Zero divisor conjecture for finite fields

Question

I believe that the most attractive "zero-divisor" conjecture is the existence of non-trivial zero-divisors in a group ring $\mathbb{C}[G]$ for a torsion free group $G$. For the sake of knowledge let me ask if the zero conjecture is known for finite fields instead of $\mathbb{C}$. More precisely:

• Let $G$ be a torsion-free group and $K$ be a finite field. Is it true that the group ring $K[G]$ has no non-trivial zero divisors?

In particular:

• Let $K$ be the field with two elements, $G$ be a torsion free group and let $rank(a)$ be the smallest number of elements in the expression of $a$ in the sum $a=\sum g_i$, $g_i\in G$. Is there a constant $R>0$ such that we know for sure that $rank(a)>R$ for every zero-divisor $a\in K[G]$?

Here is related question:

Group ring and left zero divisor.

Answer 1

@Kate: I believe that the answer to the first question is unknown and the question is considered as complicated for the field $\mathbb{F}_2$ as for $\mathbb{C}$. I do not know any reduction from the case of one field to the case of another field, though. There were several attempts to disprove Kaplansky conjecture for $\mathbb{F}_2$ but there are no promising ideas.

Answer 2

The zero-divisor conjecture over $\mathbb Q$ is equivalent to the zero-divisor conjecture over the ring $\mathbb Z$ by clearing denominators. At the same time, the zero-divisor conjecture over $\mathbb Z$ is implied by the zero-divisor conjecture over $\mathbb Z/p \mathbb Z$ for all primes $p$. Indeed, if $a,b \in \mathbb Z[G]$ are non-zero and $ab =0$, then we may assume that $a$ and $b$ are not divisible by $p$ (otherwise we divide by $p$). Hence, $\bar a,\bar b$ are non-zero in $(\mathbb Z/p\mathbb Z)[G]$.

Hence, the case $\mathbb Q$ is implied by the cases $\mathbb Z/p \mathbb Z$ for every individual prime $p$.

I believe that a similar argument shows that the case $\mathbb C$ is equivalent to the case $\overline {\mathbb Q}$ (by Hilbert's Nullstellensatz), is equivalent to the case $\mathcal O$ (the ring of algebraic integers), is implied by the case of many finite fields.

An easy way to see that the case $\mathbb C$ is implied by the cases for many finite fields is to observe that $\mathbb C$ embeds into an ultra-product of finite fields. Indeed, if the zero-divisor conjecture holds for each of the finite fields, then it holds for $\mathbb C$.

Answer 3

The answer to your second question is $R=2$.

Suppose a zero-divisor $a$ has rank 2 in a torsion-free group G.

W.l.o.g. there is a $b$ such that $a\cdot b = 0$. By multiplying with a suitable group element from the left we can achieve that $a = 1 + g$ for some $1\neq g \in G$. Similarly by multiplication from the right we can achieve that $b$ has the form $b= 1 + h_2 + \ldots + h_{k}$, where $k$ is the rank of $b$, and all $h_i$ are distinct and different from 1. For notational simplicity we define $h_1 = 1$.

Since $a \cdot b = 1 + h_2 + \ldots + h_{k} + g + g h_2 + \ldots + g h_{k} = 0$, and since $g h_i \neq g h_j$ for $i\neq j$ there is a matching that pairs every element in $A = \lbrace 1, h_2, \ldots, h_{k} \rbrace$ with an element in $B= \lbrace g, g h_2, \ldots, g h_{k}\rbrace$. The elements that are paired are equal (i.e., $h_i = g h_j$ if $h_i$ is paired with $g h_j$).

From the matching we conclude that there is an index $i_1$, such that $g = h_{i_1}$.
Then there is an index $i_2$ (different from indices previously used) such that $g^2 = g h_{i_1} = h_{i_2}$.
Then there is an index $i_3$ (different from indices previously used) such that $g^3 = g h_{i_2} = h_{i_3}$.
And so on $\ldots$
By induction, some index $i_t$ must be equal to 1 and $g^t = h_1 = 1$ must hold for some $t\in\mathbb{N}$. Which shows that G has torsion and yields a contradiction.

For all I know, a similar statement is not known for $R = 3$. However, the Conjecture itself over $\mathbb{Q}$ is equivalent to $R = \infty$. I have been researching this question for $R= 3$, and the proof does not seem to adapt easily. However it is possible to show statements of the following form: If $a\cdot b = 0$ then $a$ must have rank larger than $R_1\in \mathbb{N}$ or $b$ must have rank larger than $R_2\in \mathbb{N}$. Such a statement can be reduced to a finite case analysis (potentially involving undecidable torsion-freeness questions), which is still doable by hand for $R_1=4$ and $R_2 = 4$. However, the number of cases in the reduction (of the finite case analysis I know) grows like the double factorial.

Answer 4

For a fixed group $G$, the zero divisor conjecture over $\mathbb{C}$ is implied by the zero divisor conjecture over an algebraically closed field of positive characteristic, so in particular it would suffice to prove it for all finite characteristic $p$ fields, for your favourite prime $p$. This is due to the extension theorem for places and is exercise 21 in chapter 13 of Passman's book "The algebraic structure of group rings" ZBL0654.16001 MR798076.