+2.\end{aligned}$$

(2.50)

Similar to the proof of Theorem 2.6, (2.48) and (2.50) imply

$$\begin{aligned} (xy-yx)y^{n+1}=0. \end{aligned}$$

(2.51)

Hence, (2.47), (2.49) and (2.51) imply that in any case, one has

$$\begin{aligned} (xy-yx)y^{n+1}=0. \end{aligned}$$

(2.52)

Substitute $y+1$ for $y$ in (2.52), one gets

$$\begin{aligned} (xy-yx)(y+1)^{n+1}=0\end{aligned}$$

(2.53)

By Lemma 2.3, (2.52), and (2.53), one obtains $xy=yx$. Thus $R$ is commutative. $\square$

Following [5], an element $x$ of $R$ is called weakly clean if $x=u+e$ or $x=u-e$ for some $u\in U(R)$ and $e\in E(R)$. The ring $R$ is said to be weakly clean if all of its elements are weakly clean. Clean rings are weakly clean. But the converse is not true because of the example $Z_{(3)}\cap Z_{(5)}$ where $Z_{(p)}=\{ \frac{r}{s}| p$ does not divide $s\}$. An element $x$ of $R$ is called weakly exchange if there exists $e\in E(R)$ such that $e\in xR$ and $1-e\in (1-x)R$ or $1-e\in (1+x)R$. The ring $R$ is said to be weakly exchange if all of its elements are weakly exchange. Clearly, exchange elements are weakly exchange. Checking carefully the proof of [5, Theorem 2.1], we find that weakly clean elements are weakly exchange, so weakly clean rings and exchange rings are all weakly exchange. In fact, [5, Theorem 2.1] showed that Abel weakly exchange rings are weakly clean. In this paper, we obtain that $NLI$ weakly exchange rings are weakly clean.

Theorem 2.8

Let $R$ be an $NLI$ ring and $x\in R$.

• (1) If $x$ is weakly exchange, then $x$ is weakly clean.

• (2) If $x$ is exchange, then $x$ is clean.

• (3) If $R$ is a weakly exchange ring, then $R$ is a weakly clean ring.

• (4) If $R$ is an exchange ring, then $R$ is a clean ring.

Proof

(1) Let $e\in E(R)$ such that $e\in xR$ and $1-e\in (1-x)R$ or $1-e\in (1+x)R$. Write $e=xy$ for some $y=ye\in R$. If $1-e\in (1-x)R$, then let $1-e=(1-x)z$ for some $z=z(1-e)\in R$. By computing, we have $(x-(1-e))(y-z)=1-(1-e)y-ez$. Since $R$ is a $NLI$ ring and $(1-e)y=(1-e)ye\in N(R)$, $(1-e)yez-ez(1-e)y\in N(R)$, that is, $(1-e)yz-ezy\in N(R)$. Hence there exists $n\ge 1$ such that $((1-e)yz-ezy)^n=0$. By computing, we have $((1-e)yz)^n+(-1)^n(ezy)^n=0$, this leads to $((1-e)yz)^n=(ezy)^n=0$. Since $(ez+(1-e)y)^2=ezy+(1-e)yz$, $(ez+(1-e)y)^{2n}=(ezy)^n+((1-e)yz)^n=0$. Hence $1-ez-(1-e)y\in U(R)$, that is, $(x-(1-e))(y-z)\in U(R)$. Let $u\in R$ such that $((x-(1-e))(y-z))u=1$. Let $g=((y-z)u)(x-(1-e))$. Then $(x-(1-e))g=x-(1-e)$ and $g^2=g$. Let $h=(x-(1-e))-g(x-(1-e))$. Then $hg=h$, $gh=0$ and $h^2=0$. Since $R$ is an $NLI$ ring, $(y-z))uh-h(y-z))u\in N(R)$, that is, $(y-z))uh-(1-g)\in N(R)$. Hence there exists $n\ge 1$ such that $((y-z))uh-(1-g))^n=0$, this gives $1-g=d(y-z)uh$ for some $d\in R$. Thus $1-g=d(y-z))uh=d(y-z))uhg=(1-g)g=0$, so $((y-z)u)(x-(1-e))=g=1$. Hence $x-(1-e)\in U(R)$. If $1-e\in (1+x)R$, then let $1-e=(1+x)w$ for some $w=w(1-e)\in R$. By computing, we have $(x+(1-e))(y+w)=1+(1-e)y-ew$. Similarly, we obtain $1+(1-e)y-ew\in U(R)$, this gives $x+(1-e)\in U(R)$. We are done. (2) It has been have shown in (1).

(3) and (4) are immediate corollaries of (1) and (2), respectively.$\square$

A ring $R$ is called a generalized CN ring if for any $a\in N(R)$ and $b\in R$, $ab=0$ implies $aRb=0$ or there exists $c\in R$ such that $0\ne acb\in Z(R)$. Clearly, $CN$ rings are generalized $CN$. But the converse is not true. For example, the ring $R$ in Example 2.5 is a generalized $CN$ ring, but $R$ is not $CN$.

A ring $R$ is called left SF if every simple left $R-$module is flat. In [13, Remark 3.13], it is shown that if $R$ is a reduced left $SF$ ring, then $R$ is strongly regular. We can generalize this result as follows.

Proposition 2.1

Let $R$ be a generalized $CN$ ring. If $R$ is a left $SF$ ring, then $R$ is a strongly regular ring.

Proof

Let $a\in R$ with $aRa=0$. If $a\ne 0$, then there exists a maximal left ideal $M$ of $R$ such that $r(aR)\subseteq M$. Since $R$ is a left $SF$ ring, $R/M$ is flat as left $R-$module. Since $a\in r(aR)\subseteq M$, $a=am$ for some $m\in M$. Since $R$ is a generalized $CN$ ring and $a(1-m)=0$, $aR(1-m)=0$, or there exists $c\in R$ such that $0\ne ac(1-m)\in Z(R)$. If $aR(1-m)=0$, then $1-m\in r(aR)\subseteq M$, this gives $1=(1-m)+m\in M$, a contradiction. Thus there exists $c\in R$ such that $0\ne ac(1-m)\in Z(R)$. Since $(ac(1-m))^2=0$, there exists a maximal left ideal $N$ of $R$ such that $l(ac(1-m))\subseteq N$. Since $R$ is a left $SF$ ring, $R/N$ is flat as left $R-$module. Then since $ac(1-m)\in N$, $ac(1-m)=ac(1-m)n$ for some $n\in N$. Since $ac(1-m)\in Z(R)$, $ac(1-m)=nac(1-m)$, this leads to $1-n\in l(ac(1-m))\subseteq N$, which implies $1=(1-n)+n\in N$, a contradiction. Hence $a=0$, which implies $R$ is a semiprime ring. Now let $b\in R$ with $b^2=0$. Since $R$ is a generalized $CN$ ring, either $bRb=0$ or there exists $c\in R$ such that $0\ne bab\in Z(R)$. If there exists $c\in R$ such that $0\ne bcb\in Z(R)$, then $bcbRbcb=bcbbcbR=0$, so $bcb=0$ because $R$ is semiprime, which is a contradiction. Hence $bRb=0$, also, the semiprimeness of $R$ implies $b=0$. Hence $R$ is a reduced ring. By [13, Remark 3.13], $R$ is a strongly regular ring. $\square$

Following [4], a ring $R$ is said to be semiperiodic if for each $x\in R \backslash (J(R)\cup Z(R))$, there exist $m, n \in \mathbb {Z}$, of opposite parity, such that $x^n-x^m\in N(R)$. Clearly, the class of semiperiodic rings contains all commutative rings, all Jacobson radical rings, and certain non-nil periodic rings.

Lemma 2.3

Let $R$ be a generalized $CN$ ring. If $R$ is a semiperiodic ring, then $N(R)\subseteq J(R)$

Proof

Let $a\in N(R)$ with $a^k = 0$, and let $x\in R$. If $ax\in J(R)$, then $ax$ is right quasiregular; and if $ax\in Z(R)$, then $ax$ is nilpotent and again $ax$ is right quasiregular. Suppose, then, that $ax\notin J(R)\cup Z(R)$, in which case [4, Lemma 2.3(iii)] gives $q\in \mathbb {Z}^+$ and an idempotent $e$ of form $axy$ such that $(ax)^q = (ax)^qe$. Since $e=axy=eaxy=ea(1-e)xy+eaexy=ea(1-e)xy+ea^2(xy)^2=ea(1-e)xy+ea^2(1-e)(xy)^2+ea^2e(xy)^2 =ea(1-e)xy+ea^2(1-e)(xy)^2+ea^3(xy)^3=\cdots =\Sigma _{i=1}^{k-1}ea^i(1-e)(xy)^i+ea^k(xy)^k=\Sigma _{i=1}^{k-1}ea^i(1-e)(xy)^i$. For any $z\in R$, $ez(1-e)\in N(R)$ and $(ez(1-e))^2=0$. Since $R$ is a generalized $CN$ ring, either $ez(1-e)Rez(1-e)=0$ or there exists $c\in R$ such that $0\ne ez(1-e)cez(1-e)\in Z(R)$. If there exists $c\in R$ such that $0\ne ez(1-e)cez(1-e)\in Z(R)$, then $ez(1-e)cez(1-e)=(ez(1-e)cez(1-e))(1-e)=(1-e)ez(1-e)cez(1-e)=0$, which is a contradiction. Hence $ez(1-e)Rez(1-e)=0$, which implies $ez(1-e)\in J(R)$ for any $z\in R$. Therefore $e=\Sigma _{i=1}^{k-1}ea^i(1-e)(xy)^i\in J(R)$, this leads to $e=0$ and $(ax)^q=0$, which shows that $ax$ is right quasi-regular. Thus $a\in J(R)$.$\square$

Theorem 2.9

If $R$ is a generalized $CN$ semiperiodic ring, then $R/J(R)$ is commutative.

Proof

By [4, Theorem 4.3], $R$ is either commutative or periodic, so we may assume $R$ is periodic. Since $J(R)$ contains no nonzero idempotents, $J(R)$ is contained in $N(R)$ and hence $J(R)=N(R)$ by Lemma 2.3. Thus $R/J(R)=R/N(R)$ is reduced; and since $R/N(R)$ is also semiperiodic, it is commutative by [4, Theorem 4.4]. $\square$

Theorem 2.10

Let $R$ be a generalized $CN$ semiperiodic ring. Then

• (1) $N(R)$ is an ideal of $R$.

• (2) If $J(R)\ne N(R)$, then $R$ is commutative.

Proof

In the proof of Theorem 2.9, we obtain that if $R$ is not commutative, then $J(R)=N(R)$. Hence $(2)$ holds and $(1)$ also holds for noncommutative ring $R$. But also if $R$ is commutaive, $N(R)$ is an ideal; hence $(1)$ holds in any case. $\square$