Group ring and left zero divisor. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:19:13Z http://mathoverflow.net/feeds/question/58267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58267/group-ring-and-left-zero-divisor Group ring and left zero divisor. Kate Juschenko 2011-03-12T15:00:35Z 2011-03-13T09:40:38Z <p>Let $K$ be a finite field and $G$ be a discrete group. </p> <blockquote> <p>Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?</p> </blockquote> <p>It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields?</p> http://mathoverflow.net/questions/58267/group-ring-and-left-zero-divisor/58272#58272 Answer by Jack Schmidt for Group ring and left zero divisor. Jack Schmidt 2011-03-12T17:04:27Z 2011-03-13T09:40:38Z <p>Let G be non-abelian of order 6, with x of order 2 and y of order 3. In such a group yxy = x, since both x and xy have order 2. Let K be a field with 2 elements. Then (x+y)⋅(1+xy) = x+y + y+yxy = x+y + y+x = 0, but (1+xy)⋅(x+y) = x+y + xyx+xyy = x+y + yy + xyy ≠ 0.</p> <p>You may be thinking of the property: if a⋅b = 0 then there is some non-zero c such that c⋅a = 0. This holds in all (two-sided) Artinian rings (because elements are either units or zero-divisors). I believe this is true for two-sided self-injective rings as well. I don't know if it is possessed by group rings of infinite groups over finite fields.</p> <p>(Thanks to Greg Marks:) The classification of finite group rings over fields where ab=0 implies ba=0 is given in:</p> <blockquote> <p>Gutan, Marin; Kisielewicz, Andrzej. "Reversible group rings." J. Algebra <strong>279</strong> (2004), no. 1, 280–291. MR<a href="http://www.ams.org/mathscinet-getitem?mr=2078399" rel="nofollow">2078399</a> DOI:<a href="http://dx.doi.org/10.1016/j.jalgebra.2004.02.011" rel="nofollow">j.jalgebra.2004.02.011</a>.</p> </blockquote> <p>In particular, K is a field of order 2<sup>2n-1</sup> and G is the quaternion group of order 8, or G is abelian. Li and Parmenter (2007) extend this to finite group rings over commutative rings with 1 in MR<a href="http://www.ams.org/mathscinet-getitem?mr=2372321" rel="nofollow">2372321</a>.</p> http://mathoverflow.net/questions/58267/group-ring-and-left-zero-divisor/58278#58278 Answer by Vladimir Sotirov for Group ring and left zero divisor. Vladimir Sotirov 2011-03-12T18:23:22Z 2011-03-12T18:23:22Z <p>Using the miniscule amounts of representation theory I know, you can construct an example as follows.</p> <p>Let $G=S_3=D_6=\left&lt; r,s\mid r^3=s^2=srsr=e \right>$ (rotation $r$ and flip $s$), and let $K=\mathbb F_5[\omega]$ where $\omega^3=1$ (finite field!). Then an irreducible representation of $G$ is given by the $2$-dimensional vector space $V$ with basis ${v, sv}$ and action of $G$ on $V$ $\rho\colon G\to Aut(V)$ generated by by $\rho(s)(v)=sv$, $\rho(s)(sv)=v$, $\rho(r)v=\omega v$ and $\rho(r)(sv)=\omega^2 sv$.</p> <p>Hence, we have $\rho(e)=\left[\begin{matrix}1&amp;0\newline 0&amp;1\end{matrix}\right]$, $\rho(r)=\left[\begin{matrix}\omega&amp;0\newline 0&amp;\omega^2\end{matrix}\right]$, $\rho(r^2)=\left[\begin{matrix}\omega^2&amp;0\newline 0&amp;\omega\end{matrix}\right]$, $\rho(s)=\left[\begin{matrix}0&amp;1\newline 1&amp;0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&amp;\omega^2\newline \omega&amp;0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&amp;\omega\newline \omega^2&amp;0\end{matrix}\right]$.</p> <p>The first four matrices are linearly independent over $\mathbb K$ and hence generate $End_{\mathbb K}(V)$ which is $4$-dimensional. Now we know that if $\bar K$ is the algebraic closure of $K$, then $\bar K[G]=\bigoplus_W End_{\bar{\mathbb K}}(W)$ where $W$ are irreducible representations over $\bar K$. Character theory tells us that the projection of $\bar K[G]$ onto $End_{\bar {\mathbb K}}(W)$ is given by left-multiplication by the idempotent $\frac{\dim W}{|G|}\sum_{g\in G}\chi_W(g^{-1})g$ where $\chi_W$ is the trace of $\rho_W\colon G\to Aut(W)$.</p> <p>So in particular we have the idempotent $\phi=\frac13(2e-r-r^2)\in D_6$, which projects $\bar K[G]$ onto $End_{\bar{\mathbb K}}(V)$, and hence must project $K[G]$ into $End_{\mathbb K}(V)$. But we see that $\phi(e)=\frac13(2e-r-r^2)$, $\phi(r)=\frac13(-e+2r-r^2)$, $\phi(r^2)=\frac13(-e-r+2r^2)$, and $\phi(s)=\frac13(2s-rs-r^2s)$ are four linearly independent elements in the image of $\psi$ and thus span $End_{\mathbb K}(V)$.</p> <p>So take your favorite $2\times 2$ matrices $A$ and $B$ (over $K)$ such that $AB=0$ but $BA\neq 0$, find $a_g$ and $b_g$ such that $A=\sum_{g\in {e,r,r^2,s}} a_g\rho(g)$ and $B=\sum_{g\in {e,r,r^2,s}} b_g\rho(g)$, and then $a=\sum_{g\in {e,r,r^2,s}}a_g\phi(g)$ and $b=\sum_{g\in {e,r,r^2,s}}b_g\phi(g)$ will be such that $ab=0$ but $ba\neq 0$.</p> http://mathoverflow.net/questions/58267/group-ring-and-left-zero-divisor/58297#58297 Answer by Greg Marks for Group ring and left zero divisor. Greg Marks 2011-03-12T22:39:46Z 2011-03-12T22:59:54Z <p>The condition $ab=0 \Rightarrow ba=0$ defines what are often called <i>reversible</i> rings, which, for example, have the property that the set of nilpotent elements is an ideal that coincides with the prime radical.&#160; A full matrix ring can't have this property, so you can construct counterexamples by taking any finite field $K$ and nonabelian group $G$ to which Maschke's theorem applies.&#160; An alternative example of a non-reversible group algebra is $K[G]$ where $K$ is the field of two elements and $G$ is the dihedral group of order $8$.&#160; Here the set of nilpotent elements does coincide with the prime radical (the ring is local artinian), but one can find elements $a,b \in K[G]$ with $ab=0$ but $ba \neq 0$.</p>