1 Introduction
Throughout, R is an associative ring with unity, and
$U(R)$
,
${\textrm {idem}}(R)$
and
${\textrm {nil}}(R)$
denote, respectively, the group of units, the set of idempotents and the set of nilpotents in R. In the literature, an extensive knowledge has been developed for rings R satisfying, respectively,
$R={\textrm {idem}}(R)+U(R)$
,
$R={\textrm {idem}}(R)+{\textrm {nil}}(R)$
and
$R\backslash (0)=U(R)+{\textrm {nil}}(R)$
. A ring R with
$R={\textrm {idem}}(R)+U(R)$
is called a clean ring, a notion first appeared in 1977 in the prominent paper [Reference Nicholson10] by Nicholson. A ring R with
$R={\textrm {idem}}(R)+{\textrm {nil}}(R)$
is called a nil-clean ring, introduced by Diesl [Reference Diesl5] in 2013. A ring R with
$R\backslash (0)=U(R)+{\textrm {nil}}(R)$
is called a fine ring, introduced by Cǎlugǎreanu and Lam more recently in [Reference Cǎlugǎreanu and Lam3]. These notions can be defined elementwise: an element
$a\in R$
is called a clean element if
$a\in {\textrm {idem}}(R)+U(R)$
, and one defines nil-clean elements and fine elements in a similar manner.
All these notions have natural multiplicative duals. An element in a ring is unit-regular if it is a product of a unit and an idempotent (in either order), and a ring is unit-regular if each of its elements is unit-regular. Thus, unit-regular elements and unit-regular rings are multiplicative duals of clean elements and clean rings. In other words, clean elements and clean rings are additive duals of unit-regular elements and unit-regular rings. An element in a ring is a UN-element if it is a product of a unit and a nilpotent, and a ring is a UN-ring if every nonunit is a product of a unit and a nilpotent. Thus, UN-elements and UN-rings are multiplicative duals of fine elements and fine rings. Unit-regular rings have been well studied in the literature, and UN-rings is a topic discussed recently by Cǎlugǎreanu in [Reference Cǎlugǎreanu2].
While nil-clean rings are widely investigated (for example, see [Reference Breaz, Cǎlugǎreanu, Danchev and Micu1, Reference Diesl5, Reference Kosan, Lee and Zhou6, Reference Kosan, Wang and Zhou7, Reference Matczuk8, Reference McGovern, Raja and Sharp9, Reference Sahinkaya, Tang and Zhou12]), there has been no discussion of their multiplicative dual. Our interest is to fill up what is missing. As a multiplicative dual of a nil-clean element, an element
$a\in R$
is called dual nil-clean if
$a=be$
where b is a nilpotent and e is an idempotent. Because a unit cannot be dual nil-clean, we define a ring R to be dual nil-clean if every nonunit of R is dual nil-clean. We will see that the order of the factors in the product does not matter for a dual nil-clean ring, but matters for a single dual nil-clean element. Here, we completely determine dual nil-clean rings, and our main result states that a ring is dual nil-clean if and only if it is either a local ring with nil Jacobson radical or a
$2\times 2$
matrix ring over a division ring.
For a ring R, the Jacobson radical of R is denoted by
$J(R)$
. We write
${\mathbb M}_n(R)$
for the ring of
$n\times n$
matrices over R. For an element a in a ring R,
$a^{\perp }$
(resp.,
$^{\perp }a$
) denotes the right (resp., left) annihilator of a in R. A ring is called abelian if each of its idempotents is central.
2 The result
A ring R is called dual nil-clean if every nonunit a in R is dual nil-clean, i.e.,
$a=be$
where
$b\in {\textrm {nil}}(R)$
and
$e^2=e\in R$
.
Lemma 2.1 Let R be a dual nil-clean ring. If
$a^{\perp }=0$
or
$^\perp a=0$
, then
$a\in U(R)$
.
Proof Assume
$a^\perp =0$
and
$a\notin U(R)$
, and write
$a=be$
where
$b\in {\textrm {nil}}(R)$
and
$e^2=e\in R$
. Then
$a(1-e)=be(1-e)=0$
, so
$1-e\in a^{\perp }$
, and hence
$e=1$
. So
$a=b$
is nilpotent. Choose
$n\ge 1$
such that
$a^n\not =0$
but
$a^{n+1}=0$
. Then
$0\not = a^n\in a^{\perp }$
, a contradiction.
Assume
$^\perp a=0$
and
$a\notin U(R)$
, and write
$a=be$
where
$b\in {\textrm {nil}}(R)$
and
$e^2=e\in R$
. Then
$b\not = 0$
. Let us say
$b^{n+1}=0$
but
$b^n\not = 0$
. Then
$b^na=b^{n+1}e=0$
, so
$0\not = b^n\in {}^{\perp}a$
, a contradiction. ▪
Lemma 2.2
[Reference Robson11] Let R be a ring and
$n\ge 2$
. Then R is isomorphic to some
$n\times n$
matrix ring if and only if R contains elements
$a_1, \ldots , a_n$
and f such that
$1=\sum _{i=1}^{n}f^{i-1}a_if^{n-i}$
and
$f^n=0$
.
Dual nil-clean rings can be completely determined.
Theorem 2.3 A ring R is dual nil-clean if and only if R is either a local ring with
$J(R)$
nil or the
$2\times 2$
matrix ring over a division ring.
Proof
$(\Leftarrow )$
. If R is a local ring with
$J(R)$
nil, then R is clearly dual nil-clean. Let D be a division ring and let
$A\in {\mathbb M}_2(D)$
be a nonunit. By Gaussian elimination, there is a unit
$U\in {\mathbb M}_2(D)$
such that
$UA=\begin{pmatrix}a&b\\ 0&0\end {pmatrix}$
. Thus,
$UAU^{-1}=\begin {pmatrix}x&y\\ 0&0\end {pmatrix}$
for some
$x,y\in D$
. If
$y\not = 0$
, then
$\begin {pmatrix}x&y\\ 0&0\end {pmatrix}=\begin {pmatrix}0&y\\ 0&0\end {pmatrix}\begin {pmatrix}0&0\\ y^{-1}x&1\end {pmatrix}$
is a product of a sqaure-zero matrix and an idempotent. If
$y= 0$
, then
$\begin {pmatrix}x&y\\ 0&0\end {pmatrix}=\begin {pmatrix}0&x\\ 0&0\end {pmatrix}\begin {pmatrix}1&0\\ 1&0\end {pmatrix}$
is a product of a square-zero matrix and an idempotent. Therefore, in
${\mathbb M}_2(D)$
,
$UAU^{-1}$
is a product of a square-zero matrix and an idempotent, and so is A. Hence,
${\mathbb M}_2(D)$
is a dual nil-clean ring.
$(\Rightarrow )$
. First assume that R is an abelian ring. Let
$a\in R$
be a nonunit, and let
$x\in aR$
. Since R is abelian, x is a nonunit. Write
$x=be$
where
$b\in {\textrm {nil}}(R)$
and
$e^2=e\in R$
. So
$x^n=b^ne$
for all
$n\ge 1$
. As b is nilpotent, x is nilpotent. Thus,
$aR$
is nil and hence
$a\in J(R)$
. It follows that R is local with
$J(R)$
nil.
Next assume that R is not abelian. Then R has a noncentral idempotent e. With
$e'=1-e$
, we show:
-
(1) There exist
$x_0\in eRe'$
and
$y_0\in e'Re$
such that
$x_0y_0=e$
. -
(2) Whenever
$xy=e$
,
$x\in eRe'$
and
$y\in e'Re$
, we have
$yx=e'$
.
Proof of (1). The Peirce decomposition of R with respect to e gives
$R=\begin {pmatrix}eRe&eRe'\\ e'Re&e'Re'\end {pmatrix}$
. Let
$A:=\begin {pmatrix}e&0\\ 0&0\end {pmatrix}$
and write
$A=BE$
where
$B=(b_{ij})$
is a nilpotent and
$E=(e_{ij})$
is an idempotent. Then
$A=AE$
and it follows that
$e_{11}=e$
and
$e_{12}=0$
. From
$A=BE$
it follows that
$b_{11}=e-b_{12}e_{21}$
,
$b_{21}=-b_{22}e_{21}$
. Thus,
$B=\begin {pmatrix}e-b_{12}e_{21}&b_{12}\\ -b_{22}e_{21}&b_{22}\end {pmatrix}$
, so
$1-B=\begin {pmatrix}b_{12}e_{21}&-b_{12}\\ b_{22}e_{21}&e'-b_{22}\end {pmatrix}$
. Hence,
$C:=(1-B)\begin {pmatrix}e&0\\ e_{21}&e'\end {pmatrix}=\begin {pmatrix}0&-b_{12}\\ e_{21}&e'-b_{22}\end {pmatrix}$
, which is an invertible matrix with inverse, say
$Y:=(y_{ij})$
. So,
$1=YC$
. That is,
$$ \begin{align*} \begin{pmatrix}e&\!0\\ 0&\!e'\end{pmatrix}=\begin{pmatrix}y_{12}e_{21}&\!-y_{11}b_{12}+y_{12}(e'-b_{22})\\ y_{22}e_{21}&\!-y_{21}b_{12}+y_{22}(e'-b_{22})\end{pmatrix}. \end{align*} $$
Thus,
$x_0y_0=e$
where
$x_0=y_{12}\in eRe'$
and
$y_0=e_{21}\in e'Re$
.
Proof of (2). Suppose that
$xy=e$
,
$x\in eRe'$
, and
$y\in e'Re$
. By
$(1)$
, with e replaced by
$e'$
we have
$y'x'=e'$
where
$x'\in eRe'$
and
$y'\in e'Re$
. Let
$U=\begin {pmatrix}0&x'\\ y&0\end {pmatrix}$
. If
$UX=0$
where
$X=(x_{ij})\in R$
, then
$0=\begin {pmatrix}0&x'\\ y&0\end {pmatrix}\begin {pmatrix}x_{11}&x_{12}\\ x_{21}&x_{22}\end {pmatrix}=\begin {pmatrix}x'x_{21}&x'x_{22}\\ yx_{11}&yx_{12}\end {pmatrix}$
, so
$$ \begin{align*} x'x_{21}=0,\,\,\,x'x_{22}=0,\,\,\, yx_{11}=0,\,\,\,yx_{12}=0. \end{align*} $$
Thus,
$x_{11}=ex_{11}=xyx_{11}=0$
,
$x_{12}=ex_{12}=xyx_{12}=0$
,
$x_{21}=e'x_{21}=y'x'x_{21}=0$
, and
$x_{22}=e'x_{22}=y'x'x_{22}=0$
. So the right annihilator of U in R is zero. Hence,
$U\in R$
is a unit by Lemma 2.1. Let
$V=(v_{ij})$
be the inverse of U. Then
$$\begin{align*}UV=\begin{pmatrix}0&\!x'\\ y&\!0\end{pmatrix}\begin{pmatrix}v_{11}&\!v_{12}\\ v_{21}&\!v_{22}\end{pmatrix}=\begin{pmatrix}x'v_{21}&\!x'v_{22}\\ yv_{11}&\!yv_{12}\end{pmatrix},\end{align*}$$
so
$e'=yv_{12}$
. But we have
$v_{12}=ev_{12}=xyv_{12}=xe'=x$
, and hence
$yx=e'$
.
Next we show that R is a
$2\times 2$
matrix ring over a division ring. Consider the nonunit
$A:=\begin {pmatrix}e&0\\ 0&0\end {pmatrix}\in R$
and write
$A=BE$
where
$B=(b_{ij})\in R$
is nilpotent and
$E=(e_{ij})\in R$
is an idempotent. Then
$A=AE$
and it follows that
$E=\begin {pmatrix}e&0\\ e_{21}&e_{22}\end {pmatrix}$
. From
$A=BE$
it follows that
$B=\begin {pmatrix}e-b_{12}e_{21}&b_{12}\\ -b_{22}e_{21}&b_{22} \end {pmatrix}$
, so
$1-B=\begin {pmatrix}b_{12}e_{21}&-b_{12}\\ b_{22}e_{21}&e'-b_{22}\end {pmatrix}$
. Hence,
$$\begin{align*}C:=(1-B)\begin{pmatrix}e&0\\ e_{21}&e'\end{pmatrix}=\begin{pmatrix}0&-b_{12}\\ e_{21}&e'-b_{22}\end{pmatrix},\end{align*}$$
which is an invertible matrix with inverse, say
$Y:=(y_{ij})$
. Thus,
$1=YC$
. That is,
$$ \begin{align*} \begin{pmatrix}e&0\\ 0&e'\end{pmatrix}=\begin{pmatrix}y_{12}e_{21}&-y_{11}b_{12}+y_{12}(e'-b_{22})\\ y_{22}e_{21}&-y_{21}b_{12}+y_{22}(e'-b_{22})\end{pmatrix}. \end{align*} $$
It follows that
$y_{12}e_{21}=e$
. By
$(2)$
,
$e_{21}y_{12}=e'$
. Therefore,
$1=e+e'=y_{12}e_{21}+e_{21}y_{12}$
with
$e_{21}^{2}=0$
. So, by Lemma 2.2, R is a
$2\times 2$
matrix ring. Write
$R={\mathbb M}_2(S)$
for some ring S. We verify that S is a division ring. If
$x\in S$
is not a unit, then
$A:=\begin {pmatrix}1&0\\ 0&x\end {pmatrix}\in {\mathbb M_2(S)}$
is not a unit, so it is dual nil-clean in R. Write
$A=BE$
, where
$B=(b_{ij})\in {\mathbb M}_2(S)$
is nilpotent and
$E=(e_{ij})\in {\mathbb M}_2(S)$
is an idempotent. We have
$A=AE=\begin {pmatrix}e_{11}&e_{12}\\ xe_{21}&xe_{22}\end {pmatrix}$
, so
$$ \begin{align} e_{11}=1,\,\,\,e_{12}=0,\,\,\, xe_{21}=0\,\,\,{\text{and}}\,\,\, x=xe_{22}. \end{align} $$
Thus,
$E=\begin {pmatrix}1&0\\ e_{21}&e_{22}\end {pmatrix}$
. From
$A=BE$
, we have
$\begin {pmatrix}1&0\\ 0&x\end {pmatrix}=\begin {pmatrix}b_{11}+b_{12}e_{21}&b_{12}e_{22}\\ b_{21}+b_{22}e_{21}&b_{22}e_{22}\end {pmatrix}$
, so
$b_{11}=1-b_{12}e_{21}$
and
$b_{21}=-b_{22}e_{21}$
. Thus,
$B=\begin {pmatrix}1-b_{12}e_{21}&b_{12}\\ -b_{22}e_{21}&b_{22}\end {pmatrix}$
. As B is nilpotent,
$I_2-B$
is invertible. So
$(I_2-B)\begin {pmatrix}1&0\\ e_{21}&1\end {pmatrix}=\begin {pmatrix}0&-b_{12}\\ e_{21}&1-b_{22}\end {pmatrix}$
is invertible. It follows that
$e_{21}\in U(R)$
. So, by (2.1),
$x=0$
. Therefore, S is a division ring. ▪
Corollary 2.4 Let
$n\ge 2$
be a fixed integer. The following are equivalent for a ring R:
-
(1) For each nonunit
$a\in R$
,
$a=be$
where
$b^n=0$
and
$e^2=e$
. -
(2) R is either a local ring with
$j^n=0$
for all
$j\in J(R)$
or the
$2\times 2$
matrix ring over a division ring.
Proof
$(1)\Rightarrow (2)$
. Assume that R is not the
$2\times 2$
matrix ring over a division ring. Then, by Theorem 2.3, R is a local ring. For
$j\in J(R)$
,
$j=be$
where
$b^n=0$
and
$e^2=e$
. As R is local,
$e=0$
or
$e=1$
. It follows that
$j^n=0$
.
$(2)\Rightarrow (1)$
. We may assume that
$R={\mathbb M}_2(D)$
where D is a division ring. Let
$A\in {\mathbb M}_2(R)$
be a nonunit. Then, by Gaussian elimination, there is a unit
$U\in {\mathbb M}_2(R)$
such that
$UA=\begin {pmatrix}a&b\\ 0&0\end {pmatrix}$
. Thus,
$UAU^{-1}=\begin {pmatrix}x&y\\ 0&0\end {pmatrix}$
for some
$x,y\in R$
. If
$y\not = 0$
, then
$\begin {pmatrix}x&y\\ 0&0\end {pmatrix}=\begin {pmatrix}0&y\\ 0&0\end {pmatrix}\begin {pmatrix}0&0\\ y^{-1}x&1\end {pmatrix}$
is a product of a sqaure-zero matrix and an idempotent. If
$y= 0$
, then
$\begin {pmatrix}x&y\\ 0&0\end {pmatrix}=\begin {pmatrix}0&x\\ 0&0\end {pmatrix}\begin {pmatrix}1&0\\ 1&0\end {pmatrix}$
is a product of a square-zero matrix and an idempotent. Therefore, in
${\mathbb M}_2(R)$
,
$UAU^{-1}$
is a product of a square-zero matrix and an idempotent, and so is A. ▪
By Theorem 2.3, for a ring R, every element of R is a product of a nilpotent and an idempotent if and only if every element of R is a product of an idempotent and a nilpotent. We end with an example of an element a in a ring R such that
$a=be$
where b is nilpotent and
$e^2=e$
, but
$a\not = fc$
for any nilpotent c and any idempotent f in R.
Example 2.5 Let
$R=\begin {pmatrix}\mathbb Z&\mathbb Z\\ 4\mathbb Z&\mathbb Z\end {pmatrix}$
and
$A=\begin {pmatrix}-4&-2\\ 0&0\end {pmatrix}$
. We see that
$A=\\ \begin {pmatrix}1&0\\ 0&0\end {pmatrix}\begin {pmatrix}-4&-2\\ 8&4\end {pmatrix}$
, a product of an idempotent and a nilpotent. Assume that
$A=BE$
where
$B\in R$
is nilpotent and
$E^2=E\in R$
. It is clear that E can not be trivial, so
$E=\begin {pmatrix}a&b\\ c&1-a\end {pmatrix}$
where
$bc=a-a^2$
(see [Reference Chen, Yang and Zhou4, Lemma 1.5]). Thus,
$A=AE=\begin {pmatrix}-4a-2c&-2+2a-4b\\ 0&0\end {pmatrix}$
, and it follows that
$-4a-2c=-4$
and
$-2+2a-4b=-2$
. That is,
$a=2b$
and
$c=2-4b$
. As
$c\in 4\mathbb Z$
, we deduce that
$2=4b+c$
is divided by
$4$
. This is a contradiction.
Acknowledgment
This research was supported by a Discovery Grant (RGPIN-2016-04706) from NSERC of Canada.
