# Torsion subgroups, solvability and the Engel condition in associative rings

Orest Artemovych, Victor Bovdi  Department of Applied Mathematics, Cracow
University of Technology, Cracow, Poland<EMAIL_ADDRESS>UAEU, United
Arab Emirates<EMAIL_ADDRESS>
###### Abstract.

The connections between the properties of associative rings that are Lie-
solvable (Engel, n-Engel, locally finite, respectively) and the properties of
their adjoin subgroups are investigated.

###### Key words and phrases:

Engel ring, Lie nilpotent ring, Lie solvable ring, torsion group, Engel group,
nilpotent group, locally finite ring, Artinian ring, Noetherian ring, Goldie
ring, local ring, derivation, solvable group, semiprime ring, $\pi$-regular
ring

###### 1991 Mathematics Subject Classification:

20C05, 16S34, 20F45, 20F19, 16W25

##  1\.  Introduction

Let $R=(R,+,\cdot)$ be an associative ring (not necessary with unity). The set
of all elements of $R$ forms a semigroup with respect to the circle operation
“$\circ$” defined by the rule $a\circ b=a+b+a\cdot b$ for each $a,b\in R$. The
set

$R^{\circ}=\\{a\in R\mid a\circ b=0=b\circ a\quad\text{for some}\quad b\in
R\\}$

is a group (so-called the adjoint group of $R$). If $R$ has unity and $U(R)$
is the unit group of $R$, then

$R^{\circ}\ni a\mapsto 1+a\in U(R)$

is a group isomorphism. If $R=R^{\circ}$, then $R$ is called radical.

We study properties of associative rings and their adjoint groups which are
connected with solvability, Engel conditions and periodicity.

We always assume that $p$ is a prime number, $\mathbb{N}$ is the set of
positive integers, $\mathbb{F}$ is a field, ${\mathbb{Z}}_{n}$ is the ring of
integers modulo $n$. Let $n,k\in\mathbb{N}$, $m\in\mathbb{Z}$ and let $x,g\in
R$. We introduce the following notation:

$\mu_{n}(x)=\sum\limits_{k=1}^{n}\binom{n}{k}x^{k}$,

$[x,{}_{1}g]=[x,g]=x\cdot g-g\cdot x,\qquad[x,{}_{n+1}g]=[[x,{}_{n}g],g]$,

$C_{R}(g)=\\{x\in R\mid xg=gx\\}$ is the centralizer of $g$ in $R$,

$J(R)$ is the Jacobson radical of $R$,

$N(R)$ is the set of all nilpotent elements of $R$,

$g^{(m)}$ is the $m$-th power of $g$ in $R^{\circ}$,

$F(R)=\\{x\in R\mid x\ \text{is of finite order in}\ R^{+}\\}$ is the torsion
part of the additive group $R^{+}$,

$\operatorname{char}R$ is the characteristic of $R$,

$Z(R)$ is the center of $R$,

${\mathbb{P}}(R)$ is the prime radical of $R$ (i.e., the intersection of all
prime ideals of $R$),

$N^{*}(R)$ is the nil radical of $R$ (i.e., the sum of all nil ideals),

$N_{r}(R)$ is the sum of all nil right ideals of $R$ (moreover, $N_{r}(R)$ is
the sum of all nil left ideals of $R$ and, therefore, $N_{r}(R)$ is a two-
sided ideal of $R$),

$[A,B]$ is the additive subgroup of $R^{+}$ generated by all $[a,b]$, where
$a\in A,b\in B$ and $A,B\subseteq R$,

$C(R)$ is the commutator ideal of $R$ (i.e., an ideal of $R$ generated by all
$[g,x]$),

$D=\operatorname{Der}R$ is the set of all derivations of $R$,

$\Delta(R)$ is the ideal of $R$ generated by all $\delta(R)$, where
$\varnothing\neq\Delta\subseteq\operatorname{Der}R$ and $\delta\in\Delta$,

$\gamma_{1}R=[R,R]$ and $\gamma_{n+1}R=[\gamma_{n}R,R]$,

$\delta_{0}R=R$ and $\delta_{n+1}R=[\delta_{n}R,\delta_{n}R]$,

${\langle X\rangle}_{\rm rg}$ is a subring of $R$ generated by $X\subseteq R$
(if $X=\varnothing$, then ${\langle X\rangle}_{\rm rg}=0$).

If $\Delta=(\Delta,+[-,-])$ is a Lie ring, then

$\gamma_{1}\Delta:=\Delta,\
\ldots,\gamma_{k+1}\Delta:=[\gamma_{k}\Delta,\Delta],\ldots\qquad(k\in\mathbb{N}).$

and
$\Delta^{(1)}:=\Delta,\ldots,\quad\Delta^{(n+1)}:=[\Delta^{(n)},\Delta^{(n)}]$.

Let $\tau(G)$ be the set of all torsion elements of a group $G$. Recall that a
ring $R$ is called:

  * $\bullet$

nil if each $x\in R$ is nilpotent, i.e., there exists $n=n(x)\in\mathbb{N}$
such that $x^{n}=0$; if there exists $n\in\mathbb{N}$ such that $x^{n}=0$ for
any $x\in N(R)$, then $R$ is of bounded index of nilpotency (of bounded index
for short),

  * $\bullet$

local if $R\ni 1$ and $R/J(R)$ is a simple ring,

  * $\bullet$

right Artinian in case for each ascending chain

$I_{1}\supseteq I_{2}\supseteq\cdots\supseteq I_{n}\supseteq\cdots$

of right ideals $I_{j}$ of $R$ $(j=1,2,\ldots)$, there exists $n\in\mathbb{N}$
such that $I_{n+1}=I_{n}$,

  * $\bullet$

right Noetherian in case for each descending chain

$I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{n}\subseteq\cdots$

of right ideals $I_{j}$ of $R$ $(j\in\mathbb{N})$, there exists
$n\in\mathbb{N}$ such that $I_{n+1}=I_{n}$,

  * $\bullet$

semilocal if $R/J(R)$ is a left Artinian ring,

  * $\bullet$

right Goldie  if it has no infinite direct sum of left ideals and has the
ascending chain condition on right annihilators,

  * $\bullet$

Lie nilpotent of class $n$ if $n$ is a minimal positive integer such that
$\gamma_{n+1}R=0$,

  * $\bullet$

locally nilpotent if each its finitely generated subring is nilpotent,

  * $\bullet$

locally Lie nilpotent if each finitely generated subring of $R$ is Lie
nilpotent,

  * $\bullet$

Lie soluble of length at most $n$ if $\delta_{n}R=0$,

  * $\bullet$

Lie metabelian if $\delta_{2}R=0$,

  * $\bullet$

Lie centrally metabelian if $\delta_{2}R\subseteq Z(R)$,

  * $\bullet$

Engel (or equivalently $R$ satisfies the Engel condition) if, for each $x,y\in
R$, there exists $n=n(x,y)\in\mathbb{N}$ such that $[x,{}_{n}y]=0$,

  * $\bullet$

$n$-Engel (or equivalently $R$ satisfies the $n$-Engel condition or $R$ is
bounded Engel) if $[x,{}_{n}y]=0$ for any $x,y\in R$,

  * $\bullet$

semiprime if it has no nonzero nilpotent ideals,

  * $\bullet$

prime if the product of each two nonzero ideals is nonzero,

  * $\bullet$

simple if $R^{2}\neq 0$ and $0$, $R$ are the only ideals of $R$,

  * $\bullet$

reduced if $N(R)=0$,

  * $\bullet$

$2$-primal if ${\mathbb{P}}(R)=N(R)$,

  * $\bullet$

right quasi-duo if each maximal right ideal is two-sided,

  * $\bullet$

of stable range $1$ if $R\ni 1$ and, for each $a,b,x,y\in R$ with $ax+by=1$,
there exists $h\in R$ such that $a+bh\in U(R)$,

  * $\bullet$

abelian if its all idempotents are central,

  * $\bullet$

$\pi$-regular if, for each $a\in R$, there exists $b\in R$ such that
$a^{n}=a^{n}ba^{n}$ for some $n\in\mathbb{N}$,

  * $\bullet$

right weakly $\pi$-regular if, for each $a\in R$, there exists
$n=n(a)\in{\mathbb{N}}$ such that $a^{n}R=(a^{n}R)^{2}$.

We shall use freely the following well known facts: Right Artinian rings are
$\pi$-regular and $\pi$-regular rings are weakly $\pi$-regular. An associative
ring $R$ is Lie nilpotent (Engel, respectively) if and only if $R^{L}$ is
nilpotent (Engel, respectively). Each (Lie or associative) locally nilpotent
ring is Engel. Each Lie nilpotent ring of nilpotency class $n$ is $n$-Engel. A
radical ring $R$ is Lie nilpotent if and only if its adjoint group $R^{\circ}$
is nilpotent [35].

In certain papers (see, for example, [5, 12, 55, 69] and others) by many
authors was investigated properties of Lie solvable rings and its relations
with groups. Each radical ring $R$ with solvable adjoint group $R^{\circ}$ is
Lie solvable [4, Theorem A]. As a consequence, if $R$ is a semilocal ring with
the solvable unit group $U(R)$, then $R$ is Lie solvable.

Each $2$-torsion-free Lie solvable ring $R$ has a nilpotent ideal $I$ such
that $R/I$ is Lie centre-by-metabelian (and so $R^{\circ}$ is solvable) (see
[69] and [55]). There exists a Lie center-metabelian (and so it is Lie
solvable) total $(2\times 2)$-matrices ring $M_{2}(R)$ over an infinite
commutative domain $R$ of characteristic $2$, but its adjoint group
$M_{2}(R)^{\circ}$ is non-solvable. The group of units $U(R)$ of a Lie
metabelian unitary ring $R$ is metabelian (see [56] and [40, Theorem 1]).

Our result is the following.

######  Theorem 1.

Let $R$ be a ring. The following statements hold:

  * $(i)$

if $R$ is Lie nilpotent, then the adjoint group $R^{\circ}$ is nilpotent-by-
abelian;

  * $(ii)$

if $R^{\circ}$ is solvable-by-finite, then $[J(R),R]\subseteq{\mathbb{P}}(R)$
and $J(R)C(R)\subseteq{\mathbb{P}}(R)$;

  * $(iii)$

if $R$ is $2$-torsion-free Lie solvable, then:

    * $(a)$

$C(R)\subseteq{\mathbb{P}}(R)=N(R)$ (i.e., $R$ is $2$-primal and right quasi-
duo), $N(R)$ is the locally nilpotent ideal of $R$ and $R^{\circ}$ is (locally
nilpotent)-by-abelian. Moreover, if $R$ has unity, then it is abelian;

    * $(b)$

if $R^{\circ}$ is torsion, then it is locally nilpotent;

    * $(c)$

if $N(R)^{+}$ is torsion-free divisible, then $R^{\circ}$ is nilpotent-by-
abelian.

In the local case we obtain the following.

######  Theorem 2.

Let $R$ be a local ring. The following statements hold:

  * $(i)$

if the unit group $U(R)$ is solvable, then:

    * $(a)$

$C(R)\subseteq L(R)=N(R)$ is a locally nilpotent ideal of $R$, where $L(R)$ is
the Levitzki radical of $R$;

    * $(b)$

$R$ is Lie solvable;

  * $(ii)$

if $R$ is a Lie solvable $\mathbb{Q}$-algebra, then $R^{\circ}$ is nilpotent-
by-abelian.

Recall that the Levitzki radical $L(R)$ of $R$ is its unique maximal locally
nilpotent ideal.

An additive map $\delta:R\to R$ is called a derivation of $R$ if
$\delta(ab)=\delta(a)b+a\delta(b)$ for all $a,b\in R$. The set $D$ of all
derivations of $R$ is a Lie ring. Properties of a ring $R$ which induced by
the Engel condition of the derivation ring $D$ gives the following.

######  Theorem 3.

Let $R$ be a ring. If $\varnothing\neq\Delta\subseteq D$, then

  * $(i)$

if $\Delta$ is a nilpotent Lie ring and $R\Delta\subseteq\Delta$ modulo
$C(R)$, then $\Delta(R)^{m}\subseteq C(R)$ (in particular, if $D$ is Lie
nilpotent, then $D(R)^{m}\subseteq C(R)$) for some $m\in\mathbb{N}$;

  * $(ii)$

if $\Delta$ is a nilpotent Lie ring of the nilpotent length $n$ and
$R\Delta\subseteq\Delta$ modulo $C(R)$, then $d^{n}(R)d(R)\subseteq C(R)$ for
some $d\in\Delta^{(n-1)}$;

  * $(iii)$

if $\Delta$ is an Engel Lie ring and $R\Delta\subseteq\Delta$ modulo $C(R)$,
then, for each $a\in R$ and $\delta\in\Delta$, there exists
$n=n(\delta,a)\in{\mathbb{N}}$ such that $\delta^{n}(a)\delta(R)\subseteq
C(R)$.

If the unit group $U(R)$ of a semilocal ring $R$ is $m$-Engel, then $U(R)$ is
locally nilpotent and, furthermore, $R$ is $n$-Engel provided that $R$ is
generated by $U(R)$ (bibliography in this way see in [6, 47]). Moreover, a
local ring $R$ is Lie nilpotent if and only if $U(R)$ is nilpotent [61] and
that is this case the classes of nilpotency of both structures coincide. We
prove the next.

######  Theorem 4.

Let $R$ be a $n$-Engel local ring. If $F(R)=0$, then $R$ is Lie nilpotent.

A ring $R$ is called locally finite if each finite subset of $R$ generates a
finite semigroup multiplicatively. The class of locally finite rings is closed
under formation of subrings, homomorphic images and direct sums (see [34,
Proposition 2.1]). A finite subset of a locally finite ring generates a finite
subring (not necessary with unity) [34, Theorem 2.2] and a locally finite ring
is strongly $\pi$-regular [34, Lemma 2.4(ii)]. Recall that a ring $R$ is
called strongly $\pi$-regular if, for each $a\in R$, there exist
$n=n(a)\in{\mathbb{N}}$ and $b\in R$ such that $a^{n}=a^{n+1}b$. A ring $R$ is
strongly $\pi$-regular [8] if and only if it satisfies the descending chain
condition on principal right ideals of the form

$aR\supseteq a^{2}R\supseteq\cdots\supseteq
a^{n}R\supseteq\cdots\qquad(\forall a\in R).$

Local rings with the nil Jacobson radical and semilocal rings with the nil
Jacobson radical of bounded index are strongly $\pi$-regular [34, Lemma 3.1
and Corollary 3.3]. The Jacobson radical $J(R)$ of a locally finite ring $R$
is a locally nilpotent ring (in view of [2, Corollaries 1 and 4] and Lemma
16$(ii)$). We precise [34, Propositions 2.5, 2.10 and 2.11] in the following

######  Proposition 5.

Let $R$ be a $2$-torsion-free locally finite ring with unity. The following
statements hold:

  * $(i)$

each prime ideal of $R$ is maximal as a right ideal (and so $R$ is
$\pi$-regular);

  * $(ii)$

$R$ is an abelian exchange ring of stable range $1$,
$C(R)\subseteq{\mathbb{P}}(R)=N(R)=J(R)$ and $R/J(R)$ is a subdirect product
of locally finite fields (so each element of $R$ is a sum of a unit and a
central element).

Each absolute field (i.e., a field in which each nonzero element is a root of
$1$) is a locally finite ring. In a locally finite field $\mathbb{F}$ every
finite subset $X\subseteq{\mathbb{F}}$ generates a finite subfield. Since the
unit group $U({\langle X\rangle}_{\rm rg})$ is cyclic, we deduce that a
locally finite field is absolute. A locally finite right Noetherian ring is
right Artinian (see Proposition 22).

Rings with torsion adjoint groups were intensively studied in [2, 27, 28, 29,
41, 42, 57, 62] and others. It is well known [28, Theorem 8] that a division
ring $D$ with the torsion multiplicative group $D^{*}$ is commutative.
Moreover, a torsion normal subgroup of the multiplicative group $U(D)$ of a
skew field $D$ is central [68, Lemma 10]. Each torsion subgroup of a linear
group over a field is locally finite by classical results of W. Burnside and
I. Schur. A torsion subgroup of the unit group $U(R)$ of a unitary $PI$-ring
is locally finite by results of C. Procesi and A.I. Shirshov. Each locally
finite subgroup of the adjoint group $R^{\circ}$ of a radical ring $R$ is
locally nilpotent [2, Corollary 1].

We have the following.

######  Proposition 6.

Let $R$ be a ring such that $R^{\circ}$ is torsion and $F(R)=0$. The following
statements hold:

  * $(i)$

$R$ is commutative or without zero-divisors,

  * $(ii)$

if $R$ is prime with unity, then $R$ is a domain such that $J(R)=0$ and the
unit group $U(R)$ is finite of one of the following types:

    * $(a)$

$U(R)$ is a cyclic group of order $2$ such that $\langle
U(R)\rangle_{rg}\cong{\mathbb{Z}}$;

    * $(b)$

$U(R)$ is a cyclic group of order $4$ such that $\langle
U(R)\rangle_{rg}\cong{\mathbb{Z}}[i]$ is the ring of Gaussian integers;

    * $(c)$

$U(R)$ is a cyclic group of order $6$ such that $\langle
U(R)\rangle_{rg}\cong{\mathbb{Z}}[\zeta_{3}]$ is the subring of integer
elements of the Eisenstein field ${\mathbb{Q}}[i\sqrt{3}]$;

    * $(d)$

$U(R)$ is the quaternion group of order $8$ such that $\langle
U(R)\rangle_{rg}\cong{\mathbb{Z}}[i,j]$ is the ring of quaternions with
integer coefficients;

    * $(e)$

$U(R)=\langle a,b\mid a^{3}=b^{2}=(ab)^{2}\rangle$ is the dicyclic group of
order $12$ such that $\langle U(R)\rangle_{rg}\cong{\mathbb{Z}}\cdot
1+{\mathbb{Z}}\cdot\alpha+{\mathbb{Z}}\cdot\beta+{\mathbb{Z}}\cdot\gamma$ is
the ring with the following Cayley table of multiplication:

$\begin{array}[]{||l||r|r|r||}\hline\cr{\cdot}&\alpha&\beta&\gamma\\\
\hline\cr\hline\cr\alpha&-1&\gamma&-\beta\\\
\hline\cr\beta&-\alpha-\gamma&-1-\beta&\alpha\\\
\hline\cr\gamma&1+\beta&-\alpha-\gamma&-1\\\ \hline\cr\end{array}$

    * $(f)$

$U(R)=\langle a,b\mid a^{3}=b^{3}=(ab)^{2}=1\rangle$ is the binary tetrahedral
group of order $24$ such that $\langle U(R)\rangle_{rg}$ is a subring of the
skew field ${\mathbb{Q}}(i,j)$ of quaternions which is generated by $i,j$ and
$\frac{1+i+j+k}{2}$.

Any unexplained terminology is standard as in [22, 52].

##  2\. Semiprime rings

The unit group $U(R)$ of a unitary $n$-Engel ring $R$ is $m$-Engel for some
$m=m(n)\in{\mathbb{N}}$ depending on $n$ (see [51, Corollary 1] and [54,
Corollary]). The adjoint group $R^{\circ}$ of a radical ring $R$ is $n$-Engel
if and only if $R$ is an $m$-Engel ring for some $m=m(n)\in{\mathbb{N}}$ [4,
Main Theorem]. Each $n$-Engel Lie algebra is locally nilpotent and each
$n$-Engel Lie algebra over a field of characteristic zero is nilpotent [70,
71, 72].

If a group $G$ contains a non-trivial $p$-element and the unit group
$U({\mathbb{F}}[G])$ of the group algebra ${\mathbb{F}}[G]$ is bounded Engel,
then ${\mathbb{F}}[G]$ is bounded Engel [13] (see also [14, 15]). An unitary
associative bounded Engel algebra $A$ over a field of prime characteristic has
the bounded Engel group $U(A)$ [54], which is locally nilpotent [54, Remark].
In the case of zero characteristic, $U(A)$ is nilpotent and $A$ is Lie
nilpotent (see [37] and [51]). Each bounded Engel subgroup of the adjoint
group $R^{\circ}$ of a radical ring $R$ is locally nilpotent [2, Corollary 1].

It is known that $N_{r}(R)\subseteq R^{\circ}$, $N^{*}(R)\subseteq
J(R)\subseteq R^{\circ}$,

${\mathbb{P}}(R)\subseteq L(R)\subseteq N^{*}(R)\subseteq N_{r}(R)\subseteq
N(R).$

If $R$ is an $n$-Engel ring, then its commutator ideal $C(R)$ is nil (for
example, see [25, Application 2]) and, additionally, $C(R)\subseteq L(R)$ [4,
Lemma 3.1].

We use the following.

######  Lemma 7.

[44, Theorem 1] Let $L\not=0$ be a left ideal of a prime ring $R$. Let $D$ be
the Lie ring of all derivations of $R$. Let $k,n\in\mathbb{N}$ and let
$0\neq\delta\in D$. If

(2.1) $[\delta(x^{k}),{}_{n}x^{k}]=0,\qquad(\forall x\in L)$

then $R$ is commutative.

As a consequence, we have the next.

######  Corollary 8.

Let $R$ be an $n$-Engel ring. The following statements hold:

  * $(i)$

$C(R)\subseteq{\mathbb{P}}(R)=N(R)\subseteq J(R)$ (i.e., $R$ is $2$-primal and
right quasi-duo);

  * $(ii)$

if $R\ni 1$, then $R$ is abelian;

  * $(iii)$

the adjoint group $R^{\circ}$ is (locally nilpotent)-by-abelian and $N(R)$ is
a locally nilpotent ideal of $R$;

  * $(iv)$

if $F(N(R))=0$, then $R^{\circ}$ is nilpotent-by-abelian.

###### Proof.

$(i)$ The quotient ring $R/P$ is $n$-Engel for each prime ideal $P$ of $R$ and
each inner derivation $\delta$ of $R/P$ satisfies $(\ref{PIP})$, so we
conclude that $R/P$ is commutative by Lemma 7. That yields
$C(R)\subseteq{\mathbb{P}}(R)=N(R)$ and $N(R)$ is locally nilpotent as a
$PI$-ring. Thus, each maximal right ideal of $R$ is two-sided.

$(ii)$ If $R$ has unity, then it is abelian by [66, 3.20].

$(iii)$ The set $N(R)$ is an ideal in view of the part $(i)$, $N(R)^{\circ}$
is $m$-Engel for some $m\in\mathbb{N}$ [4, Main Theorem] and $N(R)^{\circ}$ is
locally nilpotent [4, Lemma 2.2]. Then $N(R)$ is locally nilpotent by [2,
Lemma 3].

$(iv)$ If $F(N(R))=0$, then $N(R)^{\circ}$ is torsion-free. Since
$N(R)^{\circ}$ is $m$-Engel for some $m\in\mathbb{N}$ [4, Main Theorem], it is
nilpotent [71]. ∎

######  Lemma 9.

(see [19, Theorem 3]) Let $R$ be a ring with $N_{r}(R)=0$. If, for given
$x,y\in R$ there exist positive integers $m=m(x,y)$, $n=n(x,y)$ and $k=k(x,y)$
such that

(2.2) $[x^{m},{}_{k}y^{n}]=0,$

then $R$ is commutative.

An unitary associative $PI$-algebra $R$ with the Engel condition over a field
of any characteristic has the nil commutator ideal by [50, Proposition 2.3].
Furthermore, each nil ring is Engel by [63, Proposition 4.2], but there exist
nil rings $R$ (which also are algebras over arbitrary fields) such that their
adjoint groups $R^{\circ}$ are not Engel [58, Theorem 1.5]. Inasmuch as an
Engel ring $R$ satisfies (2.2), we obtain the following.

######  Corollary 10.

Let $R$ be an Engel ring. The following statements hold:

  * $(i)$

$C(R)\subseteq N_{r}(R)=N(R)\subseteq J(R)$, $N(R)$ is an ideal of $R$ (i.e.,
$R$ is right (left) quasi-duo) and $R^{\circ}$ is Engel-by-abelian;

  * $(ii)$

if $R$ is locally Lie nilpotent, then the adjoint group $R^{\circ}$ is
(locally nilpotent)-by-abelian;

  * $(iii)$

if $R$ has unity, then it is abelian.

###### Proof.

The part $(i)$ follows from Lemma 9. Assume that $R$ is locally Lie nilpotent.
If $x,y\in R$, then the subring $\langle x,y\rangle_{\rm rg}$ is Lie nilpotent
and so there exist $n=n(x,y)\in{\mathbb{N}}$ such that $[x,{}_{n}y]=0$. This
means that $R$ is Engel, the adjoint group $N(R)^{\circ}$ is locally nilpotent
in view of [3, Main Theorem] and so $R^{\circ}$ is (locally nilpotent)-by-
abelian. All maximal right ideals in unitary Engel ring $R$ are two-sided so
$R$ is abelian by [66, 3.20].∎

Every domain of characteristic $0$ that is Engel (as a Lie ring) is
commutative [10, Theorem 4]. If the unit group $U(D)$ is $m$-Engel, then a
division ring $D$ is commutative by [6, Lemma 4.1]. We obtain an affirmative
answer on [21, Question1.2].

######  Proposition 11.

Each Engel division ring is commutative.

###### Proof.

The assertion holds from Corollary 10. ∎

Our next result confirm a conjecture of [21, Hypothesis 1.1].

######  Proposition 12.

An Engel adjoint group $R^{\circ}$ of a right Artinian ring $R$ is nilpotent.

###### Proof.

The Jacobson radical $J(R)$ is a nilpotent ideal of $R$, so $J(R)^{\circ}$ is
a nilpotent group. We can assume that $J(R)^{2}=0$. Since $R$ is a right
Noetherian [22, Theorem 18.3], $J(R)$ is a finite direct sum of minimal ideals
of $R$ and $J(R)\subseteq Z(R)$ by [21, Lemma 2.1]. Thus $R^{\circ}$ is a
nilpotent group by a well-known Ph. Hall’s Theorem. ∎

##  3\. Torsion subgroups

It is well known the following.

######  Lemma 13.

Let $R$ be a ring with $1$. The following statements hold:

  * $(i)$

$U(R)$ is a torsion group if and only if the Jacobson radical $J(R)$ is nil
with the torsion additive group $J(R)^{+}$ and $U(R/J(R))$ is torsion [41,
Lemma 1.1];

  * $(ii)$

if $F(R)=0$ and $U(R)$ is torsion, then $J(R)=0$ and $R$ is reduced [41,
Corollary 1.2];

  * $(iii)$

if $F(R)=0$, then $U(R)$ is torsion if and only if $U(R)$ is locally finite
([41, Theorem 3.3] and [57, Proposition 2]).

We precise [67, Corollary 2.10] as the following.

######  Proposition 14.

Let $R$ be a ring with the additive $p$-group $R^{+}$. The set $N(R)$ is a
subring of $R$ if and only if $N(R)^{\circ}$ is a normal Sylow $p$-subgroup of
$R^{\circ}$.

###### Proof.

Clearly, $pR$ is an ideal of $R$ such that $pR\subseteq N(R)$. The groups
$(R/pR)^{\circ}$ and $R^{\circ}/(pR)^{\circ}$ are isomorphic, so we can assume
that $pR=0$.

$(\Rightarrow)$ Suppose that $N(R)$ is a subring. Then $N(R)^{\circ}$ is a
$p$-subgroup of $R^{\circ}$ in view of [1, Lemma 2.4] and $N(R)^{\circ}$ is
contained in some maximal (Sylow) $p$-subgroup $S$ of $R^{\circ}$. If $g\in
S\setminus N(R)^{\circ}$, then

$0=g^{(p^{n})}=\mu_{p^{n}}(g)=g^{p^{n}},\qquad\qquad(\text{for some}\quad
n\in\mathbb{N})$

so $g\in N(R)$, a contradiction. Hence $N(R)^{\circ}$ is a Sylow $p$-subgroup
of $R^{\circ}$. If $S_{1}$ is a maximal (Sylow) $p$-subgroup of $R^{\circ}$
and $h\in S_{1}$ has order $p^{m}$, then $0=h^{(p^{m})}=h^{p^{m}}$, so $h\in
N(R)$ and $S_{1}=S$. Consequently, $S$ is normal in $R^{\circ}$.

$(\Leftarrow)$ Since $N(R)$ is closed under the circle operation ”$\circ$”,
$N(R)$ is a subring of $R$ by [67, Theorem 2.1]. ∎

Let $Dz(R)$ be the set of all left and right zero divisors and $0\in Dz(R)$.

###### Proof of Proposition 6.

$(i)$ Let $R$ be not commutative and let $g\in N(R)$. If $g^{2}=0$, then

$0=g^{(n)}=\mu_{n}(g)=ng$

for some $n\in\mathbb{N}$ and so $g=0$. This means that $N(R)=0$. It follows
that if $ab=0$ for some nonzero $a,b\in R$, then $(ba)^{2}=0$ what implies
that $ba=0$. Therefore $Dz(R)$ is commutative and, consequently, $Dz(R)$ is an
ideal of $R$ by [39, Theorem 5.5]. Moreover, $Dz(R)^{2}C(R)=0$ [39, Lemma
5.4], so $C(R)^{3}=0$. Thus $C(R)=0$, which is a contradiction.

$(ii)$ Indeed, $J(R)=N(R)=0$ by Lemma 13. If $ac=0$ for some $a,c\in R$, then
$(ca)^{2}=c(ac)a=0$ implies that $ca=0$. As a consequence, $0=a(cR)=cRa$ and
$c=0$ or $a=0$ by the primeness of $R$. Hence $R$ is a domain. The rest
follows from [57, Proposition 4]. ∎

######  Proposition 15.

Let $R$ be an unitary domain of characteristic $p>0$. If $U(R)$ is torsion,
then it is a $p^{\prime}$-group and the following statements hold:

  * $(i)$

$R^{\circ}\setminus\\{0\\}\subseteq U(R)$ and $I\cap R^{\circ}=0$ for any
proper right (left) ideal $I$ of $R$; in particular, $J(R)=0$;

  * $(ii)$

if $S$ is a subring of $R$ and $S\cap R^{\circ}\neq 0$, then $1\in S$.

###### Proof.

Obviously, $N(R)=0$. Since $g^{(p)}=\mu_{p}(g)=g^{p}$ for $g\in R^{\circ}$, we
deduce that $g^{(p)}\neq 0$ and $R^{\circ}$ is a $p^{\prime}$-group.

$(i)$ Let $I$ be a proper right (left) ideal of $R$. If $0\neq b\in I\cap
R^{\circ}$, then

$\textstyle
0=b^{(n)}=\mu_{n}(b)=b\big{(}n+\sum\limits_{k=2}^{n}\binom{n}{k}b^{k-1}\big{)}$

for some $n\in\mathbb{N}$. If the great common divisor $\text{GCD}(n,p)=1$,
then $n\in I$ and there exist $u,v\in\mathbb{Z}$ such that $r=(nr)u+(pr)v\in
I$ for any $r\in R$. Consequently, $I=R$, a contradiction. This implies that
$I\cap R^{\circ}=0$.

Inasmuch as $x\in(Rx)^{\circ}\cap R^{\circ}$ ($x\in(xR)^{\circ}\cap
R^{\circ}$, respectively) for each $x\in R^{\circ}$, we conclude that
$xR=R=Rx$, so each nonzero quasi-invertible element is invertible in $R$.

$(ii)$ If $S$ is a nonzero subring of $R$ and $0\neq b\in S\cap R^{\circ}$,
then, as above, $n\in S$ and consequently $1\in S$.∎

According to Proposition 15, we can ask the following questions:

  * Q1.

Does there exist a unitary infinite non-commutative simple ring $R$ of
characteristic $p>0$ with the torsion unit group $U(R)$?

  * Q2.

Does there exist a unitary (infinite) non-commutative ring $R$ which is not a
skew field, such that $R^{\circ}\setminus\\{0\\}\subseteq U(R)$?

##  4\. Locally finite rings

We start with some properties of locally finite rings.

######  Lemma 16.

If $R$ is a locally finite ring with unity, then the following statements
hold:

  * $(i)$

$R^{+}$ is a torsion $\pi$-group for some set $\pi$ of primes;

  * $(ii)$

$U(R)$ is locally finite;

  * $(iii)$

$1+J(R)$ is a locally nilpotent $\pi$-group.

###### Proof.

$(i)$ Obviously.

$(ii)$ If $X$ is a finite subset of $U(R)$, then $\langle
X\rangle\subseteq{\langle X\rangle}_{\rm rg}$ and so the subgroup $\langle
X\rangle$ is finite.

$(iii)$ The unipotent subgroup $1+J(R)$ of $U(R)$ is locally nilpotent [2,
Corollary 1]. Since $J(R)$ is nil, $1+J(R)$ is a $\pi$-group in view of [1,
Lemma 2.4]. ∎

######  Lemma 17.

If $P$ is a minimal prime ideal of a $2$-torsion-free ring $R$, then
$\operatorname{char}R/P\neq 2$.

###### Proof.

Let $X=\\{2^{n}a\mid a\in R\setminus P\text{ and
}n\in\mathbb{N}\cup\\{0\\}\\}$. Clearly, $X$ is non-empty, $0\notin X$ and $X$
is an $m$-system (in the sense of [46]). Therefore, there exists a two-sided
ideal $M$ of $R$ which is maximal to being disjoint from $X$ (then $M$ is
prime by [46, Lemma 4] and $M\subseteq R\setminus X$). Since $R\setminus
P\subseteq X$, we conclude that $M\subseteq P$ and consequently $M=X$. Hence
$\operatorname{char}R/P\neq 2$. ∎

If for each $x\in R$ there exists $n\in\mathbb{N}$ with $x^{n}=x$, then $R$ is
commutative by a well-known theorem of N. Jacobson. A ring $R$ is called
periodic if, for each $x\in R$, there exist different positive integers $m$
and $n$, such that $x^{m}=x^{n}$.

######  Lemma 18.

Let $R$ be a locally finite ring with unity. The following statements hold:

  * $(i)$

$R$ is periodic;

  * $(ii)$

if $R$ is $2$-torsion-free semiprime, then it is commutative;

  * $(iii)$

if $R$ is prime of $\operatorname{char}R\not=2$, then it is a field;

  * $(iv)$

if $R$ is $2$-torsion-free, then $C(R)\subseteq{\mathbb{P}}(R)=N(R)=J(R)$
(i.e., $R$ is $2$-primal and right quasi-duo).

###### Proof.

$(i)$ For the proof, see [33, Corollary 2].

Let $R$ be a $2$-torsion-free ring.

$(ii)$ It holds in view [11, Thereom 4.5] and the part $(i)$.

$(iii)$ It follows from the part $(i)$ and the fact that any periodic domain
is a field.

$(iv)$ It is a consequence of parts $(ii)-(iii)$ and Lemma 17. ∎

###### Proof of Proposition 5..

$(i)$ The quotient ring $R/P$ is a field for each prime ideal $P$ of $R$ by
[34, Corollary 2.6] and Lemma 18$(ii)$. Thus the part $(i)$ holds.

$(ii)$ Since $R$ is strongly $\pi$-regular and $J(R)\subseteq N(R)$ by [32,
Theorem 1], we conclude that $J(R)=N(R)$ in view of Lemma 18$(iii)$. Hence $R$
is exchange of stable range $1$ by [66, Theorem 5.23 and Proposition 5.6].
Moreover, $R$ is abelian by [66, 3.20(3)], $R/J(R)$ is a subdirect product of
fields and so each element of $R$ is a sum of an invertible and a central
elements by [66, Thereom 6.29]. ∎

######  Corollary 19.

A locally finite $2$-torsion-free ring $R$ is right (left) Ore, i.e., there
exists the classical right (left) quotient ring $Q(R)$.

###### Proof.

The assertion holds in view of Lemma 18$(iii)$, Proposition 5$(ii)$ and [38,
Theorem 2.1 and Proposition 1.9(5)]. ∎

Since each $2$-torsion-free locally finite ring is abelian $\pi$-regular, we
provide the following.

######  Corollary 20.

An abelian $\pi$-regular ring $R$ satisfies the Köthe’s conjecture, i.e., the
sum of two nil left ideals is always nil.

###### Proof.

The set $N(R)$ of nilpotent elements is an ideal of $R$ by [9, Theorem 2]. The
rest follows from [38, Lemma 1.4(2) and Theorem 2.1(2)]. ∎

######  Lemma 21.

[22, Lemma 18.34B] Let $R$ be a right Noetherian ring. If $R/P$ is an Artinian
ring for each prime ideal $P$ of $R$, then $R$ is a prime ring or $R$ is a
right Artinian ring.

######  Proposition 22.

A ring $R$ is locally finite right Noetherian if and only if it is a locally
finite right Artinian.

###### Proof.

$(\Leftarrow)$ Each right Artinian ring is right Noetherian by [22, Theorem
18.13].

$(\Rightarrow)$ Since $R/P$ is a field for each prime ideal $P$ of $R$ (see
Lemma 18$(iii)$), we deduce that $R$ is right Artinian in view of Lemmas 21
and 18$(iii)$. ∎

######  Proposition 23.

Let $R$ be a semilocal ring. The following conditions are equivalent:

  * $(i)$

$R$ is a locally finite ring;

  * $(ii)$

the unit group $U(R)$ is locally finite;

  * $(iii)$

$R^{+}$ is a torsion group, $J(R)$ is a locally nilpotent ideal and
$R/J(R)={\sum\limits_{i=1}^{n}}^{\oplus}M_{m_{i}}(D_{i})$ is a finite direct
sum of rings of $m_{i}\times m_{i}$ matrices over locally finite fields
$D_{i}$ with $i=1,\ldots,n$.

###### Proof.

$(i)\Rightarrow(ii)$ It follows from Lemma 16$(ii)$.

$(ii)\Rightarrow(i)$ It is clear that $R/J(R)$ is a finite direct ring sum and
each direct summand is a locally finite field or a finite total matrix ring.
The unipotent group $1+J(R)$ is locally nilpotent group. That yields the
subring ${\langle J(R)^{\circ}\rangle}_{\rm rg}=J(R)$ is locally nilpotent by
[2, Lemma 3].

Let $X$ be a finite subset of $R$. There exists an additive group isomorphism

$({\langle X\rangle}_{\rm rg}+J(R))/J(R)\cong{\langle X\rangle}_{\rm
rg}/({\langle X\rangle}_{\rm rg}\bigcap J(R))$

and

$({\langle X\rangle}_{\rm rg}+J(R))/J(R)={\langle{X}+J(R)\rangle}_{\rm rg}$

is a finite subring of $R/J(R)$. The subring $B:={\langle X\rangle}_{\rm
rg}\cap J(R)$ is finitely generated by [45, Theorem 2]. Clearly, it is
nilpotent and so $B/B^{2}$ is a finitely generated ${\mathbb{Z}}_{n}$-module
for some $n\in\mathbb{N}$. It implies that the subring $B^{2}$ is finitely
generated by [45, Theorem 2]. Using induction on the nilpotency index of $B$,
we obtain that $B$ (and consequently ${\langle X\rangle}_{\rm rg}$) is finite.

$(ii)\Rightarrow(iii)$ For each $D_{i}$ $(i=1,\ldots,n)$ there exists a chain
$F_{1}\subseteq F_{2}\subseteq\cdots$ such that $D_{i}=\cup_{j}F_{j}$ and each
$F_{j}$ is a finite subfield of the field $F_{j+1}$. Thus $(R/J(R))^{\circ}$
is locally finite. Moreover, the adjoint group $J(R)^{\circ}$ is locally
finite, $(R/J(R))^{\circ}\cong R^{\circ}/J(R)^{\circ}$ and so $R^{\circ}$ is
locally finite.

$(iii)\Rightarrow(ii)$ It is obvious. ∎

######  Corollary 24.

A locally finite semilocal ring is semiperfect.

##  5\. Properties induced by derivations

######  Proposition 25.

Let $R$ be a commutative ring with unity. If $R$ has a derivation $\delta$
with the finite kernel $\operatorname{Ker}\delta$, then $R$ is a locally
finite ring. The prime radical ${\mathbb{P}}(R)$ has finite index in $R$ and
$\delta(R)\subseteq{\mathbb{P}}(R)$.

###### Proof.

Assume that $R$ is infinite. Obviously, $\delta(1)=0$ and
$\operatorname{Ker}\delta\not=0$. This implies that $nR=0$ for some
$n\in\mathbb{N}$ and $R$ is a finite direct ring sum of $p$-components
$F_{p}$, where the prime $p$ divides $n$ and

$F_{p}=\\{r\in R\mid p^{k}r=0\quad\text{for some}\quad
k=k(r)\in\mathbb{N}\bigcup\\{0\\}\\}.$

Consequently, without loss of generality, we can assume that $n=p^{s}$ for
some $s\in{\mathbb{N}}$. Since

$\varphi:R/pR\ni a+pR\mapsto p^{k}a+p^{k+1}R\in p^{k}R/p^{k+1}R$

is an additive group isomorphism, $p^{k}R/p^{k+1}R$ is infinite for any
$k=0,\ldots,s-1$. If $\delta(R)\subseteq pR$, then $\delta(p^{s-1}R)=0$ and so
$p^{s-1}R\subseteq\operatorname{Ker}\delta$, a contradiction. Hence
$\delta(R)\nsubseteq pR$. Inasmuch as $\delta(pR)\subseteq pR$, the rule

$\Delta:R/pR\ni a+pR\mapsto\delta(a)+pR\in R/pR$

determines a nonzero derivation $\Delta$ of $R/pR$ and
$\operatorname{Ker}\Delta$ is finite. Then $\Delta(\beta^{p})=0$ for any
$\beta\in R/pR$ and so the set $\\{\alpha^{p}\mid\alpha\in R/pR\\}$ is finite.
If $\alpha,\beta\in R/pR$ are distinct elements and $\alpha^{p}-\beta^{p}=0$,
then $\alpha-\beta\in{\mathbb{P}}(R/pR)$. This implies that the index
$|R/pR:{\mathbb{P}}(R/pR)|<\infty$. However $pR\subseteq{\mathbb{P}}(R)$ and
so $|R:{\mathbb{P}}(R)|<\infty$.

Since ${\mathbb{P}}(R)$ is nil with the torsion additive group
${\mathbb{P}}(R)^{+}$, we conclude that the adjoint group
${\mathbb{P}}(R)^{\circ}$ is locally finite. Thus $R$ is a semiperfect ring
with the torsion unit group $U(R)$, so $R$ is locally finite by Proposition
23. ∎

###### Proof of Theorem 3..

$(i)$ Assume that $R$ is commutative, $d\in\Delta$, $\delta\in Z(\Delta)$ and
$a\in R$. Then

(5.1) $0=[\delta,ad]=\delta(a)d+a[\delta,d]=\delta(a)d.$

The ideal of $R$ generated by the set $\\{\mu(R)\mid\mu\in Z(\Delta)\\}$ we
denote by $\Delta_{1}(R)$. Then $\Delta_{1}(R)^{2}=0$ in view of $(\ref{A*})$.
Since $d(\delta(R))=\delta(d(R))\subseteq\delta(R)$, we conclude that
$\Delta_{1}(R)$ is a $\Delta$-ideal of $R$ and so

(5.2) $\overline{d}:R/\Delta_{1}(R)\ni a+\Delta_{1}(R)\mapsto
d(a)+\Delta_{1}(R)\in R/\Delta_{1}(R)$

is a derivation of $B:=R/\Delta_{1}(R)$. Then
$\overline{\Delta}=\\{\overline{d}\mid d\in\Delta\\}$ is a subring of the Lie
ring $\operatorname{Der}B$ and a left $B$-module. As before,
$\overline{\Delta}_{1}(B)$ (and its inverse image in $R$) is nilpotent, where
$\overline{\Delta}_{1}:=\\{\overline{d}\mid d\in\Delta_{2}\\}$ and
$\Delta_{2}$ is an inverse image of $Z(\Delta/Z(\Delta))$ in $\Delta$. Thus
$\Delta(R)$ is nilpotent according to the induction on nilpotent length of
$\Delta$.

Now, assume that $R$ is not necessary commutative. If $\delta\in\Delta$, then
$\delta(C(R))\subseteq C(R)$ and the rule

$\overline{\delta}:R/C(R)\ni x+C(R)\mapsto\delta(x)+C(R)\in R/C(R)$

determines a derivation of $R/C(R)$. Since
$\overline{\Delta}=\\{\overline{\delta}\mid\delta\in\Delta\\}$ is a left
$(R/C(R))$-module, $\overline{\delta}(R/C(R))^{m}=\overline{0}$ for some
$m\in\mathbb{N}$ and consequently $\Delta(R)^{m}\subseteq C(R)$.

$(ii)$ Suppose that $\gamma_{n+1}(\Delta)=0$ and $R$ is commutative. If
$d\in\Delta^{(n-1)}$, $a\in R$, then

$\begin{array}[]{rccl}d(a)d&=&[d,ad]&\in\gamma_{2}(\Delta),\\\
d^{2}(a)d&=&[d,d(a)d]&\in\gamma_{3}(\Delta),\\\ {}\hfil&\vdots&{}\hfil&{}\\\
d^{n-1}(a)d&=&[d,d^{n-2}(a)d]&\in\gamma_{n}(\Delta),\\\
d^{n}(a)d&=&[d,d^{n-1}(a)d]&\in\gamma_{n+1}(\Delta)=0.\end{array}$

This implies that $d^{n}(R)d=0$. Since $d(C(R))\subseteq C(R)$, the result can
be obtain similar to that of the part $(i)$.

$(iii)$ Let $\delta\in\Delta$. If $R$ is commutative, then $a\delta\in\Delta$
for any $a\in R$ and so

$(-1)^{n}\delta^{n}(a)\delta=(-1)^{n}[a\delta,\underbrace{\delta,\ldots,\delta}_{n\quad\text{\tiny
times}}]=0$

for some $n\in\mathbb{N}$. Hence $\delta^{n}(a)\delta=0$.

Now, assume that $R$ is not necessary commutative. Then $\delta(C(R))\subseteq
C(R)$ and, by the same argument as in the part $(i)$, there exist
$n=n(\delta,a)\in{\mathbb{N}}$ such that $\delta^{n}(a)\delta(R)\subseteq
C(R)$ and the assertion holds. ∎

If $x\in R$, then the rule $\partial_{x}:R\ni a\mapsto(ax-xa)\in R$ determines
a derivations $\partial_{x}$ of $R$; this derivation is called an inner
derivation of $R$ (induced by $x$). The set $\operatorname{IDer}R$ of all
inner derivations of $R$ is an ideal of the Lie ring $D$.

######  Proposition 26.

Let $R$ be a ring. The following statements hold:

  * $(i)$

$R$ is Lie solvable (Lie nilpotent, $n$-Engel, Engel, locally Lie nilpotent,
locally Lie solvable, respectively) if and only if the Lie ring
$\operatorname{IDer}R$ is solvable (nilpotent, $n$-Engel, Engel, locally
nilpotent, locally solvable, respectively);

  * $(ii)$

if $D$ is solvable (nilpotent, $n$-Engel, Engel, locally nilpotent, locally
solvable, respectively), then $R$ is Lie solvable (Lie nilpotent, $n$-Engel,
Engel, locally Lie nilpotent, locally Lie solvable, respectively);

  * $(iii)$

if $R$ is $2$-torsion-free semiprime and $D$ ($\operatorname{IDer}R$,
respectively) is solvable, then $D=0$ ($R$ is commutative, respectively);

  * $(iv)$

if $R$ is with unity of characteristic $0$
($\operatorname{char}(R/{\mathbb{P}}(R))=0$, respectively) and $D$ is
solvable, then $C(R)\subseteq D(R)\subseteq{\mathbb{P}}(R)$;

  * $(v)$

if $R$ consists from countable many elements and $D$ ($\operatorname{IDer}R$,
respectively) is Lie solvable, then $D=0$ ($R$ is commutative, respectively);

  * $(vi)$

if $R$ is commutative and $D$ is locally nilpotent, then, for each $a\in R$
and $\delta\in D$ there exists $n=n(a,\delta)\in{\mathbb{N}}$ such that
$\delta^{n}(a)\delta=0$. Moreover, if $R$ is reduced, then $\delta^{n}(a)=0$;

  * $(vii)$

if $R$ is a semiprime ring with the $n$-Engel derivation ring $D$
($\operatorname{IDer}R$, respectively), then $R$ is commutative and, for each
$\delta\in D$ ($\delta\in\operatorname{IDer}R$, respectively), there exists
$n=n(\delta)\in\mathbb{N}$ such that $\delta^{n}(R)=0$. Moreover, if $F(R)=0$,
then $D=0$.

###### Proof.

$(i)$ Since
$[\partial_{x_{1}},\partial_{x_{2}},\cdots,\partial_{x_{n}}]=\partial_{[x_{1},x_{2},\cdots,x_{n}]}$
for any $x_{1},x_{2},\ldots,x_{n}\in R$ and

$\operatorname{IDer}\ni\partial_{x}\mapsto x+Z(R)\in R^{L}/Z(R)$

is a Lie ring isomorphism, the result is obvious.

$(ii)$ It is immediately.

$(iii)$ If $A\subseteq R$, then by $\Delta_{A}$ we denote the set
$\\{\partial_{a}\mid a\in A\\}$. If $A$ is a Lie ideal of $R$, then
$\Delta_{A}$ is an ideal of $\operatorname{IDer}R$. Assume that $D$ is
solvable of length $n>1$. If $D^{(n-1)}\cap\operatorname{IDer}R\neq 0$, then

$I_{D^{(n-1)}}:=\\{a\in R\mid\partial_{a}\in
D^{(n-1)}\bigcap\operatorname{IDer}R\\}$

is a Lie ideal of $R$ and

$0=[D^{(n-1)},D^{(n-1)}]\supseteq[\Delta_{I_{D^{(n-1)}}},\Delta_{I_{D^{(n-1)}}}]=\Delta_{[I_{D^{(n-1)}},I_{D^{(n-1)}}]}$

what gives that $[I_{D^{(n-1)}},I_{D^{(n-1)}}]\subseteq Z(R)$. Thus
$I_{D^{(n-1)}}\subseteq Z(R)$ by [27, Lemma 1] and consequently
$\Delta_{I_{D^{(n-1)}}}=0$, a contradiction. Hence
$D^{(n-1)}\cap\operatorname{IDer}R=0$. This implies that $R$ is commutative,
$D(R)^{m}=0$ for some $m\in\mathbb{N}$ by Theorem 3$(i)$ and $D(R)=0$ by the
semiprimeness of $R$. We conclude the result.

$(iv)$ Each minimal prime ideal of $R$ is closed with respect to each
$\delta\in D$ by [23, Proposition 1.3]. The map

(5.3) $\overline{\delta}:R/P\ni a+P\mapsto\delta(a)+P\in R/P$

is a derivation of a prime ring $R/P$ of characteristic ${}\neq 2$ by Lemma 17
and $\operatorname{Der}(R/P)=0$ by the part $(iii)$. Thus $C(R)\subseteq
D(R)\subseteq{\mathbb{P}}(R)$ ($C(R)\subseteq{\mathbb{P}}(R)$, respectively).

$(v)$ There exists a collection of prime ideals $P_{\beta}$ $(\beta\in\Gamma)$
by [18, Theorem 2.1] such that

$\bigcap_{\beta\in\Gamma}P_{\beta}=0\quad\text{and}\quad\delta(P_{\beta})\subseteq
P_{\beta}\qquad(\forall\delta\in D\text{ and/or
}\forall\delta\in\operatorname{IDer}R,\text{ respectively}).$

Consequently, $\overline{\delta}$ defined by the rule $(\ref{PPI1})$ is a
derivation of $R/P$ for any $P=P_{\beta}$ and $\operatorname{Der}(R/P)=0$ by
the part $(iii)$. Hence

$C(R)\subseteq D(R)\subseteq\bigcap_{\beta\in\Gamma}P_{\beta}=0$

and the assertion holds.

$(vi)$ The subring of $D$ generated by derivations $d$ and $ad$ $(a\in R\
\text{and}\ d\in D)$ is nilpotent and the result holds by the same argument as
in the proof of Theorem $\ref{GGX}(ii)$.

$(vii)$ Since $R$ is $n$-Engel it is commutative by Corollary 8. If $\delta\in
D$ ($\delta\in\operatorname{IDer}R$, respectively), then by the same argument,
as in the proof of Theorem 3$(ii)$, we obtain that $\delta^{n}(R)=0$ for some
$n\in\mathbb{N}$.

Let $F(R)=0$. We prove that $\delta=0$ using induction by $n$. If
$\delta^{2}(R)=0$, then $\delta=0$ by [16, Corollary 1]. Let $n>2$ and suppose
that $\delta^{n-1}(R)=0$ implies that $\delta=0$. Assuming $\delta^{n}(R)=0$
we see that

$0=\delta^{n}(a\delta^{n-2}(b))=(n-1)\delta^{n-1}(a)\delta^{n-1}(b)\qquad(\forall
a,b\in R)$

what implies that $(\delta^{n-1}(R))^{2}=0$ and hence $\delta^{n-1}(R)=0$ by
the semiprimeness of $R$. Thus $\delta=0$. ∎

Recall that the commutator ideal $C(R)$ of a $2$-torsion-free Lie solvable
ring $R$ is nil (see [55, Theorem 2.1] and [69, Theorem]). Proposition 26$(v)$
precise this result in the countable case.

######  Corollary 27.

Let $R$ be an algebra over a field ${\mathbb{F}}$ of characteristic $0$. If
the Lie ${\mathbb{F}}$-algebra $D$ ($\operatorname{IDer}R$, respectively) is
Engel, then $R$ is Lie nilpotent.

###### Proof.

Since $\operatorname{IDer}R$ is an Engel Lie algebra over $Z(R)$ and
$\operatorname{IDer}R$ and $R/Z(R)$ are isomorphic as Lie algebras, $R$ is Lie
Engel and so it is nilpotent by [49, Theorem B]. ∎

##  6\. Solvability

######  Lemma 28.

Each Lie solvable nil ring $R$ is locally nilpotent. Moreover, if $R$ is a
${\mathbb{Q}}$-algebra, then it is Lie nilpotent.

###### Proof.

The ring $R$ contains a nilpotent ideal $I$ such that $R/I$ satisfies the
identity

$[x_{1},[x_{2},x_{3}],[x_{4},x_{5}]]=0$

(see [55, Theorem 2.1] and [69, Theorem]). This implies that every finitely
generated subring of $R/I$ is nilpotent (see e.g. [53, Theorems 6.3.3 and
6.3.39]), so $R$ is a locally nilpotent ring. If $R$ is a
${\mathbb{Q}}$-algebra, then $R$ is locally nilpotent as an algebra and, by
[49, Theorem B], it is Lie nilpotent. ∎

###### Proof of Theorem 1..

$(i)$ The quotient ring $R/{\mathbb{P}}(R)$ is semiprime and so its adjoint
group $(R/{\mathbb{P}}(R))^{\circ}$ is abelian by Corollary 8. Moreover,
$(R/{\mathbb{P}}(R))^{\circ}\cong R^{\circ}/{\mathbb{P}}(R)^{\circ}$ and
${\mathbb{P}}(R)^{\circ}$ is nilpotent by [35].

$(ii)$ Let $P$ be a prime ideal of $R$. If $J(R/P)$ is nonzero, then
$[J(R),R]\subseteq P$ in view of [17, Theorem A]. This implies that
$[J(R),R]\subseteq{\mathbb{P}}(R)$.

Since $R/{\mathbb{P}}(R)$ is semiprime, then $J(R)\subseteq{\mathbb{P}}(R)$ or
$J(R/{\mathbb{P}}(R))$ is commutative in view of [17, Theorem B]. From this it
follows that $J(R/{\mathbb{P}}(R))^{2}\cdot C(R/{\mathbb{P}}(R))=0$ and
consequently $J(R/{\mathbb{P}}(R))\cdot C(R/{\mathbb{P}}(R))=0$ what gives
that

$J(R)\cdot C(R)\subseteq{\mathbb{P}}(R).$

$(iii)$ Let $R$ be a $2$-torsion-free Lie solvable ring.

$(a)$ First, assume that $R$ is prime of solvable length $n>1$. Since
$[R^{(n-1)},R^{(n-1)}]=0$, we conclude that $R^{(n-1)}\subseteq Z(R)$ by [27,
Lemma 1]. But then $[R^{(n-2)},R^{(n-2)}]\subseteq R^{(n-1)}$ and we obtain a
contradiction in view of [27, Lemma 1]. Hence $R$ is commutative. This implies
that $C(R)\subseteq{\mathbb{P}}(R)=N(R)$ in general case. If $R$ has unity,
then $R$ is abelian in view of [66, 3.20] and $N(R)$ is a locally nilpotent
ring by Lemma 28.

$(b)$ If $R^{\circ}$ is torsion, then it is locally finite (and so it locally
nilpotent by [2, Corollary 2]).

$(c)$ If $N(R)^{+}$ is torsion-free divisible, then $N(R)$ is a locally
nilpotent $\mathbb{Q}$-algebra and so it satisfies the Engel condition. The
algebra $N(R)$ is Lie nilpotent by [49, Theorem B] and $N(R)^{\circ}$ is
nilpotent by [35], as required. ∎

######  Corollary 29.

Let $R$ be a right Goldie ring (or $R$ satisfies the ascending chain condition
on both left and right annihilators) with unity. If $R$ satisfies one of the
conditions:

  * $(i)$

$R$ is Engel as a Lie ring;

  * $(ii)$

$R$ is $2$-torsion-free locally finite,

then $R$ is Lie solvable and the unit group $U(R)$ is nilpotent-by-abelian.

###### Proof.

We have that $C(R)\subseteq N(R)$ by Corollary 10 (by Lemma 18, respectively),
and so $N(R)$ is an ideal of $R$. Since $N(R)$ is nilpotent by [43, Theorem 1]
(by [30, Theorem 1], respectively) and $R/N(R)$ is commutative, $R$ is Lie
solvable and the unit group $U(R)$ is nilpotent-by-abelian. ∎

######  Remark 30.

If $R$ is a right Goldie $n$-Engel ring of prime characteristic $p>0$ and
$n<p$, then it is Lie nilpotent in view of [31].

##  7\. Local rings

###### Proof of Theorem 2..

$(i)$ The group unit $U(R/J(R))$ is solvable what gives that $[R,R]\subseteq
J(R)$ by [68, Theorem 2]. Since $J(R)^{\circ}$ is solvable, we deduce that
$J(R)$ (and consequently $R$) is Lie solvable by [5, Theorem A]. Moreover, the
Levitzki radical $L(R)$ of $R$ is a $PI$-ring by [5, Theorem B(2)] and so it
is locally nilpotent. If $B=R/L(R)$, then $J(B)$ is commutative in view of [5,
Theorem B] what implies that $C(B)^{3}=\overline{0}$ and consequently $B$ is
commutative. Hence $C(R)\subseteq L=N(R)$.

$(ii)$ Since $N(R)$ is a locally nilpotent ideal of $R$, the
$\mathbb{Q}$-algebra $N(R)$ is Lie nilpotent by [49, Theorem B] and so the
adjoint group $N(R)^{\circ}$ is nilpotent by [35].∎

######  Proposition 31.

Let $R$ be a local ring. The following statements are equivalent:

  * $(i)$

the unit group $U(R)$ is torsion;

  * $(ii)$

$J(R)$ is nil and $R/J(R)$ is an absolute field of characteristic $p>0$;

  * $(iii)$

$U(R)\cong(1+J(R))\rtimes U(R/J(R))$, where $1+J(R)$ is a $p$-group and
$U(R/J(R))$ is a $p^{\prime}$-group.

Therefore, a local ring $R$ with the torsion unit group $U(R)$ is a locally
finite ring.

###### Proof.

$(i)\Rightarrow(ii)$ Since $R/J(R)$ is a skew field and the unit group
$U(R/J(R))$ is torsion, we deduce that $R/J(R)$ is commutative and
$p(R/J(R))=0$ for some prime $p$. Hence $pR\subseteq J(R)$ and $p^{k}R=0$ for
some $k\in{\mathbb{N}}$ in view of Lemma $\ref{krempa}(i)$. Since $U(R/J(R))$
is torsion, $R/J(R)$ is an absolute field.

$(ii)\Rightarrow(iii)$ and $(iii)\Rightarrow(i)$ are obviously. ∎

######  Lemma 32.

Let $R$ be a local ring which is Engel as a Lie ring. If $F(R)=0$, then either
$R$ is a Lie nilpotent $\mathbb{Q}$-algebra or $pR\subseteq J(R)$ for some
prime $p$, $C(R)\subseteq N_{r}(R)=N(R)$ and $N(R)^{\circ}$ is a torsion-free
group. Moreover, if in the last case, $R$ is $n$-Engel, then it is Lie
solvable and $N(R)^{\circ}$ is nilpotent (the $R^{\circ}$ is nilpotent-by-
abelian and $R$ is Lie solvable).

###### Proof.

If $R^{+}$ is a divisible group, then $R$ is a $\mathbb{Q}$-algebra and so it
is Lie nilpotent by [49, Theorem B]. Therefore, we assume in the next that
$pR$ is proper in $R$ for some prime $p$. Then $pR\subseteq J(R)$ and
$C(R)\subseteq N_{r}(R)=N(R)$ by Corollary 10. Consequently, $N(R)$ is an
ideal of $R$ and the adjoint group $N(R)^{\circ}$ is torsion-free by [1, Lemma
2.4].

Assume that $R$ is $n$-Engel. Since $N(R)^{\circ}$ is $m$-Engel for some
$m\in\mathbb{N}$ by [4, Main Theorem], it is nilpotent by the theorem of
Zelmanov [71]. Then $N(R)$ is Lie nilpotent by [35] and consequently $R$ is
Lie solvable. ∎

###### Proof of Theorem 4..

In view of Lemma 32, we assume that $pR\subseteq J(R)$ for some prime $p$.
Since $C(R)\subseteq N(R)\subseteq J(R)$ and $(J(R)/C(R))^{\circ}\cong
J(R)^{\circ}/C(R)^{\circ}$ is abelian, we deduce that $N(R)$ is an ideal of
$R$ and $J(R)^{\circ}$ is a solvable group. Moreover, $J(R)^{\circ}$ is
$m$-Engel group for some $m\in\mathbb{N}$ depending on $n$ by [4, Main
Theorem]. Then the adjoint group $J(R)^{\circ}$ is locally nilpotent by [24,
Theorem 1].

If $0\neq a\in\tau(J(R)^{\circ})$, then
$0=a^{(n)}=a(n+\sum\limits_{k=2}^{n}\binom{n}{k}a^{k-1})$ for some
$n\in\mathbb{N}$. Hence $n\in J(R)$. Obviously, the order of each element of
$U(R/J(R))$ is relatively prime with $p$ and, thus, $1\in J(R)$, a
contradiction. Hence $\tau(J(R)^{\circ})=0$ and, by theorem of Zelmanov [71],
$J(R)^{\circ}$ is nilpotent as a locally nilpotent $m$-Engel torsion-free
group. Since $pR\subseteq J(R)$, $\gamma_{n+1}(pR)=0$ for some integer $n\geq
0$ and

$\begin{array}[]{rclrlrl}\gamma_{1}(pR)&=&pR&{}\hfil&{}&=&p\gamma_{1}(R),\\\
\gamma_{2}(pR)&=&[pR,pR]&=&p^{2}[R,R]&=&p^{2}\gamma_{2}(R),\\\
&\vdots&{}&\vdots&{}&\vdots&{}\\\
\gamma_{n+1}(pR)&=&[\gamma_{n}(pR),\gamma_{n}(pR)]&=&p^{n+1}[\gamma_{n}(R),\gamma_{n}(R)]&=&p^{n+1}\gamma_{n+1}(R).\end{array}$

Thus we conclude that $\gamma_{n+1}(R)=0$, i.e., $R$ is Lie nilpotent. ∎

######  Lemma 33.

Let $R$ be a local ring with the nil Jacobson radical $J(R)$. If $R$ is Engel
and $\operatorname{char}R/J(R)=0$, then $R$ is Lie nilpotent.

###### Proof.

If $0\neq a\in R$ and $pa=0$ for some prime $p$, then $a\cdot pR=0$ and
therefore $pR\subseteq J(R)$, a contradiction. Hence $F(R)=0$. If $qR\neq R$
for some prime $q$, then $qR\subseteq J(R)$ and, for any $x\in R$, there
exists $k=k(x)\in\mathbb{N}$ such that $q^{k}x^{k}=0$, a contradiction. Hence
$R^{+}$ is a divisible group. As a consequence, $R$ is an algebra over the
rational numbers field $\mathbb{Q}$ and $R$ is Lie nilpotent by [49, Theorem
B]. ∎

##  8\. Corollaries

There are large number of articles which extend the Cohen’s Theorem [20] and
Kaplansky Theorem [36, Theorem 12.3] (see e.g. [48, 65] and others). We also
present the following generalizations of these theorems.

######  Corollary 34.

Let $R$ be a ring with unity. If the commutator ideal $C(R)$ is nil (in
particular, $R$ is Engel), then the following statements hold:

  * $(i)$

if prime ideals of $R$ are finitely generated as right ideals, then the
quotient ring $R/C(R)$ is a commutative Noetherian ring;

  * $(ii)$

if $R$ is a right Noetherian ring and each its maximal right ideal is
principal, then $R/C(R)$ is a commutative principal ideal ring;

  * $(iii)$

if each prime ideal of $R$ is principal as a right ideal, then $R/C(R)$ is a
commutative principal ideal ring;

  * $(iv)$

if $R$ is right Noetherian and $R/{\mathbb{P}}(R)$ is finite, then $R/C(R)$ is
finite.

###### Proof.

$(i)$–$(iii)$ Since $I+C(R)$ is an ideal of $R$ for its any right ideal $I$,
the result follows from Cohen’s Theorem [20] and Kaplansky Theorem [36,
Theorem 12.3].

$(iv)$ Without loss of generality, assume that ${\mathbb{P}}(R)\neq 0$.
Obviously, $C(R)$ is nilpotent by [43, Theorem 1] and Corollary 8 and
$C(R)\subseteq{\mathbb{P}}(R)\subseteq P$ for any nonzero prime ideal $P$ of
$R$. Finally, $R$ is Artinian by Lemma 21 and so $R/C(R)$ is finite by [7,
Lemma 22]. ∎

######  Remark 35.

If $R$ is a Lie nilpotent algebra over a field of characteristic $0$ and
$I=I^{2}$ is a f.g. ideal (as a one-sided ideal) of $R$, then $I=eR$ for some
central idempotent $e\in R$.

In fact, this follows from [64, Theorem 1] and Corollary 8$(ii)$.

######  Proposition 36.

Let $R$ be a nil ring. The following statements hold:

  * $(i)$

if $R$ is $n$-Engel as a Lie ring and $F(R)=0$, then $R^{\circ}$ is nilpotent
(and so $R$ is Lie nilpotent);

  * $(ii)$

if $R$ is $n$-Engel of bounded index, then $R/F(R)$ is Lie nilpotent (and so
the adjoint group $R^{\circ}$ is locally nilpotent and torsion-by-(torsion-
free nilpotent)).

###### Proof.

$(i)$ The adjoint group $R^{\circ}$ is $m$-Engel for some $m\in\mathbb{N}$ by
[4, Main Theorem] and, therefore, it is locally nilpotent by [4, Lemma 2.2].
As a consequence, $R^{\circ}$ is nilpotent by [71] (see e.g. [2, Corollary 1])
and $R$ is Lie nilpotent by [35].

$(ii)$ In view of the result of Levitzky [26, Lemma 1.1], the ring $R$ is
locally nilpotent and so $R^{\circ}$ is a locally nilpotent group. If
$F(R)=0$, then $R^{\circ}$ is nilpotent by [71] and hence $R$ is Lie nilpotent
by [35]. ∎

######  Proposition 37.

Let $R$ be a ring such that $Dz(R)$ is commutative. The following holds:

  * $(i)$

either $Dz(R)^{2}\neq 0$ and $R$ is Lie metabelian (then the adjoint group
$R^{\circ}$ is metabelian) or $Dz(R)^{2}=0$ (i.e., $R$ is $2$-primal); in the
last case $R$ is commutative or $Dz(R)$ is completely prime ideal of $R$;

  * $(ii)$

if $R$ is Engel ($2$-torsion-free Lie solvable, $2$-torsion-free locally
finite, respectively) with unity, then it is Lie metabelian and $C(R)^{3}=0$.

###### Proof.

Assume that $R$ is non-commutative.

$(i)$ The set $N(R)$ is an ideal of $R$ by [39, Theorem 5.7]. If
$Dz(R)^{2}\neq 0$, then $R$ is Lie metabelian by [39, Theorem 5.7].

Now, assume that $Dz(R)^{2}=0$. Thus $Dz(R)=N(R)$ and $ab\in Dz(R)$ implies
that $a\in Dz(R)$ or $b\in Dz(R)$ for each $a,b\in R$ what means that $Dz(R)$
is completely prime.

$(ii)$ If $Dz(R)^{2}=0$, then $N(R)^{2}=0$ and therefore $R$ is Lie metabelian
in view of Corollary 10 (Theorem 1$(iii)$, Lemma 18$(iv)$, respectively).

From [39, Theorem 5.7] it follows that $R$ is Lie metabelian also in the case
$Dz(R)^{2}\neq 0$.

If $D(R)^{2}\neq 0$, then $D(R)^{2}C(R)=0$ by [39, Lemma 5.4] and so
$C(R)^{3}=0$.

Finally, $Dz(R)^{2}C(R)=0$ by [39, Lemma 5.4] and so $C(R)^{3}=0$. ∎

######  Remark 38.

Let $R$ be a ring such that the set $N(R)$ is commutative. If $R$ is an Engel
($2$-torsion-free Lie solvable, $2$-torsion-free locally finite,
respectively), then $N(R)$ is an ideal of $R$, $C(R)\subseteq N(R)$ and
$N(R)^{2}\cdot C(R)=0$.

In fact, $C(R)\subseteq N(R)$ by Corollary 10 (by Theorem 1, Lemma 18$(iv)$,
respectively) what gives that $N(R)$ is an ideal of $R$ and the result
follows.

######  Corollary 39.

Let $R$ be a ring with unity. The following statements hold:

  * $(i)$

if $R$ is right Noetherian $\pi$-regular and satisfies the $n$-Engel condition
(is $2$-torsion-free Lie solvable, respectively), then it is a field or right
Artinian;

  * $(ii)$

if $R$ is an Engel ring of bounded index and $J(R)$ is nil, then

$C(R)\subseteq{\mathbb{P}}(R)=N(R)=J(R),$

$N(R)$ is a locally nilpotent ideal of $R$ and $U(R)$ is (locally
nilpotent)-by-abelian.

###### Proof.

$(i)$ The ring $R$ is abelian by Corollary 8 (by Theorem 1, respectively) and,
therefore, $R/P$ is a field for any prime ideal $P$ by [9, Theorem 5]. Hence
each prime ideal is a maximal right ideal. In view of Lemma 21 and Corollary
8$(i)$, $R$ is a field or right Artinian.

$(ii)$ We obtain that $C(R)\subseteq N(R)=J(R)$ by Corollary 10. Then $R/P$ is
commutative for each prime ideal $P$ of $R$ and so
$C(R)\subseteq{\mathbb{P}}(R)$. The ideal $N(R)$ is locally nilpotent in view
of [26, Lemma 1.1] and so the unit group $U(R)$ is (locally nilpotent)-by-
abelian. ∎

Finitely generated non-commutative radical rings need not be nil [59]. Each
nil ring is Engel [63, Proposition 4.2] and the adjoint group $R^{\circ}$ of a
nil algebra $R$ is locally graded (i.e., each finitely generated infinite
subgroup of $R^{\circ}$ contains a proper subgroup of finite index) [60,
Theorem 1]. We extend this result in the following way.

######  Corollary 40.

For each Engel ($2$-torsion-free Lie solvable, $2$-torsion-free locally
finite, respectively) algebra $R$, its adjoint group $R^{\circ}$ is locally
graded.

###### Proof.

The set $N(R)$ is an ideal of $R$ by Corollary 10 (by Theorem 1$(iii)$ and
Proposition 5$(ii)$, respectively) and $N(R)^{\circ}$ is locally graded by
[60, Theorem 1]. Inasmuch as $R^{\circ}/N(R)^{\circ}$ is abelian, we deduce
that $R^{\circ}$ is locally graded. ∎

######  Proposition 41.

If $R$ is a ring of bounded index, then the following statements hold:

  * $(i)$

${\mathbb{P}}(R)=N_{r}(R)$ is a locally nilpotent ring (and so
${\mathbb{P}}(R)^{\circ}$ is a locally nilpotent group);

  * $(ii)$

if $R$ is Engel ($2$-torsion-free locally solvable, $2$-torsion-free locally
finite, respectively), then $R^{\circ}$ is a (locally nilpotent)-by-abelian
group;

  * $(iii)$

if $R$ is $n$-Engel and $F(R)=0$, then $N(R)$ is nilpotent and $R^{\circ}$ is
(torsion-free nilpotent)-by-abelian;

  * $(iv)$

if $R$ is $n$-Engel, then $N(R)^{\circ}$ is (torsion locally
nilpotent)-by-(torsion-free nilpotent).

###### Proof.

$(i)$ Obviously, ${\mathbb{P}}(R)\subseteq N_{r}(R)$ and so
$N_{r}(R/{\mathbb{P}}(R))=0$ in view of [26, Lemma 1.1]. Thus
${\mathbb{P}}(R)=N_{r}(R)$. Since each nonzero homomorphic image of
${\mathbb{P}}(R)$ contains a nonzero nilpotent ideal, ${\mathbb{P}}(R)$ is
locally nilpotent. If $S=\langle g_{1},\ldots,g_{n}\rangle$ for
$g_{1},\ldots,g_{n}\in N_{r}(R)$, then $S\subseteq(\langle
g_{1},\ldots,g_{n}\rangle_{rg})^{\circ}$. Hence $S$ is nilpotent and
${\mathbb{P}}(R)^{\circ}$ is locally nilpotent.

$(ii)$ Inasmuch as $C(R)\subseteq N(R)={\mathbb{P}}(R)$ by Corollary 10
(Theorem 1$(iii)$, Lemma 18$(iv)$, respectively), the ideal $N(R)$ is locally
nilpotent in view of [26, Lemma 1.1] and the result follows.

$(iii)$ We get that $C(R)\subseteq{\mathbb{P}}(R)=N(R)$ by Corollary 8 and
$N(R)^{\circ}$ is a torsion-free locally nilpotent $m$-Engel group for some
$m\in\mathbb{N}$ in view of the part $(i)$ and [4, Main Theorem].
Consequently, $N(R)^{\circ}$ is nilpotent in view of [71] and
$(R/N(R))^{\circ}$ is abelian.

$(iv)$ The quotient group $N(R)^{\circ}/F(N(R))^{\circ}$ is torsion-free
locally nilpotent $m$-Engel for some $m\in\mathbb{N}$ and so it is nilpotent.
Futhermore, $F(N(R))^{\circ}$ is locally nilpotent by [2, Corollary 2]. ∎

Finally, we have also the following.

######  Proposition 42.

If the set $N(R)$ of nilpotent elements of a ring $R$ is finite and
$C(R)\subseteq N(R)$, then the following statements hold:

  * $(i)$

$R$ is an $FC$-ring (i.e., the centralizer $C_{R}(a)=\\{r\in R\mid ra=ar\\}$
is of finite index in the additive group $R^{+}$ for any $a\in R$ [7]);

  * $(ii)$

$R^{\circ}$ is a finite-by-abelian group with the finite commutator subgroup
(i.e., $R^{\circ}$ is a $BFC$-group (see e.g. [52])).

###### Proof.

$(i)$ The quotient group $R^{+}/\ker\partial_{x}$ is isomorphic to the image
$\operatorname{Im}\partial_{x}$ for any $x\in R$. Inasmuch as
$\operatorname{Im}\partial_{x}\subseteq C(R)$, we conclude that $R$ is an
$FC$-ring.

$(ii)$ Since $N(R)^{\circ}$ is a finite normal subgroup of $R^{\circ}$ and
$R^{\circ}/N(R)^{\circ}\cong(R/N(R))^{\circ}$, the assertion is true. ∎

## References

  * [1] B. Amberg and O. Dickenschied.  On the adjoint group of a radical ring.  Canad. Math. Bull., 38(3):262–270, 1995.
  * [2] B. Amberg, O. Dickenschied, and Y. P. Sysak.  Subgroups of the adjoint group of a radical ring.  Canad. J. Math., 50(1):3–15, 1998.
  * [3] B. Amberg and Y. Sysak.  Associative rings whose adjoint semigroup is locally nilpotent.  Arch. Math. (Basel), 76(6):426–435, 2001.
  * [4] B. Amberg and Y. P. Sysak.  Radical rings with Engel conditions.  J. Algebra, 231(1):364–373, 2000.
  * [5] B. Amberg and Y. P. Sysak.  Radical rings with soluble adjoint groups.  J. Algebra, 247(2):692–702, 2002.
  * [6] B. Amberg and Y. P. Sysak.  Associative rings with metabelian adjoint group.  J. Algebra, 277(2):456–473, 2004.
  * [7] O. Artemovych.  $FC$-rings.  Miskolc Math. Notes, 18(2):623–637, 2017.
  * [8] G. Azumaya.  Strongly $\pi$-regular rings.  J. Fac. Sci. Hokkaido Univ. Ser. I., 13:34–39, 1954.
  * [9] A. Badawi.  On abelian $\pi$-regular rings.  Comm. Algebra, 25(4):1009–1021, 1997.
  * [10] H. Bell and A. Klein.  On rings with Engel cycles.  Canad. Math. Bull., 34(3):295–300, 1991.
  * [11] H. Bell and A. Yaqub.  On commutativity of semiperiodic rings.  Results Math., 53(1-2):19–26, 2009.
  * [12] L. Bokut, I. Lvov, and V. Kharchenko.  Noncommutative rings.  In Current problems in mathematics. Fundamental directions, Vol. 18 (Russian), Itogi Nauki i Tekhniki, pages 5–116. Akad. Nauk SSSR, Vsesoyuz. Inst. Nauchn. i Tekhn. Inform., Moscow, 1988.
  * [13] A. Bovdi.  Group algebras with an Engel group of units.  J. Aust. Math. Soc., 80(2):173–178, 2006.
  * [14] A. Bovdi and I. Khripta.  The Engel property of the multiplicative group of a group algebra.  Mat. Sb., 182(1):130–144, 1991.
  * [15] V. Bovdi.  Group algebras whose unit group is locally nilpotent.  J. Aust. Math. Soc., 109(1):17–23, 2020.
  * [16] L. Carini.  Derivations on Lie ideals in semiprime rings.  Rend. Circ. Mat. Palermo (2), 34(1):122–126, 1985.
  * [17] F. Catino, M. Miccoli, and Y. Sysak.  On the adjoint group of semiprime rings.  Comm. Algebra, 35(1):265–270, 2007.
  * [18] C.-L. Chuang and T.-K. Lee.  Semiprime rings with prime ideals invariant under derivations.  J. Algebra, 302(1):305–312, 2006.
  * [19] C.-L. Chuang and J.-S. Lin.  On a conjecture by Herstein.  J. Algebra, 126(1):119–138, 1989.
  * [20] I. S. Cohen.  Commutative rings with restricted minimum condition.  Duke Math. J., 17:27–42, 1950.
  * [21] R. Y. Evstafev.  On Artinian rings satisfying the Engel conditions.  Ukraïn. Mat. Zh., 58(9):1264–1270, 2006.
  * [22] K. Feĭs.  Algebra: koltsa, moduli i kategorii. II.  “Mir”, Moscow, 1979.  Translated from the English by L. A. Koĭfman, A. V. Mikhalev, T. S. Tolskaya, G. M. Tsukerman and M. P. Dorofeeva, Edited by L. A. Skornjakov.
  * [23] K. R. Goodearl and R. B. Warfield, Jr.  Primitivity in differential operator rings.  Math. Z., 180(4):503–523, 1982.
  * [24] K. Gruenberg.  Two theorems on Engel groups.  Proc. Cambridge Philos. Soc., 49:377–380, 1953.
  * [25] I. Herstein.  A remark on rings and algebras.  Michigan Math. J., 10:269–272, 1963.
  * [26] I. Herstein.  Topics in ring theory.  The University of Chicago Press, Chicago, Ill.-London, 1969.
  * [27] I. Herstein.  On the Lie structure of an associative ring.  J. Algebra, 14:561–571, 1970.
  * [28] I. Herstein.  Multiplicative commutators in division rings.  Israel J. Math., 31(2):180–188, 1978.
  * [29] I. Herstein.  Multiplicative commutators in division rings. II.  Rend. Circ. Mat. Palermo (2), 29(3):485–489 (1981), 1980.
  * [30] I. Herstein and L. Small.  Nil rings satisfying certain chain conditions.  Canad. J. Math., 16:771–776, 1964.
  * [31] P. Higgins.  Lie rings satisfying the Engel condition.  Proc. Cambridge Philos. Soc., 50:8–15, 1954.
  * [32] Y. Hirano.  Some studies on strongly $\pi$-regular rings.  Math. J. Okayama Univ., 20(2):141–149, 1978.
  * [33] Y. Hirano.  On periodic P.I. rings and locally finite rings.  Math. J. Okayama Univ., 33:115–120, 1991.
  * [34] C. Huh, N. K. Kim, and Y. Lee.  Examples of strongly $\pi$-regular rings.  J. Pure Appl. Algebra, 189(1-3):195–210, 2004.
  * [35] S. Jennings.  Radical rings with nilpotent associated groups.  Trans. Roy. Soc. Canada. Sect. III. (3), 49:31–38, 1955.
  * [36] I. Kaplansky.  Elementary divisors and modules.  Trans. Amer. Math. Soc., 66:464–491, 1949.
  * [37] A. Kemer.  Nonmatrix varieties.  Algebra i Logika, 19(3):255–283, 382, 1980.
  * [38] N. K. Kim, T. K. Kwak, and Y. Lee.  On a generalization of right duo rings.  Bull. Korean Math. Soc., 53(3):925–942, 2016.
  * [39] A. Klein and H. Bell.  Rings with commuting nilpotents and zero divisors.  Results Math., 51(1-2):73–85, 2007.
  * [40] A. Krasilnikov.  On the group of units of a ring whose associated Lie ring is metabelian.  Uspekhi Mat. Nauk, 47(6(288)):217–218, 1992.
  * [41] J. Krempa.  Rings with periodic unit groups.  In Abelian groups and modules (Padova, 1994), volume 343 of Math. Appl., pages 313–321. Kluwer Acad. Publ., Dordrecht, 1995.
  * [42] J. Krempa.  Rings with periodic groups of units. II.  In Groups St. Andrews 1997 in Bath, II, volume 261 of London Math. Soc. Lecture Note Ser., pages 503–511. Cambridge Univ. Press, Cambridge, 1999.
  * [43] C. Lanski.  Nil subrings of Goldie rings are nilpotent.  Canad. J. Math., 21:904–907, 1969.
  * [44] C. Lanski.  An Engel condition with derivation for left ideals.  Proc. Amer. Math. Soc., 125(2):339–345, 1997.
  * [45] J. Lewin.  Subrings of finite index in finitely generated rings.  J. Algebra, 5:84–88, 1967.
  * [46] N. McCoy.  Prime ideals in general rings.  Amer. J. Math., 71:823–833, 1949.
  * [47] E. Picelli.  Semilocal rings with Engel conditions.  Arch. Math. (Basel), 87(4):289–294, 2006.
  * [48] M. Reyes.  Noncommutative generalizations of theorems of Cohen and Kaplansky.  Algebr. Represent. Theory, 15(5):933–975, 2012.
  * [49] D. Riley.  Algebras with collapsing monomials.  Bull. London Math. Soc., 30(5):521–528, 1998.
  * [50] D. Riley.  Generalised nilpotence conditions in $n$-Engel Lie algebras.  Comm. Algebra, 28(10):4619–4634, 2000.
  * [51] D. M. Riley and M. C. Wilson.  Associative rings satisfying the Engel condition.  Proc. Amer. Math. Soc., 127(4):973–976, 1999.
  * [52] D. Robinson.  A course in the theory of groups, volume 80 of Graduate Texts in Mathematics.  Springer-Verlag, New York-Berlin, 1982.
  * [53] L. Rowen.  Ring theory. Vol. II, volume 128 of Pure and Applied Mathematics.  Academic Press, Inc., Boston, MA, 1988.
  * [54] A. Shalev.  On associative algebras satisfying the Engel condition.  Israel J. Math., 67(3):287–290, 1989.
  * [55] R. Sharma and J. Srivastava.  Lie solvable rings.  Proc. Amer. Math. Soc., 94(1):1–8, 1985.
  * [56] R. Sharma and J. Srivastava.  Lie centrally metabelian group rings.  J. Algebra, 151(2):476–486, 1992.
  * [57] A. Shlyafer.  Periodic multiplicative subgroups of rings and algebras.  In Abelian groups and modules, No. 9 (Russian), pages 136–152, 159. Tomsk. Gos. Univ., Tomsk, 1990.
  * [58] A. Smoktunowicz.  On Engel groups, nilpotent groups, rings, braces and the Yang-Baxter equation.  Trans. Amer. Math. Soc., 370(9):6535–6564, 2018.
  * [59] A. Smoktunowicz and E. Puczył owski.  A polynomial ring that is Jacobson radical and not nil.  Israel J. Math., 124:317–325, 2001.
  * [60] A. Sozutov and I. Alexandrova.  On some properties of adjoint groups of associative nil algebras.  Zh. Sib. Fed. Univ. Mat. Fiz., 10(4):503–508, 2017.
  * [61] B. Stolz.  Nilpotenzbedingungen für radikale und lokale rings.  Diplomarbeit Johannes Gutenberg-Universität Mainz, pages 1–30, 2002.
  * [62] Y. Sysak.  Periodic subgroups of the adjoint group of a radical ring.  In Topics in infinite groups, volume 8 of Quad. Mat., pages 327–333. Dept. Math., Seconda Univ. Napoli, Caserta, 2001.
  * [63] Y. P. Sysak.  The adjoint group of radical rings and related questions.  In Ischia group theory 2010, pages 344–365. World Sci. Publ., Hackensack, NJ, 2012.
  * [64] J. Szigeti.  Idempotent ideals in Lie nilpotent rings.  In Methods in ring theory (Levico Terme, 1997), volume 198 of Lecture Notes in Pure and Appl. Math., pages 287–292. Dekker, New York, 1998.
  * [65] J. Szigeti and L. Van Wyk.  On Lie nilpotent rings and Cohen’s theorem.  Comm. Algebra, 43(11):4783–4796, 2015.
  * [66] A. Tuganbaev.  Semiregular, weakly regular, and $\pi$-regular rings.  J. Math. Sci. (New York), 109(3):1509–1588, 2002.  Algebra, 16.
  * [67] J. Šter.  Rings in which nilpotents form a subring.  Carpathian J. Math., 32(2):251–258, 2016.
  * [68] A. Zalesskii.  Solvable groups and crossed products.  Mat. Sb. (N.S.), 67 (109):154–160, 1965.
  * [69] A. Zalesskii and M. Smirnov.  Associative rings satisfying the identity of Lie solvability.  Vestsī Akad. Navuk BSSR Ser. Fīz.-Mat. Navuk, (2):15–20, 123, 1982.
  * [70] E. Zelmanov.  Engel Lie algebras.  Dokl. Akad. Nauk SSSR, 292(2):265–268, 1987.
  * [71] E. Zelmanov.  Some problems in the theory of groups and Lie algebras.  Mat. Sb., 180(2):159–167, 1989.
  * [72] E. Zelmanov.  Solution of the restricted Burnside problem for groups of odd exponent.  Izv. Akad. Nauk SSSR Ser. Mat., 54(1):42–59, 221, 1990.