Rings Whose Nilpotent Elements Form a Wedderburn Radical Subring

Abstract

We introduce and study a class of rings, which we call NRW rings, distinguished by the condition that the set of nilpotent elements forms a Wedderburn radical subring. This class includes symmetric and semicommutative rings, as well as many other well-known and important classes. We also define nilpotent-semicommuting rings as those satisfying the semicommutativity condition restricted to nilpotent elements and prove that every nilpotent-semicommuting ring is an NRW ring. We provide an element-wise characterization of NRW rings and show that the NRW property is symmetric with respect to one-sided principal ideals. Based on the right ideals of a ring that are NRW, for any ring R we inductively construct the ideal $\overline{\mathcal{W}}\left(R\right)$ and prove that the prime radical of R equals the intersection of $\overline{\mathcal{W}}\left(R\right)$ with the sum of all nil right ideals of R. As a consequence, a positive solution to the Köthe conjecture follows for all rings R satisfying $R=\overline{\mathcal{W}}\left(R\right)$. We also characterize when certain classical ring constructions, such as direct sums, matrix rings, and the Dorroh extension, yield an NRW ring.

1. Introduction

The rings considered in this paper are associative but not necessarily commutative or unital. In ring theory, various classes of rings have been distinguished according to properties that generalize commutativity and reflect certain symmetry conditions in multiplication. One example is the class of symmetric rings, defined as rings R in which the condition $abt=0$ implies $atb=0$ for all $a,b,t\in R$. Another is the broader class of semicommutative rings, consisting of rings R such that $ab=0$ implies $atb=0$ for all $a,b,t\in R$. Many more examples of such symmetry-related and commutativity-related classes of rings will be presented in the later part of this work.

In this paper, we introduce a new class of rings that includes all symmetric and semicommutative rings, as well as many other well-studied classes in ring theory, which will be discussed in Section 3. For reasons that will become clear shortly, we assign the name NRW rings to rings of this new type. To define NRW rings, we first need some notation. For a ring R, let $N\left(R\right)$ denote the set of nilpotent elements of R, and let $W\left(R\right)$ stand for the Wedderburn radical of R; that is, $W\left(R\right)$ is the sum of all nilpotent ideals of R. A ring R is said to be a Wedderburn radical ring if $R=W\left(R\right)$.

It is well-known that the set $N\left(R\right)$ strongly influences the properties of a ring R. For example, a ring R is a subdirect product of domains if and only if $N\left(R\right)=\left\{0\right\}$, i.e., if R is reduced. The Wedderburn radical also plays a significant role in ring theory; for instance, it appears as the first step in the inductive construction of the prime radical of a ring.

If R is a commutative ring, then $N\left(R\right)$ is a subring of R. Thus, rings R in which $N\left(R\right)$ is a subring, called NR rings (with NR standing for nilpotents form a ring, [1]), are a generalization of commutative rings. It is well known that both symmetric and semicommutative rings are NR.

In general, for a ring R, the set $N\left(R\right)$ is neither closed under addition nor under multiplication. Indeed, if R is the ring of $2\times 2$ matrices over a ring with unity, then for the matrices $a={(}_{-1-1}^{11})$ and $b={(}_{10}^{00})$, we have $a,b\in N\left(R\right)$, but neither $a+b$ nor $ab$ belong to $N\left(R\right)$. Thus, in general, a ring need not be an NR ring. Furthermore, even when a ring R is NR, the subring $N\left(R\right)$ need not be Wedderburn radical. For example, if A is any ring such that $W(A/W(A\left)\right)\ne \left\{0\right\}$ (for references to examples of such A, see [2] (p. 112)) and R is the ideal of A satisfying $W(A/W(A\left)\right)=R/W\left(A\right)$, then $N\left(R\right)=R$, and thus R is an NR ring. However, $W\left(N\right(R\left)\right)=W\left(R\right)=W\left(A\right)⊊R=N\left(R\right)$.

As indicated above, for a ring R, the set $N\left(R\right)$ is typically not a subring of R, and even when it is, this subring need not be Wedderburn radical. Nevertheless, there exist many important classes of rings (a wide range of them will be presented in Section 3) in which the nilpotent elements form a Wedderburn radical subring. The aim of this paper is to investigate such rings systematically. We call them NRW rings, since they are NR rings, and the additional letter “W” indicates that the subring formed by nilpotent elements is Wedderburn radical.

Definition 1.

A ring R is called an NRW ring if the set $N\left(R\right)$ of nilpotent elements of R is a Wedderburn radical subring of R.

Note that if R is a commutative ring and $a,b\in N\left(R\right)$, then the ideals generated by a and b are both nilpotent, so their sum is nilpotent as well, which implies that R is an NRW ring. Moreover, any reduced ring R is an NRW ring, since in this case we have $N\left(R\right)=\left\{0\right\}$, which is trivially a Wedderburn radical subring of R. Numerous further examples of well-known classes of rings that fall within the NRW framework will be presented later on.

The structure of the paper is as follows. In Section 2, Proposition 1 provides a useful element-wise characterization of NRW rings, whose immediate consequence is that the class of NRW rings is closed under isomorphisms and subrings. We also show (Lemma 1) that the NRW property, when considered for one-sided principal ideals of a ring R, is symmetric in the sense that for any $a\in R$, the left ideal of R generated by a is an NRW ring if and only if so is the right ideal generated by a.

In Section 3, we introduce a new class of nilpotent-semicommuting rings, defined as rings that satisfy the semicommutativity condition ($ab=0\Rightarrow atb=0$) when restricted to nilpotent elements. In Theorem 1, we prove that every nilpotent-semicommuting ring is an NRW ring, while Example 3 shows that the converse does not hold. Moreover, we present numerous examples of important classes of rings that are nilpotent-semicommuting, including right nil-symmetric, right distributive, Armendariz, lineal, and right filial rings. Consequently, all these rings also belong to the class of NRW rings.

In Section 4, in Proposition 2, we show that the Wedderburn radical of an NRW ring coincides with the sum $A\left(R\right)$ of all nil right ideals of R. We also answer a question posed by Šter regarding nilpotent elements in exchange rings. Moreover, for any ring R, we define the set $\mathcal{W}\left(R\right)$ as the sum of all right ideals of R that are NRW rings. Since $\mathcal{W}\left(R\right)$ turns out to be an ideal of R, we can use it to inductively construct the ideal $\overline{\mathcal{W}}\left(R\right)$, analogously to how the Wedderburn radical $W\left(R\right)$ is used in the construction of the prime radical $P\left(R\right)$ of R. In Theorem 3, we prove that for any ring R, the equality $P\left(R\right)=\overline{\mathcal{W}}\left(R\right)\cap A\left(R\right)$ holds. An immediate consequence is that the Köthe conjecture has a positive solution in the class of all rings R such that $R=\overline{\mathcal{W}}\left(R\right)$, in particular for NRW rings.

In Section 5, we consider NRW rings in the context of the following ring constructions: direct sums, matrix rings, the Dorroh extension, and the trivial extension. For each of these, we provide the conditions under which they yield an NRW ring.

In this paper, the ring of integers is denoted by $\mathbb{Z}$, and $\mathbb{N}$ denotes the set of positive integers. For an element a of a ring R, the right (resp. left) ideal generated by a is denoted by $a{R}^1$ (resp. $R^{1}a$); that is, $a{R}^1=a\mathbb{Z}+aR=\{ma+ar:m\in \mathbb{Z},r\in R\}$ and $R^{1}a=a\mathbb{Z}+Ra=\{ma+ra:m\in \mathbb{Z},r\in R\}$. A ring R is called an NI ring (with NI standing for nilpotents form an ideal) if $N\left(R\right)$ is an ideal of R. The Jacobson radical of a ring R is denoted by $J\left(R\right)$, and the right annihilator of a nonempty subset $A\subseteq R$ is denoted by $r_R\left(A\right)$. A unital ring R is said to be local if its Jacobson radical $J\left(R\right)$ is the unique maximal right ideal of R.

2. Characterization and Preliminary Properties of NRW Rings

For nonempty subsets $S_1,S_2,\dots ,S_n$ of a ring R, we denote

$$S_1{S}_2\dots S_n=\{s_1{s}_2\dots s_n:s_1\in S_1,s_2\in S_2,\dots ,s_n\in S_n\}.$$

(1)

If $S_{i+1}=S_{i+2}=\dots =S_{i+k}=A$ for some i and k, then the set (1) is written as $S_1\dots S_i{A}^k{S}_{i+k+1}\dots S_n$. If some $S_i$ is a singleton, say $S_i=\left\{a\right\}$, then (1) is written as $S_1\dots S_{i-1}a{S}_{i+1}\dots S_n$, rather than $S_1\dots S_{i-1}\left\{a\right\}{S}_{i+1}\dots S_n$. The zero ideal of R, i.e., the set $\left\{0\right\}$, is denoted by 0. For a nonempty subset S of R, we say that S is a nilpotent set if $S^n=0$ for some $n\in \mathbb{N}$.

The following proposition provides a useful element-wise characterization of NRW rings.

Proposition 1.

For a ring R, the following conditions are equivalent.

(1) 

R is an NRW ring.

(2) 

For any $a\in N\left(R\right)$, the set $aN\left(R\right)$ is nilpotent.

(3) 

For any $a\in N\left(R\right)$, the set $N\left(R\right)a$ is nilpotent.

(4) 

For any $a\in N\left(R\right)$, the set $N\left(R\right)aN\left(R\right)$ is nilpotent.

Proof. 

(1) ⇒ (2): Assume (1), and consider any $a\in N\left(R\right)$. Let I be the ideal of $N\left(R\right)$ generated by a. Since R is an NRW ring, I is nilpotent. Hence, since $aN\left(R\right)\subseteq I$, the set $aN\left(R\right)$ is nilpotent.

(2) ⇒ (3): For any $a\in R$ and $n\in \mathbb{N}$, we have that ${\left(N\left(R\right)a\right)}^{n+1}=N\left(R\right){\left(aN\left(R\right)\right)}^{n}a$, and consequently (2) implies (3).

(3) ⇒ (4): This implication follows from the observation that (3) implies that the set $N\left(R\right)$ is multiplicatively closed; thus, for any $a\in R$ and $n\in \mathbb{N}$, we have that ${\left(N\left(R\right)aN\left(R\right)\right)}^n\subseteq {\left(N\left(R\right)a\right)}^{n}N\left(R\right)$.

(4) ⇒ (1): Assume (4), and consider any $a,b\in N\left(R\right)$. By assumption there exists $n\in \mathbb{N}$ such that ${\left(N\left(R\right)aN\left(R\right)\right)}^n={\left(N\left(R\right)bN\left(R\right)\right)}^n=0.$ Clearly, ${(a+b)}^{6n}$ is a finite sum of elements of the form $c=a^{{\alpha }_1}{b}^{{\beta }_1}{a}^{{\alpha }_2}{b}^{{\beta }_2}\dots a^{{\alpha }_k}{b}^{{\beta }_k},$ where ${\alpha }_i$ and ${\beta }_i$ are nonnegative integers satisfying ${\sum }_{i=1}^k({\alpha }_i+{\beta }_i)=6n$. Then ${\sum }_{i=1}^k{\alpha }_i\ge 3n$ or ${\sum }_{i=1}^k{\beta }_i\ge 3n$; thus, $c\in {\left(N\left(R\right)aN\left(R\right)\right)}^n=0$ or $c\in {\left(N\left(R\right)bN\left(R\right)\right)}^n=0,$ which implies ${(a+b)}^{6n}=0.$ Hence $a+b\in N\left(R\right)$, i.e., the set $N\left(R\right)$ is additively closed. Thus, [3] (Theorem 2.1) implies that $N\left(R\right)$ is a subring of R. Moreover, if I is the ideal of $N\left(R\right)$ generated by a, then $I^3\subseteq N\left(R\right)aN\left(R\right)$; thus, $I^{3n}\subseteq {\left(N\left(R\right)aN\left(R\right)\right)}^n=0$, which shows that the ring $N\left(R\right)$ is Wedderburn radical. Hence R is an NRW ring. □

Corollary 1.

If R is an NRW ring, then

(i) 

Any ring isomorphic to R is an NRW ring;

(ii) 

Any subring of R is an NRW ring.

The example below shows that a homomorphic image of an NRW ring need not be an NRW ring.

Example 1.

Let F be a field, and let $R=F\langle x,y\rangle$ be the polynomial algebra over F in non-commuting indeterminates x and y. Since R is a domain, R is an NRW ring. Let I be the ideal of R generated by $x^2$ and $y^2$, and let $\overline{R}=R/I$ be the corresponding factor ring. Then $x+I\in N\left(\overline{R}\right)$ and $y+I\in N\left(\overline{R}\right)$, but $(x+I)(y+I)\notin N\left(\overline{R}\right)$, which shows that the set of nilpotent elements of $\overline{R}$ does not form a subring. Hence $\overline{R}$ is not an NRW ring.

The properties of NRW rings stated in the following lemma will be very useful in Section 4. Part (ii) of the lemma shows that the NRW property, when considered for one-sided principal ideals of a ring R, is symmetric in the sense that, for any element $a\in R$, the left ideal of R generated by a is an NRW ring if and only if so is the right ideal generated by a.

Lemma 1.

Let R be a ring, and let $a\in R$. Then

(i) 

For any right (resp. left) ideal I of R, if I is an NRW ring, then $aI$ (resp. $Ia$) is an NRW ring.

(ii) 

The right ideal $a{R}^1$ is an NRW ring if and only if the left ideal $R^{1}a$ is an NRW ring.

Proof. 

(i) Assume that I is a right ideal of R such that I is an NRW ring. Let S be a subset of R such that $N\left(aI\right)=aS$. Observe that if $s\in S$, then ${\left(as\right)}^m=0$ for some $m\in \mathbb{N}$; hence ${\left(sa\right)}^{m+1}=s{\left(as\right)}^{m}a=0$, and thus $sa\in N\left(I\right)$. Therefore, $Sa\subseteq N\left(I\right)$. Now, let $b\in N\left(aI\right)$. Then $b=at$ for some $t\in S$, and thus $ta\in N\left(I\right)$. Since I is an NRW ring, by Proposition 1 there exists $n\in \mathbb{N}$ with ${\left(taN\left(I\right)\right)}^n=0$. Hence

$${\left(bN\left(aI\right)\right)}^{n+1}={\left(ataS\right)}^{n+1}\subseteq a{\left(taSa\right)}^{n}taS\subseteq a{\left(taN\left(I\right)\right)}^{n}taS=0,$$

and Proposition 1 implies that $aI$ is an NRW ring.

The case where I is a left ideal that is an NRW ring can be proved similarly.

(ii) Note first that, for any $k\in \mathbb{N}$, $x,a,r,r_1,\dots ,r_k\in R$ and $m,m_1,\dots ,m_k\in \mathbb{Z}$, we have

$$x(ma+ra)\prod _{i=1}^k(m_{i}a+r_{i}a)=(xm+xr)\left(\prod _{i=1}^k(a{m}_i+a{r}_i)\right)a$$

(2)

and

$$\left(\prod _{i=1}^k(a{m}_i+a{r}_i)\right)(am+ar)x=a\left(\prod _{i=1}^k(m_{i}a+r_{i}a)\right)(mx+rx).$$

(3)

From (2) and (3) it follows that, for any $a,r\in R$ and $m\in \mathbb{Z}$, the following holds:

$$am+ar\in N\left(a{R}^1\right)\mathit{if}\mathit{and}\mathit{only}\mathit{if}ma+ra\in N\left(R^{1}a\right).$$

(4)

Now, assume that $a{R}^1$ is an NRW ring, and let $b\in N\left(R^{1}a\right)$. Then $b=ma+ra$ for some $m\in \mathbb{Z}$ and $r\in R$. Since $a{R}^1$ is NRW, and $am+ar\in N\left(a{R}^1\right)$ by (4), Proposition 1 implies the existence $n\in \mathbb{N}$ such that

$${\left((am+ar)N\left(a{R}^1\right)\right)}^n=0.$$

(5)

Consequently, using (2), (4) and (5), we get

$${\left(bN\left(R^{1}a\right)\right)}^{n+1}={\left((ma+ra)N\left(R^{1}a\right)\right)}^{n+1}\subseteq R{\left((am+ar)N\left(a{R}^1\right)\right)}^{n}a=0.$$

Thus, by Proposition 1, $R^{1}a$ is an NRW ring. The converse implication follows by a symmetric argument. □

3. Examples of NRW Rings

As already noted, every commutative ring and every reduced ring is an NRW ring. In this section, we present many examples of important classes of rings, well known in ring theory, each consisting entirely of NRW rings. These examples illustrate that the NRW condition captures a structural property common to a wide variety of ring classes.

Recall that a ring R is said to be semicommutative if for any elements $a,b,t\in R$,

$$ab=0\mathrm{implies}atb=0.$$

(6)

Such rings are a common generalization of commutative rings and reduced rings. Below, we introduce a much broader class of rings defined by requiring that condition (6) holds whenever the elements involved are nilpotent, rather than for arbitrary elements of the ring R. We also show that all rings from this new class are NRW rings.

Definition 2.

A ring R is said to be nilpotent-semicommuting if for any $a,b,t\in N\left(R\right)$, $ab=0$ implies $atb=0$.

Remark 1.

Since the class of rings just defined is characterized by restricting the semicommutativity condition (i.e., $ab=0\Rightarrow atb=0$) only to nilpotent elements, it would be natural to call such rings nilpotent semicommutative. However, this name is already used for a different class of rings (see Example 10); therefore, we have named rings of this new type nilpotent-semicommuting.

To illustrate Definition 2, in the example below we present two rings: one that is nilpotent-semicommuting and one that is not.

Example 2.

For a ring R, let $U_2\left(R\right)$ denote the ring of upper triangular $2\times 2$ matrices over R.

(i) 

Consider the ring $R={\mathbb{Z}}_4$ of integers modulo 4. The set of nilpotent elements in $U_2\left({\mathbb{Z}}_4\right)$ is $N\left(U_2\left({\mathbb{Z}}_4\right)\right)={(}_{02{\mathbb{Z}}_4}^{2{\mathbb{Z}}_4{\mathbb{Z}}_4})$, and it follows that $N{\left(U_2\left({\mathbb{Z}}_4\right)\right)}^3=0$. Thus, for any three nilpotent matrices $a,b,t\in U_2\left({\mathbb{Z}}_4\right)$, we have $atb=0$, regardless of whether $ab=0$ or not. Hence the ring $U_2\left({\mathbb{Z}}_4\right)$ is nilpotent-semicommuting.

(ii) 

Now let $R={\mathbb{Z}}_8$ be the ring of integers modulo 8. Consider the following three nilpotent matrices in $U_2\left({\mathbb{Z}}_8\right)$: $a={(}_{04}^{22}),b={(}_{02}^{42})$ and $t={(}_{00}^{01})$. We have $ab=0$, but $atb={(}_{00}^{04})\ne 0$. Therefore, the ring $U_2\left({\mathbb{Z}}_8\right)$ is not nilpotent-semicommuting.

In the theorem below, we show that the class of nilpotent-semicommuting rings is a subclass of NRW rings.

Theorem 1.

If R is a nilpotent-semicommuting ring, then R is an NRW ring.

Proof. 

Let R be a nilpotent-semicommuting ring. For any $n\in \mathbb{N}$, define $N_n=\{a\in R:a^n=0\}$. We claim that if $a\in N_n$ and $b\in N\left(R\right)$, then $ab\in N_n$, which obviously implies that the set $N\left(R\right)$ is closed under multiplication. To prove the claim, we proceed by induction on n. The case $n=1$ is clear. Now assume $n\ge 2$ and that $cd\in N_{n-1}$ for any $c\in N_{n-1}$ and $d\in N\left(R\right)$. Let $a\in N_n$ and $b\in N\left(R\right)$. Since $2(n-1)\ge n$, it follows that $a^2\in N_{n-1}$, and thus, by induction hypothesis, $a^2{t}_1{t}_2\dots t_m\in N_{n-1}$ for any $m\in \mathbb{N}$ and $t_1,t_2,\dots ,t_m\in N\left(R\right).$ In particular, $a^2{\left(ba\right)}^{k-1}b\in N_{n-1}$ for any $k\in \mathbb{N}$. Hence ${\left({\left(ab\right)}^{k}a\right)}^n={\left(ab\right)}^k{\left(a^2{\left(ba\right)}^{k-1}b\right)}^{n-1}a=0$, which shows that

$${\left(ab\right)}^{k}a\in N\left(R\right)\mathrm{for}\mathrm{any}k\in \mathbb{N}.$$

(7)

Now, since $a^n=0$ and R is nilpotent-semicommuting, we obtain $0=a^{n-1}a=a^{n-1}ba$. Hence $0=a^{n-2}aba$, and since R is nilpotent-semicommuting and $aba\in N\left(R\right)$ by (7), it follows that $0=a^{n-2}baba$. Thus, $0=a^{n-3}{\left(ab\right)}^{2}a$, where ${\left(ab\right)}^{2}a\in N\left(R\right)$ by (7), and applying again that R is nilpotent-semicommuting, we get $0=a^{n-3}b{\left(ab\right)}^{2}a$. Continuing this way, we obtain $ab\in N_n$, which proves the claim.

We already know that the product of any two nilpotent elements of R is nilpotent. Therefore, if $a\in R$ and $a^n=0$ for some $n\in \mathbb{N}$, then it follows directly from the definition of a nilpotent-semicommuting ring that $a{t}_{1}a{t}_2\dots a{t}_n=0$ for any $t_1,t_2,\dots ,t_n\in N\left(R\right)$. Thus, by Proposition 1, R is an NRW ring. □

In the example below, we show that an NRW ring need not be nilpotent-semicommuting. Thus, the converse of the implication in Theorem 1 does not hold (see also Example 25).

Example 3.

Let us consider the ring $U_2\left({\mathbb{Z}}_8\right)$ of upper triangular $2\times 2$ matrices over the integers modulo 8. Since $N\left(U_2\left({\mathbb{Z}}_8\right)\right)={(}_{02{\mathbb{Z}}_8}^{2{\mathbb{Z}}_8{\mathbb{Z}}_8})$, it follows that $N{\left(U_2\left({\mathbb{Z}}_8\right)\right)}^4=0$, and thus Proposition 1 implies that $U_2\left({\mathbb{Z}}_8\right)$ is an NRW ring. However, as shown in Example 2(ii), this ring is not nilpotent-semicommuting.

We now present a variety of examples of well-established classes of rings that satisfy the nilpotent-semicommuting condition and thus, according to Theorem 1, consist entirely of NRW rings. These examples, drawn from the literature, illustrate that many important classes of rings fall within the class of NRW rings. For the reader’s convenience, brief descriptions of these classes are included, together with references to the relevant sources.

Example 4.

The following well-known classes of rings provide examples of nilpotent-semicommuting rings:

-

Reduced rings, i.e., rings R such that $N\left(R\right)=0$;

-

Reversible rings, i.e., rings R in which $ab=0$ implies $ba=0$ for all $a,b\in R$;

-

Symmetric rings, i.e., rings R in which $abc=0$ implies $acb=0$ for all $a,b,c\in R$.

It is well-known (and easy to see) that any ring that is reduced, reversible, or symmetric is semicommutative (see, e.g., [4]) and, hence, automatically nilpotent-semicommuting.

Example 5.

According to [5], right symmetric rings are rings R such that for all $a,b,c\in R$, the condition $abc=0$ implies $acb=0$. Similarly, left symmetric rings are rings R such that $abc=0$ implies $bac=0$. By [5] (Proposition 2.3(3)), every right (or left) symmetric ring is semicommutative and, hence, also nilpotent-semicommuting.

Example 6.

According to [6], right (resp. left) nil-symmetric rings are rings R such that, for all $a,b\in N\left(R\right)$ and $t\in R$, the condition $abt=0$ (resp. $tab=0$) implies $atb=0$. For such a ring R, if $a,b\in N\left(R\right)$ satisfy $ab=0$, then $abt=tab=0$ for any $t\in R$; so, the nil-symmetric condition yields $atb=0$. Thus, every right or left nil-symmetric ring is nilpotent-semicommuting.

Example 7.

In the literature (see, e.g., [4]), rings in which every right (resp. left) ideal is two-sided are referred to as right (resp. left) duo rings. If R is a right duo ring, and $a,b,t\in R$ satisfy $ab=0$, then $atb\in a\left(t\left(b{R}^1\right)\right)\subseteq a\left(b{R}^1\right)=ab{R}^1=0$; so, $atb=0$. Hence every right duo ring is semicommutative. A similar argument applies to left duo rings. Therefore, both right and left duo rings are nilpotent-semicommuting.

Example 8.

In [7], the notion of a nil-semicommutative ring is used for rings R in which, for all $a,b\in N\left(R\right)$, the condition $ab=0$ implies $aRb=0$. It is clear that every nil-semicommutative ring is nilpotent-semicommuting.

Example 9.

According to [8], a ring R has the Insertion-of-Nilpotent-Factors Property (INFP) if $ab=0$ implies $atb=0$ for all $a,b\in R$ and $t\in N\left(R\right)$. In [9], a unital ring satisfying INFP is referred to as nilpotent zero-insertive (NZI). It is clear that both INFP rings and NZI rings satisfy the nilpotent-semicommuting condition.

Example 10.

In [10], right (resp. left) nilpotent semicommutative rings are introduced as rings R such that for all $a\in R$ and $b\in N\left(R\right)$ (resp. $a\in N\left(R\right)$ and $b\in R$), the condition $ab=0$ implies $aRb=0$. It is clear that every right (resp. left) nilpotent semicommutative ring is nilpotent-semicommuting.

Example 11.

In [11], rings in which every nilpotent element is central are referred to as $CN$ rings. It is obvious that every CN ring is nilpotent-semicommuting. CN rings were considered, e.g., in [12].

Example 12.

According to [13], a ring R is nilpotent elements commutative (NEC) if, for all $a,b\in N\left(R\right)$, $ab=0$ implies $ba=0$. Clearly, every NEC ring is nilpotent-semicommuting. Rings with commuting nilpotent elements have been studied, for example, in [14,15].

Example 13.

Unital rings R whose group of units $U\left(R\right)$ lies in the center of R were studied in [16] under the name unit-central rings. It was shown there that every unit-central ring is a CN ring, and thus, as noted in Example 11, it is also nilpotent-semicommuting.

Example 14.

Rings whose lattice of right ideals is distributive are called right distributive rings (see, e.g., [17]). This distributivity condition means that for all right ideals $A,B,C$ of R, the equality $(A+B)\cap C=(A\cap C)+(B\cap C)$ holds. Right distributive unital rings are a natural generalization of Prüfer domains. Below we prove that any right distributive ring R belongs to the class of INFP rings recalled in Example 9 and, hence, is nilpotent-semicommuting (in the unital case, this follows from [18] (Theorem 3.2)). Before doing so, we first establish the following characterization of right distributive rings:

Lemma 2.

For any ring R the following conditions are equivalent:

(1) 

R is right distributive;

(2) 

For any $a,b\in R$, there exist $n\in \mathbb{Z}$ and $x\in R$ such that $an+ax\in b{R}^1$ and $b-bn-bx\in a{R}^1$.

Proof. 

Assume (1), and let $a,b\in R$. Set $c=a+b$; then $c\in (a{R}^1+b{R}^1)\cap c{R}^1=(a{R}^1\cap c{R}^1)+(b{R}^1\cap c{R}^1)$. Hence $c=(cm+cy)+(cn+cx)$ for some $m,n\in \mathbb{Z}$ and $x,y\in R$ such that $cm+cy\in a{R}^1$ and $cn+cx\in b{R}^1$. Thus, $an+ax=(c-b)n+(c-b)x=(cn+cx)-bn-bx\in b{R}^1$ and $b-bn-bx=(c-a)-(c-a)n-(c-a)x=(c-cn-cx)-(a+an+ax)=(cm+cy)-(a+an+ax)\in a{R}^1$, as desired.

Now assume (2) and let $A,B,C$ be any right ideals of R. Clearly, $(A+B)\cap C\supseteq (A\cap C)+(B\cap C)$. To prove the opposite inclusion, let $c=a+b\in C$ with $a\in A$ and $b\in B$. Let $n\in \mathbb{Z}$ and $x\in R$ be as described in (2). Then $c=(c-cn-cx)+(cn+cx)$ with $c-cn-cx=(a-an-ax)+(b-bn-bx)\in (a{R}^1+a{R}^1)\cap C\subseteq A\cap C$, and $cn+cx=(an+ax)+(bn+bx)\in (b{R}^1+b{R}^1)\cap C\subseteq B\cap C$. Hence $c\in (A\cap C)+(B\cap C)$, and the proof is complete. □

Continuing Example 14, we are now in a position to prove that any right distributive ring R is an INFP ring. Let $a,b,t\in R$ be such that $ab=0$ and $atb\ne 0$. By Lemma 2, there exist $n\in \mathbb{Z}$ and $x\in R$ such that $bn+bx\in tb{R}^1$ and $tb-tbn-tbx\in b{R}^1$. Since $ab=0$, it follows that $atb-atbn-atbx=0$, and thus $atb=atbn+atbx=at(bn+bx)\in a{t}^{2}b{R}^1$, which implies that $a{t}^{2}b\ne 0$. We have shown that $ab=0$ and $atb\ne 0$ implies $a{t}^{2}b\ne 0$, and thus, by mathematical induction, we obtain that $a{t}^{2^k}b\ne 0$ for any $k\in \mathbb{N}$. Therefore, if $ab=0$ and t is nilpotent, it must be that $atb=0$. Thus, R is an INFP ring, and consequently, it is also nilpotent-semicommuting.

A left distributive ring is a ring whose lattice of left ideals is distributive. By an argument analogous to the one above, every left distributive ring is nilpotent-semicommuting.

Example 15.

Rings whose right (resp. left) ideals are linearly ordered by inclusion are known in the literature under the name right (resp. left) chain rings; see, for example, [19]. In the unital case, such rings generalize commutative valuation domains. Clearly, every right (resp. left) chain ring is right (resp. left) distributive and, hence, by Example 14, nilpotent-semicommuting.

Example 16.

In [20], the name Armendariz ring was introduced for a ring R such that, whenever polynomials $f\left(x\right)=\sum a_i{x}^i$ and $g\left(x\right)=\sum b_j{x}^j$ in $R\left[x\right]$ satisfy $f\left(x\right)g\left(x\right)=0$, it follows that $a_i{b}_j=0$ for all $i,j$. A weaker version of this condition was considered in [21], where a ring R is said to be linearly Armendariz if the same implication holds for all linear polynomials $f\left(x\right)=a_0+a_{1}x$ and $g\left(x\right)=b_0+b_{1}x$ in $R\left[x\right]$. Clearly, every Armendariz ring is linearly Armendariz.

We show that all linearly Armendariz rings belong to the class of INFP rings (for the unital case, see [22] (Theorem 2.11)), and consequently, by Example 9, both linearly Armendariz rings and Armendariz rings are nilpotent-semicommuting. To prove this, let R be a linearly Armendariz ring, and let $a,b,t$ be elements of R with $ab=0$ and $t\in N\left(R\right)$. Since t is nilpotent, there exists $n\in \mathbb{N}$ such that $t^{2^n}=0$ and, hence, $a{t}^{2^n}b=0$. Now, consider the linear polynomials $f\left(x\right)=a-a{t}^{2^{n-1}}x$ and $g\left(x\right)=b+t^{2^{n-1}}bx$ in $R\left[x\right]$. Then $f\left(x\right)g\left(x\right)=0$. Since R is linearly Armendariz, the product of the constant term of $f\left(x\right)$ and the coefficient of x in $g\left(x\right)$ must be zero, which gives $a{t}^{2^{n-1}}b=0$. Thus, we have shown that if $ab=0$ and $a{t}^{2^n}b=0$, then also $a{t}^{2^{n-1}}b=0$; that is, the exponent of t in the product decreases from $2^n$ to $2^{n-1}$, while the product remains zero. By reverse induction on n, we eventually reach the case where the exponent is $2^0=1$, which gives $atb=0$, as required.

Example 17.

The notion of a lineal ring was introduced in [23] for rings R whose right annihilators are totally ordered by inclusion; that is, for any $a,b\in R$, either $r_R\left(a\right)\subseteq r_R\left(b\right)$ or $r_R\left(b\right)\subseteq r_R\left(a\right)$. The notion is symmetric in the sense that one may equally well consider left annihilators.

We show that all lineal rings are INFP (in the unital case, it was proved in [23] (Theorem 4.1(i))) and, thus, nilpotent-semicommuting as well. Let R be a lineal ring, let $a,b\in R$ be such that $ab=0$, and let $t\in N\left(R\right)$. Then $t^n=0$ for some $n\in \mathbb{N}$. To show that $atb=0$, we proceed by induction on n. The case $n=1$ is obvious; thus, we assume that $n\ge 2$. Then ${\left(t^2\right)}^{n-1}=0$, and by induction hypothesis we have $a{t}^{2}b=0$. If $r_R\left(a\right)\subseteq r_R\left(at\right)$, then since $b\in r_R\left(a\right)$, we obtain $atb=0$, as desired. Otherwise $r_R\left(at\right)\subseteq r_R\left(a\right)$, and since $tb\in r_R\left(at\right)$, it follows that $atb=0$, which shows that R is nilpotent-semicommuting.

In the context of Examples 14, 16, and 17, let us mention that for unital rings the implications $rightdistributive\Rightarrow Armendariz\Rightarrow INFP$ were proved in [17] (Corollary 4.3) and [24] (Proposition 4.5(i)), respectively, and the implication lineal⇒ linearly Armendariz follows from [23] (Theorem 5.3).

Example 18.

In [25], rings R, in which the right (resp. left) annihilator of any element of R is comparable with every right (resp. left) ideal of R, are referred to as right (resp. left) annelidan rings. Since all right (left) annelidan rings are lineal, it follows from Example 17 that they are nilpotent-semicommuting as well.

Example 19.

Let P be a completely semiprime right ideal of a ring R (that is, $a^2\in P$ implies $a\in P$ for any $a\in R$). According to [26] (Remark 2.5), the ring R is said to satisfy weak right P-comparability if

$$for any a,b\in R,we have either a{R}^1\subseteq bR,b{R}^1\subseteq aR,or aP=bP.$$

We show that if R satisfies weak right P-comparability, then R is nilpotent-semicommuting. For a contradiction, assume there exist $a,b,t\in N\left(R\right)$ such that $ab=0$ and $atb\ne 0$.

Suppose $b{R}^1\subseteq t^{k}R$ for some $k\in \mathbb{N}$. Then $b=t^{k}c$ for some $c\in R$. Since R satisfies weak right P-comparability, $tc{R}^1\subseteq cR$, $c{R}^1\subseteq tcR$, or $cP=tcP$. If $tc{R}^1\subseteq cR$, then $atb=at\left(t^{k}c\right)=a{t}^k\left(tc\right)\in a{t}^{k}cR=abR=0$, and thus, $atb=0$, a contradiction. If $c{R}^1\subseteq tcR$, then $c{R}^1\subseteq t^{n}cR$ for any $n\in \mathbb{N}$, and thus, $c=0$, giving $b=0$, a contradiction. Therefore, it must be $cP=tcP$, and $cP=0$ follows. On the other hand, since P is completely semiprime and b is nilpotent, we deduce that $b\in P$. If $tP=0$, then $0\ne atb\in atP=0$, a contradiction. Hence $tP\ne 0=cP$, and since R satisfies weak right P-comparability, $t{R}^1\subseteq cR$ or $c{R}^1\subseteq tR$. If $t{R}^1\subseteq cR$, then $tc{R}^1\subseteq cR$, which, as we already know, leads to a contradiction $atb=0$. Hence it must be $c{R}^1\subseteq tR$, and thus, $b{R}^1=t^{k}c{R}^1\subseteq t^{k}tR\subseteq t^{k+1}R$.

We have shown that if $k\in \mathbb{N}$ and $b{R}^1\subseteq t^{k}R,$ then $b{R}^1\subseteq t^{k+1}R.$ Since t is nilpotent and $b\ne 0,$ it follows that $b{R}^1⊈tR,$ and since furthermore R satisfies weak right P-comparability, $t{R}^1\subseteq bR$ or $bP=tP.$ In both cases we have $tP\subseteq bR,$ and $atb\in atP\subseteq abR=0$ follows. Hence $atb=0,$ and this contradiction shows that any ring with weak right P-comparability is nilpotent-semicommuting.

Example 20.

In [27], a unital ring R is called a right pseudo-chain ring if R is a local ring and for any $a,b\in R$ we have either $aR\subseteq bR$, $bR\subseteq aR$, or $aJ\left(R\right)=bJ\left(R\right)$, or, equivalently, $aR\subseteq bR$ or $bJ\left(R\right)\subseteq aJ\left(R\right)$, where $J\left(R\right)$ is the Jacobson radical of R. It follows immediately from Example 19 that any right pseudo-chain ring is nilpotent-semicommuting.

Example 21.

Let P be a completely prime right ideal of a unital ring R (that is, $ab\in P$ implies $a\in P$ or $b\in P$ for any $a,b\in R$). According to [26], the ring R is said to satisfy right comparability with respect to P if for any $a,b\in R$ one of the following conditions holds: $aR\subseteq bR$, $bR\subseteq aR$ or $\left(aR\right)S^{-1}=\left(bR\right)S^{-1}$, where for any $c\in R$, $\left(cR\right)S^{-1}=\{x\in R:xs\in cRforsomes\in R\setminus P\}$.

As noted in [26] (Remark 2.5), any ring R satisfying right comparability with respect to P satisfies weak P-comparability, and thus R is nilpotent-semicommuting by Example 19.

Example 22.

In the literature (see, e.g., [28]), a ring R is called right filial if, for any right ideal J of R, if I is a right ideal of J, then I is already a right ideal of R (roughly speaking, the relation of being a right ideal is transitive). If R is a right filial ring and $a\in N\left(R\right)$, then by [28] (Theorem 2.2), we have $aR\subseteq {\sum }_{i=1}^n{a}^i\mathbb{Z}$. Therefore, for any $b\in R$ with $ab=0$, it follows that $aRb=\left(aR\right)b\subseteq \left({\sum }_{i=1}^n{a}^i\mathbb{Z}\right)b=0$. Thus, $aRb=0$, which shows that R is left nilpotent semicommutative and, therefore, nilpotent-semicommuting by Example 10.

Below, we present some additional examples of NRW rings.

Example 23.

In [29], a unital ring R is said to be strongly 2-primal if $N\left(R\right)=W\left(R\right)$. Clearly, every strongly 2-primal ring is an NRW ring.

Example 24.

Semicommutative rings (i.e., rings R such that $ab=0\Rightarrow aRb=0$ for any $a,b\in R$) are also called IFP rings, where IFP stands for the insertion of factors property. In [30], a more general concept of a quasi-IFP ring was introduced. Namely, according to [30], a ring R with unity is said to be quasi-IFP if ${\sum }_{i=0}^{n}R{a}_{i}R$ is nilpotent whenever ${\sum }_{i=0}^n{a}_i{x}^i\in R\left[x\right]$ is nilpotent. It is shown in [30] (Lemma 1.3) that a ring R with unity is quasi-IFP if and only if $N\left(R\right)=W\left(R\right)$ (thus, quasi-IFP rings are exactly the $W_1$-reduced rings in the terminology of [31], and they coincide with the strongly 2-primal rings presented in Example 23). Therefore, all quasi-IFP rings are NRW rings.

We close this section with an example of an NRW ring that is not nilpotent-semicommuting. The ring in this example is taken from [29] and offers an alternative to the ring constructed via upper triangular matrices in Example 3.

Example 25.(An NRW ring need not be nilpotent-semicommuting) Let R be the ring from the second part of the proof of Proposition 2.4 in [29]; that is, $R=F\langle a,t\mid a^2=t^2=tat=0\rangle ,$ where F is a field. As shown in [29], the ring R is strongly 2-primal and hence an NRW ring. However, we have $aa=t^2=0$ but $ata\ne 0$, which shows that R is not nilpotent-semicommuting.

4. NRW Rings and Radicals

One of the most prominent open problems in the theory of noncommutative rings is the Köthe conjecture, which asserts that if a ring contains no nonzero nil (two-sided) ideals, then it also contains no nonzero nil one-sided ideals. A comprehensive overview of this conjecture and related issues can be found in [32].

Although the Köthe conjecture remains unsolved in general, it has been confirmed for several important classes of rings, including one-sided Noetherian rings, rings with Krull dimension, algebras over uncountable fields, and PI rings. In what follows, we show that NRW rings also satisfy the conjecture and may thus be added to this list of positive cases.

Let us now fix some notation. For a ring R, denote by $P\left(R\right)$ its prime radical, by $Nil\left(R\right)$ the nilradical of R (i.e., the largest nil ideal), and by $A\left(R\right)$ the sum of all nil right ideals of R. It is easy to see that this sum coincides with the sum of all nil left ideals; hence, $A\left(R\right)$ is an ideal.

The Köthe conjecture can thus be restated as the claim that $A\left(R\right)$ is always nil; that is, that for every ring R, we have $Nil\left(R\right)=A\left(R\right)$. It is well known that $Nil\left(R\right)=A\left(R\right)$ holds for any ring R whose nilpotent elements form a subring of R (see e.g., [3] (Theorem 2.1)). Thus, the Köthe conjecture is true in the class of NRW rings. For these rings, in the following proposition we obtain much more.

Proposition 2.

If R is an NRW ring, then $W\left(R\right)=P\left(R\right)=Nil\left(R\right)=A\left(R\right)$. In particular, the Köthe conjecture has a positive solution in the class of NRW rings.

Proof. 

Let R be an NRW ring. Clearly, $W\left(R\right)\subseteq P\left(R\right)\subseteq Nil\left(R\right)\subseteq A\left(R\right)$, and since $N\left(R\right)$ is a subring of R, furthermore $A\left(R\right)\subseteq N\left(R\right)$. Let $t\in A\left(R\right)$, and let I be the ideal of R generated by t. Then $I^3\subseteq N\left(R\right)tN\left(R\right)$. Since R is an NRW ring, Proposition 1 implies that the set $N\left(R\right)tN\left(R\right)$ is nilpotent. Hence I is nilpotent, and $t\in W\left(R\right)$ follows, as desired. □

Remark 2. 

(Answer to Šter’s question) Recall that a ring R is said to be an exchange ring if for every $a\in R$ there exists an idempotent $e=e^2\in R$ and elements $r,s\in R$ such that

$$e=ra=s+a-sa.$$

(8)

In [3] (Question 2), Janez Šter asked whether, for any exchange ring R, the condition that R is NR implies that R is NI. Below, we answer the question in the negative (independently, the question was also answered by Pace P. Nielsen and Steve Szabo in [29] (p. 30)).

Note that every unital right chain ring R is an exchange ring. Indeed, if $a\in J\left(R\right)$, then (8) holds with $e=r=0$ and $s=1-{(1-a)}^{-1}$. If $a\notin J\left(R\right)$, then (8) holds with $e=s=1$ and $r=a^{-1}$. Furthermore, as pointed out in Example 15, the ring R is nilpotent-semicommuting; hence, it is an NRW ring and, thus, automatically an NR ring. However, there exists a unital right chain ring R with an ideal P such that the factor ring $\overline{R}=R/P$ (which is still a unital right chain ring) is prime but not a domain (see [19,33]). By [2] (p. 105), $N\left(\overline{R}\right)\ne 0$, and thus Proposition 2 implies $0=P\left(\overline{R}\right)=Nil\left(\overline{R}\right)N\left(\overline{R}\right)$. Therefore, $\overline{R}$ is an exchange NR ring but not an NI ring.

Let us mention that our approach to this question was inspired by a strategy proposed by Edmund Puczyłowski (see [34]): when faced with an open problem in ring theory, let us first consider its restriction to the class of chain rings. This strategy is motivated by the fact that chain rings form a well-balanced class, on the one hand having a highly regular structure and on the other hand containing subtle and nontrivial examples.

The following lemma is a minor modification of Lemma 10 in [35], with nearly the same proof. In the proof, we will apply a standard procedure for extending a ring without unity to a ring with unity, known as the Dorroh extension ([36] (p. 10)). Recall that if R is a ring (without or with unity), then the Dorroh extension of R is the ring $R^1$ whose underlying additive group is $R\times \mathbb{Z}$ with multiplication given by $(r,k)(r^{\prime },k^{\prime })=(r{r}^{\prime }+k^{\prime }r+k{r}^{\prime },k{k}^{\prime })$. The ring $R^1$ is unital, and R is isomorphic to the ideal $R\times 0$ of $R^1$.

Lemma 3.

For a ring R, let $\lambda \left(R\right)$ (resp. $\rho \left(R\right)$) denote the set of all finite sums of elements of the form $t+jt$ (resp. $t+tj$), where $t\in N\left(R\right)$ and $j\in J\left(R\right)$. Then

(i) 

$\lambda \left(R\right)=\rho \left(R\right)$ is an additive subgroup of R containing $N\left(R\right)$.

(ii) 

$J\left(R\right)\lambda \left(R\right)\subseteq \lambda \left(R\right)$ and $\lambda \left(R\right)J\left(R\right)\subseteq \lambda \left(R\right)$.

(iii) 

$\lambda \left(R\right)\cap J\left(R\right)$ is an ideal of $J\left(R\right)$.

Proof. 

(i) Clearly, $\lambda \left(R\right)$ is an additive subgroup of R, and $N\left(R\right)\subseteq \lambda \left(R\right)$. Let $R^1$ be the Dorroh extension of R to the ring with a unity. If $t\in N\left(R\right)$ and $j\in J\left(R\right)$, then $u=1+j$ is invertible in $R^1$, and $t_1=ut{u}^{-1}\in N\left(R\right)$; thus, $t+jt=ut{u}^{-1}u=t_1(1+j)=t_1+t_{1}j\in \rho \left(R\right)$. Therefore, $\lambda \left(R\right)\subseteq \rho \left(R\right)$. Since $\rho \left(R\right)\subseteq \lambda \left(R\right)$ by a similar argument, the equality $\lambda \left(R\right)=\rho \left(R\right)$ holds.

(ii) Let $x,j\in J\left(R\right)$ and $t\in N\left(R\right)$. Then $x(t+jt)=(t+(x+j+xj\left)t\right)-(t+jt)\in \lambda \left(R\right)$, and thus $J\left(R\right)\lambda \left(R\right)\subseteq \lambda \left(R\right)$. Similarly one shows that $\rho \left(R\right)J\left(R\right)\subseteq \rho \left(R\right)$. Since $\lambda \left(R\right)=\rho \left(R\right)$ by (i), the proof of (ii) is complete.

(iii) follows immediately from (i) and (ii). □

A ring is called right (resp. left) strongly prime if every nonzero ideal of the ring contains a finite subset whose right (resp. left) annihilator is zero.

Theorem 2.

Let R be an NRW ring. Then the following conditions are equivalent:

(1) 

R is a right strongly prime ring and $N\left(R\right)\subseteq J\left(R\right)$.

(2) 

R is a left strongly prime ring and $N\left(R\right)\subseteq J\left(R\right)$.

(3) 

R is a domain.

Proof. 

(1) ⇒ (3): We adapt the arguments from the proof of Proposition 11 in [35]. Suppose, for contradiction, that R is a right strongly prime NRW ring but not a domain. Since R is prime, we have $N\left(R\right)\ne 0$. Let $\lambda \left(R\right)$ be the set defined in Lemma 3. By assumption, $N\left(R\right)\subseteq J\left(R\right)$, and thus $\lambda \left(R\right)\subseteq J\left(R\right)$. Hence, by Lemma 3(iii), $\lambda \left(R\right)$ is a nonzero ideal of $J\left(R\right)$. Let $\lambda {\left(R\right)}^{*}$ denote the ideal of R generated by $\lambda \left(R\right)$. Since R is prime, it follows from the Andrunakievich lemma (see, e.g., [36] (p. 11)) that $0\ne {\left(\lambda {\left(R\right)}^{*}\right)}^3\subseteq \lambda \left(R\right)$.

By assumption, the ring R is right strongly prime; so, there exists a finite subset $F=\{a_1,a_2,\dots ,a_n\}\subseteq {\left(\lambda {\left(R\right)}^{*}\right)}^3$ such that $r_R\left(F\right)=0$. Since ${\left(\lambda {\left(R\right)}^{*}\right)}^3\subseteq \lambda \left(R\right)$, each element $a_k\in F$ is of the form $a_k=\sum (t_{ki}+j_{ki}{t}_{ki})$, with $t_{ki}\in N\left(R\right)$ and $j_{ki}\in J\left(R\right)$.

As R is an NRW ring, $N\left(R\right)$ is a subring of R. Let S be the subring of $N\left(R\right)$ generated by all elements $t_{ki}$$(k=1,2,\dots ,n)$. It follows from the NRW property of R that the ring S is nilpotent. Thus, there exists $n\in \mathbb{N}$ such that $S^n\ne 0$ and $S^{n+1}=0$. However, $0\ne S^n\subseteq r_R\left(F\right)=0$, a contradiction. This proves that (1) implies (3).

(2) ⇒ (3) can be proved similarly, whereas (3) ⇒ (1) and (3) ⇒ (2) are obvious. □

Corollary 2.

Let R be an NRW ring. Then the following conditions are equivalent:

(1) 

$J\left(R\right)$ is a right strongly prime ring.

(2) 

$J\left(R\right)$ is a left strongly prime ring.

(3) 

$J\left(R\right)$ is a domain.

Proof. 

Since R is an NRW ring, Corollary 1(ii) implies that $J\left(R\right)$ is also an NRW ring. Furthermore, as $N\left(J\right(R\left)\right)\subseteq J\left(R\right)=J\left(J\right(R\left)\right)$, the result follows from Theorem 2. □

As we will show shortly, Proposition 2 is a special case of a more general result concerning all rings. Specifically, for any ring R, we define an ideal $\overline{\mathcal{W}}\left(R\right)$ of R such that $\overline{\mathcal{W}}\left(R\right)=R$ if R is an NRW ring, and $\overline{\mathcal{W}}\left(R\right)\cap A\left(R\right)=P\left(R\right)$. To define the ideal $\overline{\mathcal{W}}\left(R\right)$, we need the following notion.

Definition 3.

For a ring R, we define $\mathcal{W}\left(R\right)$ to be the sum of all right ideals I of R such that I is an NRW ring.

Proposition 3.

Let R be a ring. Then $\mathcal{W}\left(R\right)$ is an ideal of R containing the Wedderburn radical $W\left(R\right)$ of R.

Proof. 

By Lemma 1(i), $\mathcal{W}\left(R\right)$ is an ideal of R. The containment $W\left(R\right)\subseteq \mathcal{W}\left(R\right)$ follows immediately from the definitions of $W\left(R\right)$ and $\mathcal{W}\left(R\right)$. □

Obviously, if R is an NRW ring, then $\mathcal{W}\left(R\right)=R$. Below, we show that the equality $\mathcal{W}\left(R\right)=R$ is also possible for rings that are not NRW.

Example 26.

Let S be a non-zero reduced ring and let $R=M_2\left(S\right)$ be the ring of 2 by 2 matrices over S. Then, for any $s\in S\setminus 0$, we have $a={(}_{00}^{0s})\in N\left(R\right)$ and $b={(}_{s0}^{00})\in N\left(R\right)$, but $a+b\notin N\left(R\right)$. Thus, R is not an NRW ring (see Proposition 7 for a more general result). On the other hand, the right ideal $I={(}_{00}^{SS})$ of R is an NRW ring, because $N\left(I\right)={(}_{00}^{0S})$ and $N{\left(I\right)}^2=0$. By analogous argument, the right ideal ${(}_{SS}^{00})$ is an NRW ring. Thus, ${(}_{00}^{SS})+{(}_{SS}^{00})\subseteq \mathcal{W}\left(R\right)$, and $\mathcal{W}\left(R\right)=R$ follows.

In the next example, we show that there exist rings R with $\mathcal{W}(R/\mathcal{W}(R\left)\right)\ne 0$.

Example 27.

Let A be the algebra over a field K, generated by $y,x_1,x_2,x_3,\dots$ subject to the relations $x_i{x}_j=x_i{y}^k{x}_i=0$ for any $i,j,k\in \mathbb{N}$. Let R be the subring of A consisting of all the elements of A whose constant term is equal to 0.

Note that, for any $i\in \mathbb{N}$, we have $N\left(x_i{R}^1\right)=x_i\mathbb{Z}+{\sum }_{l=1}^{\infty }{x}_i{y}^l\mathbb{Z}+{\sum }_{j=1}^{\infty }{x}_{i}R{x}_j+{\sum }_{k=1}^{\infty }{x}_{i}R{x}_i{y}^k$. Thus, ${\left(N\left(x_i{R}^1\right)\right)}^2=0$, and Proposition 1 implies that $x_i{R}^1\subseteq \mathcal{W}\left(R\right)$. Hence, by Proposition 3, for the ideal $I={\sum }_{i=1}^{\infty }{R}^1{x}_i{R}^1$ we have $I\subseteq \mathcal{W}\left(R\right)$. We claim that $\mathcal{W}\left(R\right)=I$. If not, then there exists $a\in \mathcal{W}\left(R\right)\setminus I$. Since $a\in \mathcal{W}\left(R\right)$, there exist $w_1,w_2,\dots ,w_m\in R$ such that $a=w_1+w_2+\dots +w_m$ and $w_j{R}^1$ is an NRW ring for any $j\in \{1,2,\dots ,m\}$. If $w_j\in I$ for any j, then $a\in I$, a contradiction. Thus, there exists $j_0$ such that $w_{j_0}\notin I$. Let T be the subring of R generated by y. Since $w_{j_0}\notin I$, it follows that $w_{j_0}=t+p$ for some $t\in T\setminus \left\{0\right\}$ and $p\in I$. For any $i\in \mathbb{N}$, we have that $T{x}_{i}T{x}_i=0$, and thus $w_{j_0}{x}_i\in N\left(w_{j_0}{R}^1\right)$. Since $w_{j_0}{R}^1$ is an NRW ring, Proposition 1 implies that the set $w_{j_0}{x}_{1}N\left(w_{j_0}{R}^1\right)$ is nilpotent. However, for any $n\in \mathbb{N}$, the product

$$\left(w_{j_0}{x}_1\right)\left(w_{j_0}{x}_2\right)\left(w_{j_0}{x}_1\right)\left(w_{j_0}{x}_3\right)\left(w_{j_0}{x}_1\right)\left(w_{j_0}{x}_4\right)\dots \left(w_{j_0}{x}_1\right)\left(w_{j_0}{x}_n\right)$$

belongs to ${\left(w_{j_0}{x}_{1}N\left(w_{j_0}{R}^1\right)\right)}^n$ and is non-zero. This contradiction shows that $\mathcal{W}\left(R\right)=I$. Since the factor ring $R/\mathcal{W}\left(R\right)$ is generated by the coset determined by y, and for any $k\in \mathbb{N}$ we have that $y^k\notin I$, it follows that $R/\mathcal{W}\left(R\right)$ is a nonzero reduced ring. Hence $\mathcal{W}(R/\mathcal{W}(R\left)\right)=R/\mathcal{W}\left(R\right)\ne 0$.

Definition 4.

For any ring R, we inductively define a family of ideals ${\mathcal{W}}_{\alpha }\left(R\right)$ of R. Namely, we set ${\mathcal{W}}_0\left(R\right)=0$. For any non-limit ordinal $\alpha \ge 1$, the ideal ${\mathcal{W}}_{\alpha }\left(R\right)$ is defined by $\mathcal{W}(R/{\mathcal{W}}_{\alpha -1}\left(R\right))={\mathcal{W}}_{\alpha }\left(R\right)/{\mathcal{W}}_{\alpha -1}\left(R\right)$. For a limit ordinal α, we set ${\mathcal{W}}_{\alpha }\left(R\right)={\bigcup }_{\beta <\alpha }{\mathcal{W}}_{\beta }\left(R\right)$. Finally, we define $\overline{\mathcal{W}}\left(R\right)={\bigcup }_{\alpha }{\mathcal{W}}_{\alpha }\left(R\right)$, where α runs over all ordinals.

Note that by Lemma 1(ii), if in the definition of $\mathcal{W}\left(R\right)$ (i.e., in Definition 3), we replace right ideals with left ideals of R, then we obtain the same ideal $\mathcal{W}\left(R\right)$ and, thus, also the same ideal $\overline{\mathcal{W}}\left(R\right)$.

The construction of the ideal $\overline{\mathcal{W}}\left(R\right)$ in Definition 4, based on the ideal described in Definition 3, proceeds in the same manner as the construction of the prime radical $P\left(R\right)$ of a ring R using the Wedderburn radical (see [36] (p. 184)). Recall that $P\left(R\right)$ is constructed as follows. We set $W_0\left(R\right)=0$. For a non-limit ordinal $\alpha \ge 1$, the ideal $W_{\alpha }\left(R\right)$ of R is defined by $W(R/W_{\alpha -1}\left(R\right))=W_{\alpha }\left(R\right)/W_{\alpha -1}\left(R\right)$. For a limit ordinal $\alpha$, the ideal $W_{\alpha }\left(R\right)$ is defined as $W_{\alpha }\left(R\right)={\bigcup }_{\beta <\alpha }{W}_{\beta }\left(R\right)$. Then the prime radical of R is given by $P\left(R\right)={\bigcup }_{\alpha }{W}_{\alpha }\left(R\right)$, where the union is taken over all ordinals $\alpha$.

Theorem 3.

For any ring R, we have $P\left(R\right)=\overline{\mathcal{W}}\left(R\right)\cap A\left(R\right)$.

Proof. 

We first prove that $P\left(R\right)\subseteq \overline{\mathcal{W}}\left(R\right)\cap A\left(R\right)$. Since $P\left(R\right)\subseteq A\left(R\right)$, we only need to show that $P\left(R\right)\subseteq \overline{\mathcal{W}}\left(R\right)$. For this it suffices to show that

$$W_{\alpha }\left(R\right)\subseteq {\mathcal{W}}_{\alpha }\left(R\right)$$

(9)

for any ordinal $\alpha$. The case $\alpha =0$ is clear. Let $\alpha \ge 1$, and assume that (9) holds for all $\beta <\alpha$. If $\alpha$ is a limit ordinal, then (9) follows directly from the induction assumption. If $\alpha$ is a non-limit ordinal, then $\alpha =\beta +1$ for some $\beta \ge 0$. Since $W_{\beta }\left(R\right)\subseteq {\mathcal{W}}_{\beta }\left(R\right)$ by the induction assumption, the map $\phi :R/W_{\beta }\left(R\right)\to R/{\mathcal{W}}_{\beta }\left(R\right)$, given by $\phi (x+W_{\beta }\left(R\right))=x+{\mathcal{W}}_{\beta }\left(R\right)$ for any $x\in R$, is an epimorhism, and thus $\phi \left(W(R/W_{\beta }\left(R\right))\right)\subseteq W(R/{\mathcal{W}}_{\beta }\left(R\right))$. Hence, by applying also Proposition 3, we obtain that

$$(W_{\alpha }\left(R\right)+{\mathcal{W}}_{\beta }\left(R\right))/{\mathcal{W}}_{\beta }\left(R\right)=\phi \left(W(R/W_{\beta }\left(R\right))\right)\subseteq W(R/{\mathcal{W}}_{\beta }\left(R\right))\subseteq$$

$$\subseteq \mathcal{W}(R/{\mathcal{W}}_{\beta }\left(R\right))={\mathcal{W}}_{\alpha }\left(R\right)/{\mathcal{W}}_{\beta }\left(R\right),$$

and (9) follows.

Now, we prove that $\overline{\mathcal{W}}\left(R\right)\cap A\left(R\right)\subseteq P\left(R\right)$. For this, it suffices to show that

$${\mathcal{W}}_{\alpha }\left(R\right)\cap A\left(R\right)\subseteq P\left(R\right)$$

(10)

for any ordinal $\alpha \ge 1$. The case $\alpha =0$ is trivial. Let $\alpha \ge 1$, and assume that (10) holds for all $\beta <\alpha$. If $\alpha$ is a limit ordinal, then (10) follows directly from the induction assumption.

We are left with the case when $\alpha$ is a non-limit ordinal (this part of the proof adapts some ideas of the proof of Theorem 6 in [37]). Then $\alpha =\beta +1$ for some $\beta$. We will use bar to denote images under the canonical homomorphism $R\to R/{\mathcal{W}}_{\beta }\left(R\right)=\overline{R}$. Let J be a right nil ideal of R (note that $A\left(R\right)$ is the sum of all such J’s), and let I be a right ideal of R such that ${\mathcal{W}}_{\beta }\left(R\right)\subseteq J$ and $\overline{I}$ is an NRW ring (note that ${\mathcal{W}}_{\alpha }\left(R\right)$ is the sum of all such I’s). Let $a\in J$; then, by Lemma 1(i), $\overline{a}\overline{I}$ is an NRW ring. Since furthermore $\overline{a}\overline{I}$ is a right nil ideal of $\overline{R}$, from Proposition 2 it follows that $\overline{a}\overline{I}=W\left(\overline{a}\overline{I}\right)$. Hence, for every $x\in aI$, $\overline{x}$ generates a nilpotent ideal of $\overline{a}\overline{I}$, and thus for some $n\in \mathbb{N}$ we have ${\left(xaI\right)}^n\subseteq {\mathcal{W}}_{\beta }\left(R\right)$. Since $x\in J$, we also have ${\left(xaI\right)}^n\subseteq A\left(R\right)$, and thus, ${\left(xaI\right)}^n\subseteq {\mathcal{W}}_{\beta }\left(R\right)\cap A\left(R\right)$. Now, the induction assumption implies ${\left(xaI\right)}^n\subseteq P\left(R\right)$, and $xaI\subseteq P\left(R\right)$ follows. Since x is any element of $aI$, we obtain ${\left(aI\right)}^2\subseteq P\left(R\right)$, and thus $aI\subseteq P\left(R\right)$. Since a is an arbitrary element of J, we obtain $A\left(R\right)I\subseteq P\left(R\right)$. Hence ${\left(IA\left(R\right)\right)}^2\subseteq P\left(R\right)$, and thus, $IA\left(R\right)\subseteq P\left(R\right)$. Hence ${\mathcal{W}}_{\alpha }\left(R\right)A\left(R\right)\subseteq P\left(R\right)$; so ${({\mathcal{W}}_{\alpha }\left(R\right)\cap A\left(R\right))}^2\subseteq P\left(R\right)$, and thus ${\mathcal{W}}_{\alpha }\left(R\right)\cap A\left(R\right)\subseteq P\left(R\right)$. The proof is complete. □

Corollary 3.

The Köthe conjecture has a positive solution in the class consisting of all rings R such that $R=\overline{\mathcal{W}}\left(R\right)$.

Recall that a class $\mathcal{C}$ of rings is called a radical if the following conditions are satisfied (see e.g., [36] (Definition 2.1.1)):

(1)

The class $\mathcal{C}$ is closed under homomorphic images.

(2)

For every ring R, the sum $\mathcal{C}\left(R\right)=\sum \{I⊲R:I\in \mathcal{C}\}$ is in $\mathcal{C}$.

(3)

$\mathcal{C}(R/\mathcal{C}(R\left)\right)=0$.

Obviously, the class of rings R such that $\overline{\mathcal{W}}\left(R\right)=R$ satisfies condition (3) of the above definition. However, the class is not a radical as the following example shows.

Example 28.

Let $\mathcal{C}$ be the class of rings R such that $\overline{\mathcal{W}}\left(R\right)=R$, and let Q be any ring such that $P\left(Q\right)⊊Nil\left(Q\right)$ (see e.g., [36] (Theorem 4.2.1) for a concrete example of such a ring Q). It follows from Theorem 3 that $Q\notin \mathcal{C}$. On the other hand, Q is a homomorphic image of a free ring R. Since R, being a domain, is in $\mathcal{C}$, and Q is not in $\mathcal{C}$, the class $\mathcal{C}$ is not closed under homomorphic images. Thus, $\mathcal{C}$ is not a radical.

5. NRW Rings and Ring Constructions

In this section, we investigate under what conditions some classical ring constructions yield an NRW ring.

Proposition 4.

Let ${\left\{R_i\right\}}_{i\in I}$ be a family of rings. Then the direct sum ${⨁}_{i\in I}{R}_i$ is an NRW ring if and only if all the rings $R_i$ are NRW rings.

Proof. 

If ${⨁}_{i\in I}{R}_i$ is an NRW ring, then Corollary 1 implies that $R_i$ is an NRW ring for any $i\in I$. Now, we assume that all $R_i$ are NRW rings, and we prove that $R={⨁}_{i\in I}{R}_i$ is an NRW ring. For this, we apply Proposition 1. Let $a={\left(r_i\right)}_{i\in I}\in N\left(R\right)$. Then, for any $i\in I$, we have $r_i\in N\left(R_i\right)$, and thus by Proposition 1, there exists $m_i\in \mathbb{N}$ with ${\left(r_{i}N\left(R_i\right)\right)}^{m_i}=0$. Obviously, we can assume that $m_i=1$ if $r_{i}N\left(R_i\right)=0$. Thus, since only a finite number of the $r_i$’s is nonzero, there exists $m\in \mathbb{N}$ such that $m_i\le m$ for any $i\in I$, and ${\left(r_{i}N\left(R_i\right)\right)}^m=0$ follows. Hence

$${\left(aN\left(R\right)\right)}^m\subseteq \underset{i\in I}{⨁}{\left(r_{i}N\left(R_i\right)\right)}^m=0.$$

Therefore, R is an NRW ring. □

Given a family ${\left\{R_{\lambda }\right\}}_{\lambda \in \Lambda }$ of rings, recall that the direct product ${\prod }_{\lambda \in \Lambda }{R}_{\lambda }$ is the ring whose underlying set is the Cartesian product of the rings $R_{\lambda }$, with componentwise addition and multiplication. The following example shows that a direct product of NRW rings need not be an NRW ring.

Example 29.

Let K be a field. For each $n\in \mathbb{N}$, we set $R_n=K\langle x,y\rangle /I_n$, where $K\langle x,y\rangle$ is the polynomial algebra over K in non-commuting indeterminates x and y, and $I_n$ is the ideal of $K\langle x,y\rangle$ generated by $x^2,y^2,{\left(xy\right)}^n$ and ${\left(yx\right)}^n$. Observe that if $v\in K\langle x,y\rangle$ is a monomial of degree $\ge 2n$, then in $R_n$ we have $v+I_n=0$. Hence, if $f\in K\langle x,y\rangle$ is a polynomial with constant term equal to 0, then ${(f+I_n)}^{2n}=0$, which implies that $R_n$ is a local ring with $J{\left(R_n\right)}^{2n}=0$. Thus, $R_n$ is an NRW ring for any $n\in \mathbb{N}$. However, the direct product ${\prod }_{n\in \mathbb{N}}{R}_n$ is not an NRW ring. Indeed, for the elements $a=(x+I_1,x+I_2,x+I_3,\dots )$ and $b=(y+I_1,y+I_2,y+I_3,\dots )$ of ${\prod }_{n\in \mathbb{N}}{R}_n$, we have $a^2=b^2=0$, but $ab$ is not nilpotent (because ${(xy+I_{n+1})}^n\ne 0$ in $R_{n+1}$ for any $n\in \mathbb{N}$), which shows that nilpotent elements of ${\prod }_{n\in \mathbb{N}}{R}_n$ do not form a subring.

Our next result characterizes which Dorroh extensions are NRW rings. The definition of the Dorroh extension was recalled in the paragraph preceding Lemma 3.

Proposition 5.

For any ring R, the Dorroh extension $R^1$ of R is an NRW ring if and only if R is an NRW ring.

Proof. 

If $R^1$ is an NRW ring, then Corollary 1(ii) implies that so is R. To prove the opposite implication, we apply Proposition 1. Assume that R is an NRW ring, and let $a=(r,n)\in N\left(R^1\right)=N(R\times \mathbb{Z})$. Then $r\in N\left(R\right)$ and $n=0$, and thus in R we have ${\left(rN\left(R\right)\right)}^m=0$ for some $m\in \mathbb{N}$. Hence in $R^1$ we have ${\left(aN\left(R^1\right)\right)}^m\subseteq {\left(rN\left(R\right)\right)}^m\times 0=0\times 0$, which implies that $R^1$ is an NRW ring. □

Let R be a ring and M an R-bimodule. The trivial extension of R by M is the ring $R\propto M$ whose underlying additive group is $R\times M$ with multiplication given by $(r,m)(r^{\prime },m^{\prime })=(r{r}^{\prime },r{m}^{\prime }+r^{\prime }m)$.

Proposition 6.

For any ring R and R-bimodule M, the trivial extension $R\propto M$ is an NRW ring if and only if R is an NRW ring.

Proof. 

Since R is isomorphic to the subring $R\propto 0$ of $R\propto M$, if $R\propto M$ is an NRW ring, then so is R by Corollary 1(ii).

Conversely, assume that R is an NRW ring. To show that $A=R\propto M$ is an NRW ring, we apply Proposition 1. Let $a=(r,m)\in N\left(A\right)$, i.e., $a^n=0$ for some $n\in \mathbb{N}$. It is easy to see that

$$a^n={(r,m)}^n=(r^n,\sum _{i=1}^n{r}^{n-i}m{r}^{i-1}),$$

and thus, $r^n=0$. Hence $r\in N\left(R\right)$, and since R is an NRW ring, ${\left(rN\left(R\right)\right)}^k=0$ for some $k\in \mathbb{N}$. Thus,

$${\left(aN\left(A\right)\right)}^k={\left((r,m)N(R\propto M)\right)}^k\subseteq {\left((r,m)(N\left(R\right)\propto M)\right)}^k\subseteq {\left(rN\left(R\right)\right)}^k\propto M=0\propto M.$$

Hence ${\left(aN\left(A\right)\right)}^{2k}\subseteq {(0\propto M)}^2=0$, which shows that the set $aN\left(A\right)$ is nilpotent. Consequently, $A=R\propto M$ is an NRW ring. □

We close the paper with a characterization of NRW matrix rings. For a ring R and a number $n\in \mathbb{N}$, the ring of $n\times n$ matrices over R is denoted by $M_n\left(R\right)$.

Proposition 7.

Let R be a ring, and let $n\ge 2$ be an integer. Then the following conditions are equivalent:

(1) 

$M_n\left(R\right)$ is an NRW ring.

(2) 

$M_n\left(R\right)$ is a Wedderburn radical ring.

(3) 

R is a Wedderburn radical ring.

Proof. 

For any $x\in R$ and $i,j\in \{1,2,\dots ,n\}$, let $m_{ij}\left(x\right)$ denote the matrix with $(i,j)$-entry equal to x and all other entries equal to 0.

(1) ⇒ (2): For any $a\in R$, the matrices $m_{1n}\left(a\right)$ and $m_{n1}\left(a\right)$ are nilpotent, and since $M_n\left(R\right)$ is an NRW ring, $m_{1n}\left(a\right)+m_{n1}\left(a\right)$ is nilpotent too. Since ${\left(m_{1n}\left(a\right)+m_{n1}\left(a\right)\right)}^{2k}=m_{11}\left(a^{2k}\right)+m_{nn}\left(a^{2k}\right)$ for any $k\in \mathbb{N}$, it follows that a is nilpotent. Therefore, for any $a,i$ and j, the matrix is $m_{ij}\left(a\right)$ is nilpotent. Since nilpotent elements of $M_n\left(R\right)$ form a subring, it follows that $M_n\left(R\right)$ is a nil ring, and thus, being an NRW ring, it is Wedderburn radical by Proposition 2.

(2) ⇒ (3): Since the map $\phi :R\to M_n\left(R\right)$, given by $\phi \left(a\right)=m_{11}\left(a\right)$, is a ring monomorphism, we can regard R as a subring of $M_n\left(R\right)$. Now, the implication follows from the fact that a subring of a Wedderburn radical ring is Wedderburn radical.

(3) ⇒ (1): Let $m=\left(a_{ij}\right)\in M_n\left(R\right)$, let I be the ideal of $M_n\left(R\right)$ generated by m, and let J be the ideal of R generated by all the entries $a_{ij}$ of m. Since R is a Wedderburn radical ring, $J^n=0$ for some $n\in \mathbb{N}$. Since $I^k\subseteq M_n\left(J^k\right)$ for any $k\in \mathbb{N}$, it follows that I is nilpotent. Hence the ring $M_n\left(R\right)$ is Wedderburn radical, and thus, it is an NRW ring. □

6. Conclusions

In this paper, we introduced and studied a new class of rings, called NRW rings, characterized by the property that their nilpotent elements form a Wedderburn radical subring. We also introduced and investigated nilpotent-semicommuting rings, showing in particular that every such ring is an NRW ring.

The paper presents various structural and algebraic properties of NRW rings, including their behavior under standard ring-theoretic constructions such as factor rings, direct sums, direct products, Dorroh extensions, and matrix rings. A wide range of examples of NRW rings is provided, showing that many well-known classes of rings fall into this new class.

For any ring R, based on right ideals of R that are NRW rings, we constructed a radical-like ideal $\overline{\mathcal{W}}\left(R\right)$ and proved that the classical Köthe problem has a positive solution for rings satisfying $R=\overline{\mathcal{W}}\left(R\right)$. We also applied our results to answer a question posed by Šter.

The concept of NRW rings may serve as a framework for further developments in ring theory. Future work may explore connections with other known classes of rings, investigate the behavior of NRW rings under other standard ring constructions (e.g., polynomial rings), or seek potential applications in broader algebraic contexts.