\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{Mahboubeh Sanaei, Shervin Sahebi and Hamid H. S. Javadi} \fancyhead[CO]{J-Armendariz rings} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\theorem@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} {\noindent Journal of Mathematical Extension \\ Vol. XX, No. XX, (2014), pp-pp (Will be inserted by layout editor)}\\ ISSN: 1735-8299\\ URL: http://www.ijmex.com\\ \vspace*{9mm} \begin{center} {\Large \bf J-Armendariz rings\\} %{\bf Do You Have a Subtitle? \\ If so, Write It Here} \let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)} {\bf Mahboubeh Sanaei}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Islamic Azad University, Central Tehran Branch, 13185/768, Iran,} \vspace{2mm} {\bf Shervin Sahebi $^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Islamic Azad University, Central Tehran Branch, 13185/768, Iran,} \vspace{2mm} {\bf Hamid H. S. Javadi }\\ \vspace{2mm} {\small Department of Mathematics and Computer Science, Shahed University, Tehran, Iran,} \vspace{2mm} \end{center} \vspace{4mm} {\footnotesize \begin{quotation} {\noindent \bf Abstract.} We introduce the notion of J-Armendariz rings, which are a generalization of weak Armendariz rings and investigate their properties. We show that local rings are J-Armendariz. Also, we prove that a ring R is J-Armendariz if and only if $ R[[x]] $ is J-Armendariz. It is shown that the J-Armendariz property is not Morita invariant. As a specific case, we show that the class of J-Armendariz rings lies properly between the class of one-sided quasi-duo rings and the class of perspective rings. \end{quotation} \begin{quotation} \noindent{\bf Mathematics Subject Classification 2000:} 16U20; 16S36; 16W20 \noindent{\bf Keywords and Phrases:} Armendariz ring, Weak Armendariz ring, J-Armendariz ring, Perspective ring, Quasi duo-ring \end{quotation}} \section{Introduction} \label{intro} % It is advised to give each section and subsection a unique label. Throughout this article, $ R $ denotes an associative ring with identity. For a ring $R$, $ Nil(R) $, $ M_{n}(R) $, $ T_{n}(R) $ and $e_{ij}$ denote the set of nilpotents elements in $ R $, the $ n\times n $ matrix ring over $ R $, the $n\times n $ upper triangular matrix ring over $ R $ and the matrix with $(i,j)$-entry $1$ and elsewhere $0$, respectively. In 1997, Rege and Chhawchharia introduced the notion of an Armendariz ring. They called a ring $ R $ Armendariz if whenever polynomials $ f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}$ and $ g(x)=b_{0}+b_{1}x+\cdots +b_{m}x^{m} \in R[x]$ satisfy $ f(x)g(x)=0 $ then $ a_{i}b_{j} =0$ for all $ i$ and $j $. The name "Armendariz ring" is chosen because Armendariz \cite[Lemma 1]{EP} proved that reduced rings (that is a ring without nonzero nilpotents) satisfy this condition. A number of properties of Armendariz rings have been studied in \cite{a02, EP, a03, a04, Reg}. So far Armendariz rings are generalized in several forms \cite{alpha, javadi, zhao}. Liu and Zhao \cite{zhao} called a ring $R$ weak Armendariz if whenever polynomials $f(x)=a_{0}+a_{1}x+\cdots + a_{n}x^{n}$, $g(x)=b_{0}+b_{1}x+\cdots + b_{m}x^{m} \in R[x]$ satisfy $f(x)g(x)=0$, then $a_{i}b_{j}\in Nil(R)$ for all $i$ and $j$. The Jacobson radical is an important tool for studying the structure of noncommutative rings, and denoted by $J(R)$. Motivated by the above definitions, we investigate a generalization of weak Armendariz rings. We call a ring $ R $, \textit{J-Armendariz} if whenever polynomials $ f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}$ and $ g(x)=b_{0}+b_{1}x+\cdots +b_{m}x^{m} \in R[x]$ satisfy $ f(x)g(x)=0 $ then $ a_{i}b_{j} \in J(R)$ for all $ i$ and $j $. Clearly, for an artinian ring, weak Armendariz rings and J-Armendariz rings are the same. Although $Nil(R)$ does not always lie in the $J(R)$, we show weak Armendariz rings are J-Armendariz and local rings are J-Armendariz too, but Example \ref{examp1} shows that local rings are not necessarily weak Armendariz. Thus J-Armendariz rings are a proper generalization of weak Armendariz rings. At last we study the relation of J-Armendariz rings with other classes of rings such as: right (left) quasi duo rings, perspective rings, clean rings and strongly $\pi$-regular rings. In \cite{dinesh}, Garg et al., studied the modules whose any two isomorphic summands have a common complement. They called such modules perspective. This property in rings turns out to be left-right symmetric, that is, $ R_{R} $ is perspective if and only if $ _{R}R $ is perspective and they called such ring a perspective ring. We show that a J-Armendariz ring $ R $ is perspective. However there exists a perspective ring which is not J-Armendariz. On the other hand a ring $ R $ is called right (left) quasi-duo if every maximal right (left) ideal of $ R $ is two-sided. We prove that a right (left) quasi-duo ring is J-Armendariz, but there exists a J-Armendariz ring R which is not right (left) quasi-duo. Therefore the class of J-Armendariz rings lies properly between the class of right (left) quasi-duo rings and the class of perspective rings. \section{J-Armendariz property with respect to standard constructions} \label{sec:2} In this section, J-Armendaiz rings are introduced as a generalization of weak Armendariz rings. We study J-Armendariz property with respect to some standard constructions like direct product, factor rings, subrings, matrix rings, corner rings, polynomial rings, etc. \begin{definition}\label{def 1.1} A ring $ R $ is said to be J-Armendariz if for any nonzero polynomials $ f(x)=\sum_{i=0}^{n}a_{i}x^{i}$ and $g(x)= \sum_{j=0}^{m}b_{j}x^{j} \in R[x] , $ $ f(x)g(x)=0 $, implies that $a_{i}b_{j} \in J(R)$ for each $i,j $. \end{definition} We can easily show that weak Armendariz rings are J-Armendariz. For it, let $R$ be weak Armendariz and $ f(x)=\sum_{i=0}^{n}a_{i}x^{i}$ and $g(x)= \sum_{j=0}^{m}b_{j}x^{j} \in R[x]-\{ 0\} $ such that $f(x)g(x)=0$. Hence $rf(x)g(x)=0$ for each $r\in R$ and so $ra_{i}b_{j}\in Nil (R)$ by hypothesis. This implies that $a_{i}b_{j}\in J(R)$, as desired. But Example \ref{examp1} shows that J-Armendariz rings are not necessarily weak Armendariz. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition}\label{proposition1} Let $ R $ be a ring and $ I $ an ideal of $ R $ such that $ R/I $ is J-Armendariz. If $ I\subseteq J(R) $, then $ R $ is J-Armendariz. \end{proposition} \begin{proof} It is clear after applying $J(\frac{R}{I})=\frac{J(R)}{I}$, when $I\subseteq J(R)$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{corollary}\label{corollary1} Let $ R $ be any local ring. Then $ R $ is J-Armendariz. \end{corollary} %%%%%%%%%%%%%%%%%%%%%%%555 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% One may ask whether local rings are weak Armendariz, but the following gives a negative answer. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{example}\label{examp1} Let $ F $ be a field, $R= M_{2}(F) $ and $ R_{1}=R[[t]] $. Consider the ring \begin{center} $ S =\lbrace \sum_{i=0}^{\infty}a_{i}t^{i}\in R_{1}| a_{0}\in kI~ for ~k\in F \rbrace$, \end{center} where $ I $ is the identity matrix. It is obvious that $ S $ is local and so is J-Armendariz by corollary \ref{corollary1}. Now for $ f(x)=e_{11}t-e_{12}tx $ and $ g(x)=e_{21}t +e_{11}tx \in S[x]$, we have $ f(x)g(x)=0 $, but $ (e_{11}t)^{2} $ is not nilpotent in $ S $, and so $ S $ is not weak Armendariz. \end{example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $R_{t}$ be a ring for each $t\in I$. Note that since $\prod_{t\in I}J(R_{t})=J(\prod_{t\in I}R_{t})$, then $\prod_{t\in I}R_{t}$ is J-Armendariz if and only if $R_{t}$ is J-Armendariz, for each $t\in I$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{theorem}\label{theorem4} A ring $ R$ is J-Armendariz, if and only if $ R[[x]] $ is J-Armendariz. \end{theorem} \begin{proof} Let $ R $ be a J-Armendariz ring. Since $ R\cong \frac{R[[x]]}{\langle x\rangle} $, then by proposition \ref{proposition1}, $ R[[x]] $ is J-Armendariz. Conversely, assume $ R[[x]] $ is J-Armendariz, and $ f(y)=\sum_{i=0}^{n}a_{i}y^{i}$ and $g(y)=\sum_{j=0}^{m}b_{j}y^{j} $ are polynomials in $ R[y] $, such that $ f(y)g(y)=0 $. Since $ a_{i}b_{j} \in R\subseteq R[[x]] $ and $ R[[x]] $ is J-Armendariz, then $a_{i}b_{j} \in J(R[[x]])\cap R $. Therefore $ a_{i}b_{j} \in J(R) $, and so $ R $ is J-Armendariz. \end{proof} %%%%%%%%%%%%%%%%%%%%%% The following example shows that the polynomial ring over a J-Armendariz ring need not be J-Armendariz in general and so the subring of a J-Armendariz ring is not necessarily J-Armendariz. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%55 \begin{example}\label{poly} Take $ S $ to be the ring as in Example \ref{examp1}. Then $S[x]$ is not J-Armendariz. For it, let $ f(y)= e_{11}tx-e_{12}txy $ and $ g(y)= e_{21}tx+e_{11}txy $ be polynomials in $ S[x][y] $. Then $ f(y)g(y)=0 $, but $ (e_{11}tx)^{2}$ does not belong to $ J(S[x]) $. \end{example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition}\label{theorem3} Let $R$ be a ring. \item[(1)] If $R[x]$ is J-Armendariz then $R$ is weak Armendariz and so $R$ is J-Armendariz. \item[(2)] If $R$ is a J-Armendariz ring and $J(R)[x] \subseteq J(R[x])$, then $R[x]$ is J-Armendariz. \end{proposition} \begin{proof} (1) Suppose that $ R[x] $ is a J-Armendariz ring. Let $ f(y)=\sum_{i=0} ^{n}a_{i}y^{i}$ and $ g(y)=\sum_{j=0} ^{m}b_{j}y^{j} $ be nonzero plynomials in $ R[y] $, such that $ f(y)g(y)=0 $. By the fact that $J(R[x]) = I[x]$ for some nil ideal $ I $ of $R$ \cite{amit}, $ a_{i}b_{j} \in R \cap I[x] \subseteq Nil(R)$, and so $ R $ is weak Armendariz.\\ (2) Suppose that $ R $ is J-Armendariz and $ J(R )[x] \subseteq J(R[x])$. Let $ F(y)=f_{0}+f_{1}y+\cdots + f_{n}y^{n} $ and $ G(y)=g_{0}+g_{1}y+\cdots + g_{m}y^{m}$ be polynomials in $ R[x][y]$, with $ F(y)G(y)=0 $. We also let $ f_{i}(x)=a_{i_{0}}+a_{i_{1}}x+a_{i_{2}}x^{2}+\cdots + a_{i_{\omega_{i}}}x^{\omega_{i}}$ and $g_{j}(x)=b_{j_{0}}+b_{j_{1}}x+b_{j_{2}}x^{2}+\cdots + b_{j_{\nu_{j}}}x^{\nu_{i}} \in R[x] $ for each $ 0\leq i \leq n $ and $ 0\leq j \leq m $ . Take a positive integer $ t $ that $ t\geq deg(f_{0}(x))+ deg(f_{1}(x))+\cdots + deg(f_{n}(x))+deg(g_{0}(x))+deg(g_{1}(x))+\cdots +deg(g_{m}(x))$, where the degree is as polynomials in $ x$ and the degree of zero polynomial is taken to be $ 0 $. Then $ F(x^{t})=f_{0}+f_{1}x^{t}+\cdots + f_{n}x^{tn} $ and $ G(x^{t})=g_{0}+g_{1}x^{t}+\cdots + g_{m}x^{tm} \in R[x]$ and the set of coefficients of the $ f_{i} $'s (resp. $g_{j}$'s) equals the set of coefficients of the $F(x^{t}) $ (resp. $G(x^{t}) $). Since $ F(y)G(y)=0 $, then $ F(x^{t})G(x^{t}) =0$. So $ a_{is_{i}}b_{jr_{j}}\in J(R) $, where $ 0\leq s_{i}\leq \omega_{i} $, $ 0\leq r_{j}\leq \nu_{j} $. By hypothesis we have $ J(R)[x]\subseteq J(R[x]) $, and so $ f_{i}g_{j}\in J(R[x]) $. It implies that $ R $ is J-Armendariz. \end{proof} %%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Note that, $M_{n}(R)$ is not J-Armendariz for any nonzero ring R and $ n\geq 2 $, i.e. the J-Armendariz property is not Morita invariant. \begin{example}\label{example2} Let $ R $ be a ring and $ S=M_{2}(R) $. If $ f(x)=e_{12}-e_{11}x$ and $ g(x)=e_{11}+e_{12}-(e_{21}+e_{22})x $, then $ f(x)g(x)=0 $. But $ e_{11}(e_{11}+e_{12})=e_{11}+e_{12}$ is not in $ J(S) $. Thus $ S $ is not J-Armendariz. \end{example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{corollary} Every J-Armendariz ring $ R $ is directly finite. \end{corollary} \begin{proof} If $ R $ is not directly finite, then $ R $ contains an infinite set of matrix units $\lbrace e_{11}, e_{12},e_{13},\ldots, e_{21}, e_{22},e_{23},\ldots \rbrace $ by \cite[proposition 5.5]{good}. This is a contradiction by Example \ref{example2}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The next example shows that there exists a J-Armendariz ring $ R $ such that $ R/J(R) $ is not J-Armendariz and so the homomorphic image of J-Armendariz rings need not to be J-Armendariz. \begin{example} Let $ R $ denote the localization of the ring $ \mathbb{Z} $ of integers at the prime ideal $ \langle 3 \rangle $. Consider the quaternions $ \mathbf{Q} $ over $ R $, that is a free $ R $-module with basis $ 1,~i,~j,~k $ and multiplication satisfying $ i^{2}=j^{2}=k^{2}=-1,~ij=k=-ji $. Then $ \mathbf{Q} $ is a noncommutative domain with $ J(\mathbf{Q})=3 \mathbf{Q} $, and so is J-Armendariz. But $ \mathbf{Q}/J(\mathbf{Q}) $ is isomorphic to the $ 2 $-by-$ 2 $ full matrix ring over $ \mathbb{Z}_{3} $ and is not J-Armendariz by Example \ref{example2}. \end{example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%to %\begin{rem} %The converse of proposition \ref{proposition1} is not true in general. %In %Example \ref{poly}, Let %$ I=\langle x^{2}\rangle $ be an ideal in the ring %$ S[[x]] $. %Clearly %$ I\subseteq J(S[[x]]) $, but %$ \frac{S[[x]]}{\langle x^{2}\rangle} $ is not J-Armendariz. %\end{rem} %%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $ R $ and $ S $ be two rings and $ M $ be an $ (R,S) $-bimodule. This means that $ M $ is a left $ R $-module and a right $ S $-module such that $ (rm)s=r(ms) $ for all $ r\in R $, $ m\in M $, and $ s\in S $. Given such a bimodule $ M $ we can form \begin{center} $ T =\bigl( \begin{smallmatrix} R&M\\ 0&S \end{smallmatrix} \bigr)=\big\lbrace \bigl( \begin{smallmatrix}r&m\\ 0&s \end{smallmatrix} \bigr) : r\in R , m\in M , s\in S \big\rbrace$ \end{center} and definition a multiplication on $ T $ by using formal matrix multiplication: \begin{center} $ \bigl( \begin{smallmatrix} r&m\\ 0&s \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} r^{\prime}&m^{\prime}\\ 0&s^{\prime} \end{smallmatrix} \bigr)=\bigl( \begin{smallmatrix} rr^{\prime}&rm^{\prime}+ms^{\prime}\\ 0&ss^{\prime} \end{smallmatrix} \bigr). $ \end{center} This ring construction is called triangular ring $ T $. \begin{proposition}\label{proposition2} Let $ R $ and $ S $ be two rings and $ M $ be an $ (R,S) $-bimodule. Let $ T $ be the triangular ring $ T= \bigl( \begin{smallmatrix} R&M\\ 0&S \end{smallmatrix} \bigr) $. Then the rings $ R $ and $ S $ are J-Armendariz if and only if $ T $ is J-Armendariz. \end{proposition} \begin{proof} Let $ R $ and $ S $ be J-Armendariz. Take $ I=( \begin{smallmatrix} 0&M \\ 0&0 \end{smallmatrix} \bigr) $, therefore $T/I\cong R\times S $ is J-Armendariz and since $I\subseteq J(T)=\bigl( \begin{smallmatrix} J(R) & M \\ 0 & J(S) \end{smallmatrix} \bigr)$, then $T$ is J-Armendariz by proposition \ref{proposition1}. Conversely, let $ T $ be a J-Armendariz ring, $ f_{r}(x) =r_{0}+r_{1}x+\cdots + r_{n}x^{n}$, $ g_{r}(x) =r^{\prime}_{0}+r^{\prime}_{1}x+\cdots + r^{\prime}_{m}x^{m}\in R[x] $, such that $ f_{r}(x)g_{r}(x)=0 $, and $ f_{s}(x) =s_{0}+s_{1}x+\cdots + s_{n}x^{n} $, $ g_{s}(x) =s^{\prime}_{0}+s^{\prime}_{1}x+\cdots + s^{\prime}_{m}x^{m} \in S[x] $, such that $ f_{s}(x)g_{s}(x)=0 $. If \begin{center} $ f(x)=\bigl( \begin{smallmatrix} r_{0}&0\\ 0&s_{0} \end{smallmatrix} \bigr) +\bigl( \begin{smallmatrix} r_{1}&0\\ 0&s_{1} \end{smallmatrix} \bigr)x+\cdots +\bigl( \begin{smallmatrix} r_{n}&0\\ 0&s_{n} \end{smallmatrix} \bigr)x^{n} $ and $ g(x)=\bigl( \begin{smallmatrix} r^{\prime}_{0}&0\\ 0&s^{\prime}_{0} \end{smallmatrix} \bigr) +\bigl( \begin{smallmatrix} r^{\prime}_{1}&0\\ 0&s^{\prime}_{1} \end{smallmatrix} \bigr)x+\cdots +\bigl( \begin{smallmatrix} r^{\prime}_{m}&0\\ 0&s^{\prime}_{m} \end{smallmatrix} \bigr)x^{m} \in T[x]$ \end{center} Then from $ f_{r}(x)g_{r}(x)=0 $ and $ f_{s}(x)g_{s}(x)=0 $ it follows that $ f(x)g(x)=0 $. Since $ T $ is a J-Armendariz ring, $ \bigl( \begin{smallmatrix} r_{i}&0\\ 0&s_{i} \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} r^{\prime}_{j}&0\\ 0&s^{\prime}_{j} \end{smallmatrix} \bigr)\in J (T)=\bigl( \begin{smallmatrix} J(R)&0\\ 0&J(S) \end{smallmatrix} \bigr) $. Thus $ r_{i} r^{\prime}_{j}\in J(R)$ and $ s_{i} s^{\prime}_{j}\in J(S)$ for any $ i,j $. This shows that $ R $ and $ S $ are J-Armendariz. \end{proof} Recall that a ring $ R $ is said to be \textit{abelian} if every idempotent of it is central. Armendariz rings are abelian \cite[Lemma 7]{a04}, but J-Armendariz rings need not to be abelian in general. For example, let $F$ be a field then $ R=T_{2}(F) $ is J-Armendariz by proposition \ref{proposition2}, but it is not an abelian ring. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition}\label{proposition4} Let $ R $ be a J-Armendariz ring. Then for each idempotent $ e $ of $ R $, $ eRe $ is J-Armendariz. The converse holds if $e $ is a central idempotent. \end{proposition} \begin{proof} Let $ f(x)=\sum_{i=0}^{n}a_{i}x^{i} $, $ g(x)=\sum_{j=0}^{m}b_{j}x^{j} \in (eRe)[x] $ be such that $ f(x)g(x)=0 $. Since $ R $ is J-Armendariz and $ a_{i}, b_{j} \in eRe\subseteq R $, then we have $ a_{i}b_{j} \in J(R)\cap eRe= J(eRe) $. This means that $ eRe $ is J-Armendariz. Conversely, let $ eRe $ be a J-Armendariz ring and $ f(x)=\sum_{i=0}^{n}a_{i}x^{i} $, $ g(x)=\sum_{j=0}^{m}b_{j}x^{j} \in R[x] $, such that $ f(x)g(x)=0 $. By the hypothesis, $ 0=ef(x)eg(x)e\in (eRe)[x] $, and since $ eRe $ is J-Armendariz, we have $ a_{i}b_{j}\in J(eRe)=J(R)\cap eRe $. Thus $ R $ is J-Armendariz. \end{proof} \section{The relation of J-Armendariz rings with other classes of rings} Let $ M $ be a module and $ A $, $ B $ be two summands of $ M $. We write $ A\sim B $ to denote $ A $ and $ B $ have a common complement i.e., there exists submodule $ C $ such that $ M=A\oplus C = B\oplus C $. It is clear that $ A\sim B $ implies that $ A\cong B $. A module $ M $ is perspective when $ A\cong B $ implies $ A\sim B $ for any two summands $ A $, $ B $ of $ M $. It is clear that perspective modules satisfy the internal cancellation property in the sense that complements of isomorphic summands are isomorphic (see \cite{khur}). %Recall that a ring $ R $ is called unit regular if for every element $ a\in R $ %there exists an $ u \in U(R)$ such that $ a=aua $. In this section we give a new class of rings that are J-Armendariz. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% A ring $ R $ is called right (left) quasi-duo if every maximal right (left) ideal of $ R $ is two-sided. If $ R $ is a right (left) quasi-duo ring, then $ R/J(R) $ is reduced by \cite[Proposition 4.3]{lam}. So $ R/J(R) $ is Armendariz, and hence $R$ is J-Armendariz by Proposition \ref{proposition1}. So a right (left) quasi-duo ring is J-Armendariz but there exists a J-Armendariz ring R which is not right (left) quasi-duo by Example \ref{quasi}. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In \cite[Corollary 4.8]{dinesh} it is proved that every right (left) quasi-duo ring is a perspective ring. Moreover, in this section we prove that every J-Armendariz ring is perspective. One may ask a perspective ring is J-Armendariz. The general answer is negative and so J-Armendariz rings lie properly between right (left) quasi duo rings and perspective rings. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 The following example shows that J-Armendariz rings need not to be right quasi-duo. \begin{example}\label{quasi} Take any right primitive domain $ R $ that is not a division ring (e.g. the free algebra $ R=Q\langle x,y \rangle $). Then $R $ is J-Armendariz, but $ R $ is not right quasi-duo by \cite[Proposition 4.1]{lam}. \end{example} %%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition}\label{perspective} Let $ R $ be a J-Armendariz ring, then $ R $ is perspective, but the converse is not true in general. \end{proposition} \begin{proof} Let $ R $ be a J-Armendariz ring. Then for $ a, b \in R $ $ ab=0 $ implies $ a Nil(R) B\subseteq J(R) $. In fact, for $ 0\neq c \in Nil(R) $ there exist $ n\geq 1 $ such that $ c^{n}=0 $, and so $ a(1-cx)(1+cx+\cdots + c^{n-1}x^{n-1})b=0 $. This implies that $ acb\in J(R) $. Now taking $ a=e=e^{2} $, $ b=(1-e) $ and $ c=er(1-e) $, then we have $ eR(1-e) \subseteq J(R)$. Thus by \cite[Theorem 4.7]{dinesh}, $ R $ is a perspective ring. However there exists a perspective ring which is not J-Armendariz. Let $ R $ be a field. Then $ M_{n}(R) $ is perspective by \cite[Example 5]{dinesh}. But $ M_{n}(R) $ is not J-Armendariz for $ n\geq 2 $. \end{proof} %%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% \begin{corollary}\label{abeli} Let $ R $ be a J-Armendaiz ring such that idempotents lift modulo $ J(R) $, then $ R/J(R) $ is abelian. \end{corollary} \begin{proof} Let $ \bar{e}^{2}= \bar{e} $ be an idempotent in $ \bar{R}=R/J(R) $. Since idempotents lift modulo $ J(R) $, then for each $ r\in R $, $ e(r-re) \in J(R) $ and $ (r-er)e \in J(R) $ by the proof of Proposition \ref{perspective}. Therefore $ R/J(R) $ is abelian. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Following \cite{niki}, we define an element $ x $ of a ring $ R $ to be clean if there is an idempotent $e\in R$ such that $x-e$ is a unit of $ R $. A clean ring is defined to be one in which every element is clean. Clean rings were initially developed in \cite{niki} as a natural class of rings which have the exchange property. A ring $ R $ is an exchange ring if for every right $ R $-module $ A_{R} $ and two decompositions $ A_{R}= M \bigoplus N=\underset{i\in I} {\bigoplus}A_{i}$ where $ M_{R}\cong A_{R} $, and the index set I is finite, there exist submodules $ A^{\prime}_{i}\subseteq A_{i} $ such that $ A= M\oplus(\underset{i\in I} {\bigoplus}A^{\prime}_{i})$. A ring $ R $ is an exchange ring if and only if for any $ x\in R $ there exists an idempotent $ e\in R $ such that $ (1-e)\in R(1-x) $ (cf. \cite{exch}). %%%%%%%%%%%%%%%%%%%%%%%%%%%%55 It is known \cite[Proposition 1.8]{niki} that clean rings are exchange and the two concepts are equivalent for abelian rings. A ring $ R $ is said to have stable range one provided that for any $ a,b \in R $, $ aR+bR=R $ implies that there exists some $ y\in R $ such that $ a+by$ is unit in $R$. Now we have the following: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition} Let $ R $ be a an exchange ring. If $ R $ is a J-Armendariz ring then $ R $ is clean with stable range one. \end{proposition} \begin{proof} Let $ R $ be a J-Armendariz and exchange ring. In fact $ R $ is an exchange ring if and only if $ R/J(R) $ is an exchange ring and idempotents can be lifted modulo $ J(R) $ \cite{niki}. Then $ R/J(R) $ is abelian by Corollary \ref{abeli}. Therefore $ R/J(R) $ is clean and so $ R $ is clean by\cite[Proposition 6]{han}. Clearly $ R/J(R) $ has stable rang one by\cite[Theorem 6]{hp}. Hence $ R $ has stable rang one by \cite[Theorem 22]{ln}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 Following \cite{ diesl}, an element $a\in R$ is called strongly $\pi$-regular if $a^n \in Ra^{n+1} \cap a^{n+1}R$ for some positive integer $n$. Also, an element $ r $ in a ring $ R $ is called nil clean if there is an idempotent $ e\in R $ and a nilpotent $ b\in R $ such that $ r=e+b $. The element $r$ is further called strongly nil clean if such an idempotent and nilpotent can be chosen such that $be=eb$. A ring is called nil clean (respectively, strongly nil clean) if every one of its elements is nil clean (respectively, strongly nil clean). In \cite{diesl}, it is shown that every strongly nil clean ring is strongly $\pi$-regular. Now we have the following: %%%%%%%%%%%%%%%%%%%%%%%%% \begin{proposition} Let $ R $ a nil clean ring. If $R$ is J-Armendariz and J-adically complete, then $ R $ is strongly $ \pi $-regular. \end{proposition} \begin{proof} Let $ \bar{R}=R/J(R) $. Since $R$ is J-adically complete, then idempotents lift modulo $J(R)$ by \cite[Theorem 21.31]{L}. Therefore $\bar R$ is abelian by Proposition \ref{abeli}. On the other hand, since $R$ is nil clean, then $\bar R$ is nil clean by \cite[Corollary 3.17]{diesl}. Therefore $ \bar{R} $ is strongly nil clean. Suppose that $a \in R$, then for each $ \bar{a} \in \bar{R} $, we may write $ \bar{a}=\bar{e}+\bar{b} $ for some idempotent $ \bar{e} $ and some nilpotent $ \bar{b} $ which commute. By \cite[Proposition 3.5]{diesl}, $ \bar{a}=(1-\bar{e})+(2\bar{e}-1+\bar{b}) $ is thus strongly $ \pi $-regular decomposition of $\bar a$. Following \cite[Corollary 6]{diesl2} $a$ is strongly $\pi$-regular in $ R $ and the proof is complete. \end{proof} \vspace{4mm}\noindent{\bf Acknowledgements}\\ \noindent This paper is supported by Islamic Azad University Central Tehran Branch (IAUCTB). The authors want to thank the authority of IAUCTB for their support to complete this research. Also, we are grateful to Professor Weixing Chen for many useful suggestions during this work. % BibTeX users please use one of %\bibliographystyle{spbasic} % basic style, author-year citations %\bibliographystyle{spmpsci} % mathematics and physical sciences %\bibliographystyle{spphys} % APS-like style for physics %\bibliography{} % name your BibTeX data base % Non-BibTeX users please use \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. \bibitem{amit} S. A. Amitsur, Radicals of polynomial rings, {\it Canad. J. Math.} {\bf 8} (1956), 355-361. \bibitem{a02} D. D. Anderson, V. Camillo, Armendariz rings and Gaussian rings, {\it Comm. Algebra} {\bf 26} (1998), no. 7, 2265-2272. \bibitem{EP} E. P. Armendariz, A Note on Extensions of Baer and p.p-rings, {\it J. Aust. Math. Soc.} {\bf 18} (1974), 470-473. \bibitem{diesl} A. J. Diesl, Nil clean rings, {\it J. 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Javadi } \noindent Department of Mathematics and Computer Science \noindent Associate Professor of Mathematics \noindent Shahed University \noindent Tehran, Iran \noindent E-mail: h.s.javadi@shahed.ac.ir.}\\ \end{document}