<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE ArticleSet PUBLIC "-//NLM//DTD PubMed 2.6//EN" "http://www.ncbi.nlm.nih.gov/corehtml/query/static/PubMed.dtd">
<ArticleSet>
<Article>
<Journal>
				<PublisherName>Iranian Mathematical Society (IMS)</PublisherName>
				<JournalTitle>Bulletin of the Iranian Mathematical Society</JournalTitle>
				<Issn>1017-060X</Issn>
				<Volume>38</Volume>
				<Issue>2</Issue>
				<PubDate PubStatus="epublish">
					<Year>2012</Year>
					<Month>07</Month>
					<Day>15</Day>
				</PubDate>
			</Journal>
<ArticleTitle>Rings with a setwise  polynomial-like condition</ArticleTitle><FirstPage>305</FirstPage>
			<LastPage>311</LastPage>
			<Language>en</Language>
<AuthorList>
<Author>
					<FirstName>Ali </FirstName>
					<LastName>Tavakoli</LastName>
					<Affiliation>Islamic Azad University - Majlesi Branch</Affiliation>
				</Author>
<Author>
					<FirstName>Alireza </FirstName>
					<LastName>Abdollahi</LastName>
					<Affiliation>University of Isfahan</Affiliation>
				</Author>
<Author>
					<FirstName>Howard E. </FirstName>
					<LastName>Bell</LastName>
					<Affiliation>Brock University</Affiliation>
				</Author>
</AuthorList>
			<History>
				<PubDate PubStatus="received">
					<Year>2010</Year>
					<Month>04</Month>
					<Day>20</Day>
				</PubDate>
			</History>
		<Abstract><![CDATA[Let $R$ be an infinite ring. Here we prove that if $0_R$ belongs to ${x_1x_2cdots x_n ;|; x_1,x_2,dots,x_nin X}$ for every infinite subset $X$ of $R$, then $R$ satisfies the polynomial identity $x^n=0$. Also we prove that  if $0_R$ belongs to ${x_1x_2cdots x_n-x_{n+1} ;|; x_1,x_2,dots,x_n,x_{n+1}in X}$ for every infinite subset $X$ of $R$, then $x^n=x$ for all $xin R$.]]></Abstract>
		<ObjectList>
			<Object Type="keyword">
			<Param Name="value">Primitive rings</Param>
			</Object>
			<Object Type="keyword">
			<Param Name="value">Polynomial identities</Param>
			</Object>
			<Object Type="keyword">
			<Param Name="value">Combinatorial conditions</Param>
			</Object>
		</ObjectList>
</Article>
</ArticleSet>