Group ring and left zero divisor

Question

Let $K$ be a finite field and $G$ be a discrete group.

Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?

It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields?

Answer 1

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.

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.

(Thanks to Greg Marks:) The classification of finite group rings over fields where ab=0 implies ba=0 is given in:

Gutan, Marin; Kisielewicz, Andrzej. "Reversible group rings." J. Algebra 279 (2004), no. 1, 280–291. MR2078399 DOI:j.jalgebra.2004.02.011.

In particular, K is a field of order 2^2n-1 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 MR2372321.

Answer 2

The condition $ab=0 \Rightarrow ba=0$ defines what are often called reversible rings, which, for example, have the property that the set of nilpotent elements is an ideal that coincides with the prime radical.  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.  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 $8$.  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 $ba \neq 0$.

Answer 3

Using the miniscule amounts of representation theory I know, you can construct an example as follows.

Let $G=S_3=D_6=\left< r,s\mid r^3=s^2=srsr=e \right>$ (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 $\rho(r)(sv)=\omega^2 sv$.

Hence, we have $\rho(e)=\left[\begin{matrix}1&0\newline 0&1\end{matrix}\right]$, $\rho(r)=\left[\begin{matrix}\omega&0\newline 0&\omega^2\end{matrix}\right]$, $\rho(r^2)=\left[\begin{matrix}\omega^2&0\newline 0&\omega\end{matrix}\right]$, $\rho(s)=\left[\begin{matrix}0&1\newline 1&0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&\omega^2\newline \omega&0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&\omega\newline \omega^2&0\end{matrix}\right]$.

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 $\rho_W\colon G\to Aut(W)$.

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 $End_{\mathbb K}(V)$.

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 $ba\neq 0$.

Answer 4

If $G$ is torsion-free then the question of reversibility of $K[G]$ (that is, does $ab = 0$ imply $ba = 0$) is in fact equivalent to the zero divisor conjecture, for any field $K$.

Connell showed that such a $K[G]$ is prime, so given non-zero $a, b \in K[G]$ there exists $c \in K[G]$ such that $bca \neq 0$. If it were the case that $ab = 0$ then we have $a (bc) = 0$ but $(bc) a \neq 0$ and $K[G]$ is not reversible. This trick is how you show that the unit conjecture implies the zero divisor conjecture.

(In the other direction, if a ring has no zero divisors then reversibility holds trivially.)

Connell, I. G., On the group ring, Can. J. Math. 15, 650-685 (1963). ZBL0121.03502.

Answer 5

Somewhat belatedly, here is a class of examples. Let $G$ be a nontrivial free product of two groups (nontrivial means excluding $G = Z_2 * Z_2$). Suppose that $KG$ has a zero divisor (e.g., if at least one of the groups in $G = G_2 * G_2$ contains a torsion element). Then $KG$ has a right zero divisor that is not a left zero divisor. This is a consequence of Left and right zero divisors in group algebras by D Handelman (me) and J Lawrence, Bulletin of the Australian Mathematical Society (1976) 15 (3), 453-454. This is independent of the choice of field.