Group ring and left zero divisor II - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:14:04Z http://mathoverflow.net/feeds/question/63619 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63619/group-ring-and-left-zero-divisor-ii Group ring and left zero divisor II Kate Juschenko 2011-05-01T16:32:32Z 2011-05-01T17:30:47Z <p>Let $K$ be a finite field and $G$ be a discrete group. </p> <blockquote> <p>Is it true that for every $a=e+a_1+\ldots+a_n,b=e+b_1+\ldots+b_m\in K[G]$ with $b_i\neq e,a_j\neq e$ the condition $ab=0$ implies $ba=0$?</p> </blockquote> <p>It is related to <a href="http://mathoverflow.net/questions/58267/group-ring-and-left-zero-divisor" rel="nofollow">this</a> question. The examples proposed there do not violate the condition above.</p> http://mathoverflow.net/questions/63619/group-ring-and-left-zero-divisor-ii/63623#63623 Answer by Tom Goodwillie for Group ring and left zero divisor II Tom Goodwillie 2011-05-01T17:30:47Z 2011-05-01T17:30:47Z <p>I hope I understand the question right: the $a_i$ are to be distinct elements of $G$, as are the $b_j$?</p> <p>If so, then the answer is no. Let $K$ be $\mathbb Z/2$ and suppose that $x$ and $y$ are elements of $G$ such that $x^2=e$ and $xy$ is not equal to $yx$. Let $a=e+x$ and let $b=(e+x)(e+y)=e+x+y+xy$. Then $ab=0$ but $ba=y+xy+yx+xyx$ is not $0$.</p>