1 Introduction
In this article, we study the rings of the form
$R = R_1 + \dots + R_n$
where
$R_i$
are subrings or additive subgroups of R and
$R_1 + \dots + R_n$
denotes the set of all sums
$r_1 + \dots + r_n$
with
$r_1 \in R_1, \dots , r_n \in R_n$
. We focus mainly on the case
$n=2$
and
$n=3$
. We also consider somewhat more general situation in which
$R=A+B$
, B is a subring of R, and A is a subgroup of the additive group of R.
Many interesting results concerning the subject of this article were focused on rings which are sums of two subrings (e.g., [Reference Ferrero and Puczylowski3, Reference Kegel6, Reference Kelarev and McConnell8, Reference Kepczyk and Puczyłowski10–Reference Kepczyk12, Reference Kepczyk14, Reference Kepczyk15]). In particular, Kegel proved that a ring which is a sum of two nilpotent subrings is nilpotent itself (see [Reference Kegel6]). In his next paper [Reference Kegel7] Kegel presented various generalizations of this theorem. Among others, he showed that the prime radical
$\beta $
of a ring that is a sum of two subrings contains a product of the hyperanihilators of these subrings (Theorem 1), while assuming that one of the subrings is a
$\beta $
radical he showed that the Levitzki radical of this ring contains the hyperanihilator of the second subring (Theorem 3). These results were generalized in [Reference Kepczyk and Puczyłowski11]. In addition, answers to most of the questions raised in [Reference Kegel7] are presented in [Reference Kepczyk and Puczyłowski11].
It turned out (see [Reference Kepczyk and Puczyłowski10]) that if
$R=R_1+R_2$
is a ring,
$R_1$
and
$R_2$
are nil of bounded index subrings of R (i.e., they satisfy an identity
$x^n = 0$
), then so is R. In addition, in [Reference Andruszkiewicz and Kepczyk1], it was shown that if
$R_i$
are left (right) T-nilpotent, then R is left (right) T-nilpotent, too. However, in [Reference Kelarev9, Reference Salwa16] Kelarev and independently Salwa provided examples of a rings that are the sums of two
$\beta $
-radical subrings, which itself are not
$\beta $
-radical.
Another direction of research concerning the Kegel’s theorem is related to the class of
$PI$
rings. In [Reference Kepczyk13] it was shown that a ring that is the sum of two
$PI$
rings is a
$PI$
ring. A more effective combinatorial proof in the special case of this fact can be found in the paper [Reference Felzenszwalb, Giambruno and Leal5], in which Felzenszwalb, Giambruno, and Leal showed that if
$R=A+B$
, A and B are
$PI$
rings and for some positive integer
$n\geq 1$
either
$(AB)^n \subseteq A$
or
$ (BA)^n \subseteq A$
, then R is a
$PI$
ring. In this article using the concept of one-sided accessible subrings defined in the Section 2 (cf. [Reference Felzenszwalb4]) we prove that
$R\in \mathcal {M}$
if and only if
$A^nB^m\in \mathcal {M}$
for some positive integers
$n,m$
, where
$\mathcal {M}$
is a fairly general left (right) strong class of rings containing the class of nilpotent rings (Section 2). Moreover, we will present a new proof of Kegel’s result and some it’s extensions.
On the other hand, Bokut’ proved in [Reference Bokut2] that every algebra over a field can be embedded into a simple algebra which is a sum of three nilpotent subalgebras of index three, which shows that the cases related to the sums of two and three subrings are totally different. We will present a generalization of the result from [Reference Kepczyk14] stating that a ring is nilpotent if it is a sum of three subrings with zero multiplication.
In [Reference Kelarev and McConnell8], it was proved that if
$R=A+B$
,
$A^2\subseteq B$
, B is locally nilpotent subring of R and A is a subgroup of the group
$R^+$
, then the ring R is locally nilpotent. We generalize this result proving the claim under the assumption
$A^d\subseteq B$
for some positive integer d and
$B\in S$
, where S is a N-radical contained in the class of all locally nilpotent rings. We will complete the article with open problems for further studies.
All the rings considered in this article are associative, but are not assumed to have an identity. By
$R^1$
we denote a ring R with an identity adjoined. We follow the convention that if
$S\subseteq R$
, then
$S^0=\{1\}$
. If A is an ideal (left ideal or right ideal) of a ring R, then we write
$A\lhd R$
(
$A<_l R$
or
$A <_r R$
). If it is not necessary to distinguish the side of a one-sided ideal A of R, then we simple write
$A<R$
. For any subset C, the symbols
$l_R(C)$
and
$r_R(C)$
stand for the left and a right anihilator of C in R, respectively.
The prime, Levitzki, and Jacobson radicals are denoted by
$\beta $
,
$\mathcal {L}$
, and
$\mathcal {J}$
, respectively. The symbol
$\mathcal {N}$
stands for the class of all nilpotent rings. The set of all positive integers is denoted by
$\mathbb {N}$
.
2 New propositions
Let’s start with following simple observations.
Remark 2.1 Suppose that
$R=L+B$
, where
$L<_l R$
and B is a subring of the ring R. If the rings L and B are nilpotent, then so is R. Indeed, there exists
$n\in \mathbb {N}$
such that
$L^n=\{0\}$
. Let
$I=L+LR$
. Then
$I\lhd R$
and
$I=LR^1$
, whence
$I^n=\{0\}$
. Moreover,
$R/I=(B+I)/I\cong B/(B\cap I)$
is a nilpotent ring. Consequently, the ring R is nilpotent too.
Remark 2.2 Let A and B be nilpotent subrings of a ring R such that
$R=A+B$
. If S is a subring of the ring R such that
$A<_l S$
, then S is also a nilpotent ring. Indeed, from the modularity of the lattice of subgroups of the group
$(R,+)$
and the fact that
$A\subseteq S$
and
$R=A+B$
, we get
$S=A+(S\cap B)$
. But the rings A and
$S\cap B$
are nilpotent and
$A<_l S$
, so if follows from Remark 2.2 that the ring S is nilpotent.
Let A and B be subrings of the ring R such that
$R=A+B$
. By repeatedly using Remark 2.2 we infer that if
$A^n=\{0\}$
, then
$A<_l A+RA^{n-1}<_l A+RA^{n-2}<_l\dots <_l A+RA<_l R$
. Therefore, if subrings A and B of the ring
$R=A+B$
are nilpotent, then R is also nilpotent. Thus we obtain a simple proof of the well-known Kegel theorem from [Reference Kegel6].
This provides the motivation for us to consider subrings one-sided accessible. We say that a subring A of a ring R is one-sided accessible in R if there exist subrings
$A_0=A, A_1,\dots ,A_n=R$
such that
$A_i< A_{i+1}$
for
$i=0,1,\dots ,n-1$
. It is equivalent to
$A^mRA^k\subseteq A$
for some
$m,k\in \mathbb {N}\cup \{0\}$
. Hence, for example, every nilpotent subring of the ring R is one-sided accessible in R.
Lemma 2.3 Let A and B be subrings of a ring R such that
$R=A+B$
. Then
$S=A^{k_1}B^{l_1}\cdot\ldots \cdot A^{k_s}B^{l_s}$
and
$T=A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_s}B^{l_s}A^{k_{s+1}}$
are one-sided accessible in R for all
$s,k_1,\dots ,k_s,k_{s+1},l_1,\dots ,l_s\in \mathbb {N}$
. Moreover,
$SRS\subseteq S^2$
and
$TRT\subseteq T$
.
Proof Since
$S^2\subseteq A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_{s-1}}B^{l_{s-1}}RA^{k_s}B^{l_s}\subseteq A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_{s-1}}B^{l_{s-1}}(A+B)A^{k_s}B^{l_s} =A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_s+1}B^{l_s}+A^{k_1}B^{l_1}\cdot \ldots \cdot B^{l_{s-1}+1}A^{k_s}B^{l_s}\subseteq S$
, S is a subring of R. Moreover,
$SRS=S(A+B)S=SAS+SBS\subseteq S^2$
, because of
$AS\subseteq S$
and
$SB\subseteq S$
. Hence
$S<_r S+RS<_l R$
, which means that S is one-sided accessible ring in R.
Furthermore,
$T=SA^{k_{s+1}}$
and
$A^{k_{s+1}}S\subseteq S$
, so
$T^2\subseteq S^2A^{k_{s+1}}\subseteq T$
and
$TRT\subseteq SRSA^{k_{s+1}}\subseteq SA^{k_{s+1}}=T$
, whence T is one-sided accessible ring in R too.
Throughout the article, by
$\mathcal {M}$
we denote an arbitrary homomorphically closed class of rings which satisfy the following conditions:
$\mathcal {N}\subseteq \mathcal {M}$
,
$\mathcal {M}$
is closed under extensions, i.e., if
$I\lhd H$
and I,
$H/I\in \mathcal {M}$
, then
$H\in \mathcal {M}$
,
$\mathcal {M}$
is hereditary for subrings, i.e., if I is a subring of H and
$H\in \mathcal {M}$
, then
$I\in \mathcal {M}$
and I is one-sided strong, i.e., if
$I<H$
and
$I\in \mathcal {M}$
, then the ideal of H generated by I belongs to
$\mathcal {M}$
. The classes
$\mathcal {N}$
,
$\beta $
, and
$\mathcal {L}$
can be considered as
$\mathcal {M}$
.
Recall that a ring A is said to be left T-nilpotent if for each sequence
$(a_n)$
of elements of A, there exists
$n\in \mathbb {N}$
such that
$a_1\cdot a_2\cdot \ldots \cdot a_n=0$
. Let
$l_{0}(A)=\{0\}$
and
$l_{\alpha }(A)=\{x\in A\,|\,xA\subseteq \bigcup _{\beta <\alpha }l_{\beta }(A)\}$
for any ordinal number
$\alpha>0$
. It is well known that a ring A is left T-nilpotent if and only if
$l(A)=A$
, where
$l(A)$
is the union
$\bigcup _{\alpha \geq 0}l_{\alpha }(A)$
. The ideal
$l(A)$
is called a left hiperanihilator of A. The right hiperanihilator
$r(A)$
of A is defined analogously. It is easy to see that a ring A is left T-nilpotent if and only if every non-zero homomorphic image of A has non-zero left annihilator. Clearly, the class of all left T-nilpotent rings is homomorphically closed, hereditary for subrings, closed under extensions, and by [Reference Andruszkiewicz and Kepczyk1, Lemma 2.4], one-sided strong. Therefore, the class of all left T-nilpotent rings can be considered as
$\mathcal {M}$
.
We obtain the following.
Proposition 2.4 Let
$R_1$
and
$R_2$
be subrings of a ring R such that
$R = R_1 + R_2$
. If
$R_1$
is a nilpotent ring and
$R_2\in \mathcal {M}$
, then
$R\in \mathcal {M}$
.
Proof Clearly,
$R_1<_l R_1+RR_1^{n-1}<_l R_1+RR_1^{n-2}<_l\dots <_l R_1+RR_1<_l R$
, where n is the index of nilpotency of
$R_1$
. Hence
$R_1$
is one-sided accessible in R. As in Remark 2.2, we infer that
$R\in \mathcal {M}$
.
Proposition 2.5 Let A and B be subrings of a ring R such that
$R=A+B$
. If
$A\in \mathcal {M}$
and
$B\in \mathcal {M}$
, then the following subrings of R belong to
$\mathcal {M}$
:
$Id_l(A)=\{x\in R: xA\subseteq A\}$
,
$A+l_R(A^n)$
, and
$A+Rl_n(A)$
for
$n\in \mathbb {N}$
. Moreover,
$Id_r(B)=\{x\in R: Bx\subseteq B\}$
,
$B+r_R(B^n)$
, and
$B+r_n(B)R\in \mathcal {M}$
for
$n\in \mathbb {N}$
.
Proof Clearly,
$S=Id_l(A)$
is a subring of R and
$A<_l S$
. Furthermore
$S=A+B\cap S$
,
$B\cap S\in \mathcal {M}$
, and
$\mathcal {M}$
is one-side strong, so
$S\in \mathcal {M}$
. Since
$L=l_R(A^n)<_l R$
and
$(LA)A^n\subseteq LA^n=0$
, we infer that
$L+A$
is a subring of R. But
$(L+A)A^n=A^{n+1}\subseteq A$
, so A is one-sided accessible in
$L+A$
. Hence
$L+A\in \mathcal {M}$
, and consequently
$l_R(A)\in \mathcal {M}$
. Moreover for every
$n\in \mathbb {N}$
, we have
$l_n(A)A^n=\{0\}$
, whence
$(A+Rl_n(A))A^n=A^{n+1}\subseteq A$
, i.e., A is one-sided accessible in
$A+Rl_n(A)$
. Therefore
$A+Rl_n(A)\in \mathcal {M}$
. The rest of the proof runs analogously.
Proposition 2.6 Let A and B be subrings of a ring R such that
$R=A+B$
,
$A\in \mathcal {M}$
, and
$B\in \mathcal {M}$
. Then the following conditions are equivalent:
-
(i)
$A^nB^m\in \mathcal {M}$
for some
$n,m\in \mathbb {N}$
, -
(ii)
$AB\in \mathcal {M}$
, -
(iii)
$R\in \mathcal {M}$
.
Proof
$(i)\Rightarrow (ii)$
. Let
$S=A^nB^m$
. Then by Lemma 2.3 we have
$SRS\subseteq S^2$
. Hence
$S<_r S+RS$
, so
$S+RS^2\in \mathcal {M}$
. Consequently,
$RS^2\in \mathcal {M}$
. Moreover,
$(RS)^2=R(SRS)\subseteq RS^2$
, so
$RS\in \mathcal {M}$
. Furthermore,
$RS\lhd S+RS$
, whence
$S+RS\in \mathcal {M}$
. Thus
$I=S+RS+SR+RSR\in \mathcal {M}$
whereas
$I\lhd R$
. Let
$\overline {R}=R/I$
,
$\overline {A}=(A+I)/I$
, and
$\overline {B}=(B+I)/I$
. Then
$\overline {R}=\overline {A}+\overline {B}$
, whereas
$\overline {A}$
is left T nilpotent and
$\overline {B}\in \mathcal {M}$
. Moreover,
$\overline {A}^n\overline {B}^m=0$
, so
$\overline {A}^n\overline {B}\subseteq l_{\overline {R}}(\overline {B}^{m-1})$
. Therefore, by Proposition 2.5,
$\overline {A}^n\overline {B}=(A^nB+I)/I\in \mathcal {M}$
. But
$I\in \mathcal {M}$
, so
$A^nB\in \mathcal {M}$
. The analogous reasoning based on the the right-side version of Proposition 2.5 leads to the conclusion that
$AB\in \mathcal {M}$
.
$(ii)\Rightarrow (iii)$
. Let
$P=A+AR$
. Then
$P<_rR$
and
$P=A+AB$
, since
$R=A+B$
. Notice that
$AB<_lP$
, so combining this with
$A\in \mathcal {M}$
, we infer that
$P\in \mathcal {M}$
. Consequently,
$R\in \mathcal {M}$
.
The implication
$(iii)\Rightarrow (i)$
is obvious.
Proposition 2.7 Let
$A\in \mathcal {M}$
and
$B\in \mathcal {M}$
be subrings of a ring R such that
$R=A+B$
. If there exist positive integers
$s,k_1,\dots ,k_s,k_{s+1},l_1,\dots ,l_s$
such that
$S=A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_s}B^{l_s}\in \mathcal {M}$
or
$T=A^{k_1}B^{l_1}\cdot \ldots \cdot A^{k_s}B^{l_s}A^{k_{s+1}}\in \mathcal {M}$
, then
$R\in \mathcal {M}$
.
Proof Suppose that
$S\in \mathcal {M}$
, and define
$n=\max \{k_1,l_1,\dots ,k_s,l_s\}.$
Then
$(A^nB^n)^s\subseteq S$
, so
$(A^nB^n)^s\in \mathcal {M}$
, whence
$A^nB^n\in \mathcal {M}$
. Therefore, it follows from Proposition 2.6, that
$R\in \mathcal {M}$
.
Now suppose that
$T\in \mathcal {M}$
. In view of Lemma 2.3,
$TRT\subseteq T$
, so
$T<_lT+TR$
. It follows
$T+T^2R\in \mathcal {M}$
. Hence
$T^2R\in \mathcal {M}$
. In particular,
$T^2B\in \mathcal {M}$
. Substituting
$T^2B$
for S in the first part of the proof, we obtain
$R\in \mathcal {M}$
.
In [Reference Andruszkiewicz and Kepczyk1, Theorem 2.9], it was shown that a ring which is a sum of two left T-nilpotent subrings is left T-nilpotent. The question arises whether this result can be generalized like Kegel’s theorem in the Proposition 2.4.
Problem 1 Let A and B be subrings of a ring R such that
$R=A+B$
. Is it true that if A is a left T-nilpotent and
$B\in \mathcal {M}$
, where
$\mathcal {M}$
contains the class of all left T-nilpotent rings, then
$R\in \mathcal {M}$
?
As was mentioned before, by a result of Bokut’, Kegel’s theorem cannot be generalized to rings that are sums of more than two nilpotent subrings. Nevertheless, in [Reference Kepczyk14, Theorem 3.1] it was proved that a ring which is a sum of three subrings with zero multiplication is nilpotent. Now we will show, on the basis of the results from [Reference Bokut2], that it is not possible to increase a sum of three to a sum of four subrings with zero multiplication or index nilpotency two to three in [Reference Kepczyk14, Theorem 3.1].
Definition 2.8 Let K be associative algebra over a field k of dimension
$\mathrm {dim}_{k} K=\alpha \geqslant \aleph _0$
. We say that K satisfies a condition (
$*$
) if there exists in K a countable sequence of subalgebras
$0\subset K^{(1)}\subset K^{(2)}\subset \dots $
such that
$K=\bigcup _{n=1}^{\infty }K^{(n)}$
and
$\mathrm {dim} \,K^{(n)}=\mathrm {dim} \,K^{(n+1)}/ K^{(n)}=\alpha $
for every
$n\geqslant 1$
.
Example 2.9 Let K be an algebra over a field k of dimension
$\aleph _0$
with zero multiplication. We will show that K satisfies the condition (
$*$
) for
$\alpha =\aleph _0$
. Indeed, let X be an arbitrary base of algebra K. Since
$|X|=\aleph _0$
, there exists pairwise disjoint countable subsets
$X_1,\,X_2,\,\dots $
of X such that
$X=\bigcup _{n=1}^{\infty }X_n$
. Let
$K^{(n)}$
be a k-subspace of a linear space K generated by
$X_1\cup \dots \cup X_n$
for
$n\geq 1$
. Then
$K^{(n)}$
is a subalgebra of an algebra K and
$\mathrm {dim} \,K^{(n)}=\mathrm {dim} \,K^{(n+1)}/ K^{(n)}=\aleph _0$
for every
$n\geqslant 1$
. Clearly,
$K=\bigcup _{n=1}^{\infty }K^{(n)}$
.
Theorem 2.10 [Reference Bokut2, Theorem 2]
Let
$K_1,\,K_2,\,K_3,\, K_4$
be algebras over a field k of dimension
$\alpha \geq |k|$
, where
$\alpha \geqslant \aleph _0$
which satisfy the condition (
$*$
). Then any associative k-algebra A of dimension
$\leq \alpha $
can be embedded into a certain simple associative k-algebra
$\mathcal {A}$
such that
$\mathcal {A}=A_1+A_2+A_3+A_4$
for some its subalgebras
$A_i\cong K_i$
, where
$i=1,2,3,4$
.
Using Theorem 2.10 and Example 2.9 we get straight away the following.
Corollary 2.11 Let p be a prime number and let A be a
$\mathbb {Z}_p$
-algebra with zero multiplication of dimension
$\aleph _0$
. Then, every at most countable ring R such that
$pR=0$
can be embedded in some simple ring
$\mathcal {A}$
such that
$p\mathcal {A}=0$
and
$\mathcal {A}$
is a sum of four subrings
$A_i\cong A$
. In particular, there exist a simple idempotent countable ring P such that
$pP=0$
and
$P=P_1+P_2+P_3+P_4$
for some its countable subrings
$P_1,P_2,P_3,P_4$
with zero multiplication.
Now we will present in more detail mentioned in the introduction very important result of Bokut’.
Theorem 2.12 [Reference Bokut2, Theorem 3]
Any associative k-algebra can be embedded into a simple k-algebra being an algebraic sum of its three nilpotent k-subalgebras, each of which is a free nilpotent algebra of index nilpotency three.
Corollary 2.13 Let p be a prime number and let A be a free nilpotent
$\mathbb {Z}_p$
-algebra of index nilpotency three and dimension
$\aleph _0$
. Then, every at most countable ring R such that
$pR=0$
can be embedded in some countable simple ring
$\mathcal {A}$
such that
$p\mathcal {A}=0$
and
$\mathcal {A}$
is a sum of three subrings
$A_i\cong A$
. In particular, there exist an idempotent countable simple ring P such that
$pP=0$
and
$P=P_1+P_2+P_3$
for some nilpotent countable subrings
$P_1,P_2,P_3$
(each of which is isomorphic to A).
It turns out that the following generalization of [Reference Kepczyk14, Theorem 3.1] holds.
Proposition 2.14 Let
$R_1$
,
$R_2$
, and
$R_3$
be subrings of a ring R such that
$R = R_1 + R_2+R_3$
. If
$R_1$
is a nilpotent ring,
$R_2^{2}= \{0\}$
, and
$R_3^2= \{0\}$
, then R is a nilpotent ring.
Proof We proceed by induction with respect to the index n of nilpotency of
$R_1$
. For
$n=1$
, we get
$R_1= \{0\} $
, so the assertion follows from [Reference Kegel6]. Suppose that
$n>1$
and the assertion holds for any smaller index of nilpotency.
Consider the subring
$S=R_1+R_1^{n-1}R$
of R. By the modularity of the lattice of subgroups of the group
$R^+$
, we obtain that
$S=S\cap R=R_1+(R_2+R_3)\cap S$
. Now we show that S is nilpotent. Since
$R_1$
is a nilpotent right ideal of S, the ideal I of S generated by
$R_1$
is nilpotent too. Moreover,
$S=I+(R_2+R_3)\cap S$
. First we prove that
$((R_2+R_3)\cap S)^m= \{0\}$
for some positive integer m. Let
$m=n+3$
and let
$a_i=b_i+c_i$
be elements of
$(R_2+R_3)\cap S$
, where
$b_i\in R_2$
,
$c_i\in R_3$
and
$1\leq i\leq m$
. Notice that the condition
$R_1a_i\subseteq R_1^2$
implies that for every
$r\in R_1$
, there exists
$\overline {r}\in R_1^2$
such that
$$ \begin{align} rb_i=-rc_i+\overline{r}. \end{align} $$
To show that
$a_1a_2\dots a_m=0$
, it is enough to prove that
$$ \begin{align} x_1y_2x_3y_4\cdot \ldots \cdot z_m =0, \end{align} $$
where
$z_m=x_m$
if
$m=2k+1$
or
$z_m=y_m$
if
$m=2k$
for some
$k\in \mathbb {N}$
. We have the following two cases related to (2.2)
-
(1)
$x_1, x_3, \ldots \in R_2$
and
$y_2, y_4,\ldots \in R_3$
, -
(2)
$x_1, x_3, \ldots \in R_3$
and
$y_2, y_4,\ldots \in R_2$
.
First assume (1). Clearly,
$y_2x_3 =t_1+ x + y$
for some
$t_1\in R_1$
,
$x \in R_2$
, and
$y \in R_3$
. Furthermore, by (2.1) we have
$t_1y_4 \in t_1R_2+R_1^2$
. Hence
$x_1y_2x_3y_4\cdot \ldots \cdot z_m = x_1(y_2x_3)y_4\cdot \ldots \cdot z_m= x_1t_1y_4x_5\cdot \ldots \cdot z_m \in x_1(t_1R_{2}^2 +R_1^2x_5)\cdot \ldots \cdot z_m= x_1R_1^2x_5y_6\cdot \ldots \cdot z_m$
. Therefore,
$x_1y_2x_3y_4\cdot \ldots \cdot z_m = x_1t_2x_5y_6\cdot \ldots \cdot z_m$
for some
$t_2\in R_1^2$
. Consequently, by (2.1),
$t_2x_5 \in t_2R_3+R_1^3$
, so
$x_1y_2x_3y_4\cdot \ldots \cdot z_m = x_1t_3y_6\cdot \ldots \cdot z_m$
for some
$t_3\in R_1^3$
. Similarly, after n steps, we get
$x_1y_2x_3y_4\cdot \ldots \cdot z_m = x_1t_nz_m$
, where
$t_n\in R_1^m$
, which gives
$x_1y_2x_3y_4\cdot \ldots \cdot z_m = 0$
. The proof of (2.2) for (2) runs analogously.
It follows that
$((R_2+R_3)\cap S)^m= \{0\}$
, and consequently, S is nilpotent. Therefore, a subring
$T=R_1^{n-1}+R_1^{n-1}R\subseteq S$
is nilpotent too. But T is a right ideal of R, so the ideal J of R generated by T is nilpotent. Moreover,
$R/J=(R_1+J)/J+(R_2+J)/J+(R_3+J)/J$
. It is easy to see that
$((R_1+J)/J)^{n-1}= \{0\}$
. Now we apply the induction hypothesis to infer that the ring
$R/J$
is nilpotent and consequently R is nilpotent too.
In the context of the presented results, the following question arises.
Problem 2 Let
$R_1$
,
$R_2$
, and
$R_3$
be subrings of a ring R such that
$R = R_1 + R_2+R_3$
. Is it true that if
$R_1$
and
$R_2$
are nilpotent rings,
$R_3^2= \{0\}$
, then R is a nilpotent ring?
By above proposition and [Reference Kepczyk14, Theorem 3.1] we obtain the following.
Proposition 2.15 Let
$R_1$
,
$R_2$
, and
$R_3$
be subrings of a ring R such that
$R = R_1 + R_2+R_3$
. If
$R_1$
is a nil
$PI$
ring,
$R_2^{2}= \{0\}$
and
$R_3^2= \{0\}$
, then
$R\in \beta $
.
The next two problems are naturally associated with foregoing propositions.
Problem 3 Let
$R_1$
,
$R_2$
, and
$R_3$
be subrings of a ring R such that
$R = R_1 + R_2+R_3$
. Is it true that if
$R_1$
is a nil
$PI$
ring,
$R_2^{2}= \{0\}$
, and
$R_3^2= \{0\}$
, then R is a nil
$PI$
ring?
Problem 4 Let
$R_1$
,
$R_2$
, and
$R_3$
be subrings of a ring R such that
$R = R_1 + R_2+R_3$
. Is it true that if
$R_1$
is a left (right) T-nilpotent,
$R_2^{2}= \{0\}$
, and
$R_3^2= \{0\}$
, then R is a left (right) T-nilpotent?
To present another result recall some standard properties concerning radicals. A radical
$\mathcal {R}$
is called left hereditary or right hereditary if for every ring
$S\in \mathcal {R}$
,
$I<_lS$
or
$I<_rS$
implies
$I\in \mathcal {R}$
. Furthermore,
$\mathcal {R}$
is called left strong or right strong if for every ring S,
$\mathcal {R}(S)$
contains all the left or right S-ideals of S, respectively. A radical
$\mathcal {R}$
containing
$\beta $
, left (or right) hereditary, and left (or right) strong is called N-radical (for details see [Reference Sands17]). Examples of N-radicals are
$\beta $
,
$\mathcal {L}$
, and
$\mathcal {J}$
.
Lemma 2.16 Let B and A be an
$\mathcal {R}$
-radical subring of a ring R, where
$\mathcal {R}$
is an N-radical, and a subgroup of the additive group of R, respectively. If
$R=A+B$
and
$A^d\subseteq B$
for some positive integer d, and
$\mathcal {R}(R)=0$
, then
$AB\subseteq A$
and
$BA\subseteq A$
.
Proof Since
$\mathcal {R}(R)=0$
,
$r_R(R)^2= \{0\}$
and
$r_R(R)<_rR$
we have
$r_R(R)= \{0\}$
. Note that if
$AB\nsubseteq A$
, then there exists
$x\in AB$
such that
$x=a+b$
, where
$a\in A$
and
$b\in B$
. Hence
$A^{d-1}b\subseteq B$
. Therefore,
$A\overline {b}\subseteq B$
for some
$0\neq \overline {b}\in B$
. Consequently,
$R\overline {b}\subseteq B$
, whence
$R\overline {b}\in \mathcal {R}$
because of
$R\overline {b}<_lB$
. But
$R\overline {b}<_l R$
and
$\mathcal {R}(R)=0$
, so
$Rb= \{0\}$
, and consequently
$r_R(R)\neq \{0\}$
, a contradiction. Thus
$AB\subseteq A$
. Similarly, we show that
$BA\subseteq A$
.
Proposition 2.17 Let A be a subgroup of the additive group of R, and B be a
$\mathcal {R}$
-radical subring of R, where
$\mathcal {R}$
is an N-radical such that
$\mathcal {R}\subseteq \mathcal {L}$
. If
$R=A+B$
and
$A^d\subseteq B$
for some positive integer d, then
$R\in \mathcal {R}$
.
Proof Without loss of generality we can assume that
$\mathcal {R}(R)=0$
. Moreover,
$A\neq \{0\}$
and
$B\neq \{0\}$
, so there exists
$b\in B$
such that the left ideal
$Rb=Ab+Bb$
of R is non-zero. Furthermore,
$\mathcal {R}(R)=0$
, whence
$Rb\notin \mathcal {R}$
. By Lemma 2.16,
$Ab\subseteq A$
, so
$(Ab)^d\subseteq Bb$
. Clearly,
$Bb\in \mathcal {R}$
. Suppose that
$Bb\nsubseteq \mathcal {R}(Rb)$
. Take any
$x\in B$
such that
$xb\notin \mathcal {R}(Rb)$
. Since
$\mathcal {R}\subseteq \mathcal {L}$
, the subring G generated by
$\{b,x\}$
in B is nilpotent. Hence there exists a positive integer
$n\geq 1$
which is maximal with respect to the condition
$G^nb\nsubseteq \mathcal {R}(Rb)$
. As
$b\in G$
, we get
$bG^nb\subseteq \mathcal {R}(Rb)$
, and consequently
$RbG^nb\subseteq \mathcal {R}(Rb)$
. Therefore, if
$S=Rb/\mathcal {R}(Rb)$
, then
$r_S(S)\neq \{0\}$
, a contradiction. Hence
$Bb\subseteq \mathcal {R}(Rb)$
. Whence
$S=(Ab+\mathcal {R}(Rb))/\mathcal {R}(Rb)$
. But
$(Ab)^d\subseteq Bb\subseteq \mathcal {R}(Rb)$
, so S is nilpotent. Therefore,
$Rb\in \mathcal {R}$
once again gives a contradiction.
In this way, we obtained some generalization and simpler proof of [Reference Kelarev and McConnell8, Lemma 3.9], where the assertion of Proposition 2.17 has been proved for
$\mathcal {R}=\mathcal {L}$
under the assumption
$d=2$
, and [Reference Kepczyk14, Proposition 2.4], where
$\mathcal {R}=\beta $
. Furthermore, the assertion of Proposition 2.17 has been obtained for
$\mathcal {R}=\mathcal {J}$
in [Reference Kepczyk15, Theorem 2.2], although is known that
$\mathcal {L}\subseteq \mathcal {J}$
.
This motivates the next following question.
Problem 5 Let A be a subgroup of the additive group of a ring R, and let B be an
$\mathcal {R}$
-radical subring of R, where
$\mathcal {R}$
is an N-radical. Is it true that if
$R=A+B$
and
$A^d\subseteq B$
for some positive integer d, then
$R\in \mathcal {R}$
?
Proposition 2.18 Let A be a subgroup of the additive group of a ring R, and let B be a non-zero left T-nilpotent subring of R. If
$R=A+B$
and
$l_R(R)= \{0\}$
, then
$l_R(A^n)\cap l_R(B)= \{0\}$
for every
$n\in \mathbb {N}$
.
Proof Since
$l_R(R)= \{0\}$
, it follows from Proposition 2.4 that
$A^n\neq \{0\}$
for every
$n \in \mathbb {N}$
. Let
$L=l_R(B)$
. Since
$B\neq \{0\}$
and B is left T-nilpotent,
$l_B(B)\neq \{0\}$
, so
$L\neq \{0\}$
. For
$0\neq x\in L$
we have
$0\neq xR=xA$
. Let X be the set of all positive integers n such that
$xA^n\neq \{0\}$
for every
$x\in L\setminus \{0\}$
. Clearly,
$1\in X$
.
Suppose that X is finite. Then there exists the largest number n in X. Hence for some non-zero
$x\in L$
, we have
$xA^{n+1}= \{0\}$
. But
$xA^n\neq \{0\}$
,
$l_R(R)= \{0\}$
, and
$R=A+B$
, so
$xA^nB\neq \{0\}$
. Hence
$xA^nb_1\neq \{0\}$
for some
$b_1\in B$
. Suppose that for a certain
$k\in \mathbb {N}$
there are elements
$b_1,\dots ,b_k\in B$
such that
$xA^nb_1\cdot \ldots \cdot b_k\neq \{0\}$
. Since
$xA^nBA^n\subseteq x(A+B)A^n=xA^{n+1}= \{0\}$
,
$xA^nb_1\cdot \ldots \cdot b_kA^n= \{0\}$
, whence
$xA^nb_1\cdot \ldots \cdot b_k B\neq \{0\}$
, that is
$xA^nb_1\cdot \ldots \cdot b_k b_{k+1}\neq \{0\}$
for some
$b_{k+1}\in B$
. Thus, we constructed a sequence
$(b_k)$
of elements of B such that
$xA^n b_1\cdot \ldots \cdot b_k\neq \{0\}$
, and consequently,
$b_1\cdot \ldots \cdot b_k\neq 0$
for all
$k\in \mathbb {N}$
. This contradicts the fact that the ring B is left T-nilpotent.
Thus the set X is infinite. Take any
$m\in \mathbb {N}$
. Then there exists
$k>m$
such that
$k\in X$
. Take any non-zero
$x\in L$
. Then
$0\neq xA^k\subseteq xA^m$
, so
$xA^m\neq \{0\}$
, and consequently
$m\in X$
. Hence
$X=\mathbb {N}$
. This means that
$L\cap l_R(A^m)= \{0\}$
for every
$m\in \mathbb {N}$
.
As a consequence of Proposition 2.18 we get a simple proof of [Reference Andruszkiewicz and Kepczyk1, Theorem 2.6]. Namely, we have the following.
Corollary 2.19 Let B and A be a left T-nilpotent subring of a ring R and a subgroup of the additive group of R, respectively. If
$R=A+B$
and
$A^d\subseteq B$
for some positive integer d, then the ring R is left T-nilpotent.
Proof Since
$A^d\subseteq B$
, we have
$l_R(B)\subseteq l_R(A^d)$
. But from the left T-nilpotence of B it follows that
$l_R(B)\neq \{0\}$
, so
$l_R(A^d)\cap l_R(B)\neq \{0\}$
. By Proposition 2.18,
$l_R(R)\neq \{0\}$
. Now it is easy to see that every non-zero homomorphic image of R has the non-zero left annihilator, so R is a left T-nilpotent ring.
Acknowledgements
The authors thank Mateusz Woronowicz for his helpful comments and suggestions.
