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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.

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As for the zero divisor conjecture: I am pretty sure that if we can prove it over all finite fields, then we can prove it over all fields. In fact, it is known that the zero divisor conjecture is equivalent to the idempotent conjecture (for the same group, over the same field). Idempotence is an algebraic equation, so if it is nontrivially solvable over a field, it is also nontrivially solvable over all sufficiently high finite subfields. I'm not entirely sure about this, but the proof should go as follows: (1) Find a nontrivial solution algebraic over the prime field (which may or may not ... –  darij grinberg Apr 21 '11 at 17:10
    
... be finite). If the prime field is finite, you are done, so WLOG assume that it is not. Then our field has characteristic $0$. There is thus a nontrivial solution in an algebraic extension of $\mathbb Q$. "Rationalize the denominator" and reduce modulo a prime high enough not to appear in the denominator. You get a finite field now, and the only bad thing that can happen is that our solution becomes trivial. But choosing the prime even higher can prevent this. Am I right? –  darij grinberg Apr 21 '11 at 17:12
    
@Darij: You are correct, I think, and you do not need idempotents. If $a_1,...,a_n, b_1,...,b_m$ are elements of $G$ and you want $(\sum \alpha_i a_i) (\sum \beta_j b_j)=0$, you get a condition on the coefficients which is equivalent to a disjunction of conjunctions of equations. So the set of good $n+m$-tuples $(\alpha_1,...,\alpha_n,\beta_1,...,\beta_m)$ is a union of algebraic varieties. To exclude 0, just add one equation that the product of all variables is 1. If that union of varieties has a point over $\mathbb{C}$, it has a point over $\bar\mathbb{Q}$, etc. –  Mark Sapir Apr 21 '11 at 18:03
    
Darij, I do not think that the idempotent conjecture is equivalent to the zero-divisor conjecture. There are reductions to the case $a^2=0$ with $a \neq 0$, but I have never heard about a reduction to idempotents. –  Andreas Thom Apr 21 '11 at 20:07
    
In the second part, do you mean to ask for the largest lower bound on the rank known for a zero divisor? As it is written, it would seem that you could just take $R=1$, since $1$ is a unit. –  Richard Kent Apr 21 '11 at 20:52

3 Answers 3

@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.

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thanks, Mark. is it something known about the second part of the question? –  Kate Juschenko Apr 21 '11 at 17:09
    
(second part of the second question) –  Kate Juschenko Apr 21 '11 at 17:11
    
@Kate: The second question is very interesting and I do not know the answer. Once we (with Victor Guba) tried to show that if the rank is very large, and we impose relations that force the sum of elements to be a zero divisor at "random", then we would get a torsion-free group essentially by Gromov's random group method. But it turned out that the Gromov density of that problem is not smaller than $1/2$, so one cannot conclude anything. We did not write down the calculations, so there could be an arithmetic error somewhere, but we got discouraged enough to stop trying. –  Mark Sapir Apr 21 '11 at 17:42
    
thank you for clarifications, Mark. –  Kate Juschenko Apr 22 '11 at 12:58

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.

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Thanks, Pascal. –  Kate Juschenko Apr 28 '11 at 6:52

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 $\mathbb C$.

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I don't get it. Yes, you can suppose, WLOG, that $a, b \in \mathbb{Z}[G]$ are not divisible by $p$, but what if their product $ab$ is divisible by $p$? It sounds rather that you have argued the converse, that the Kaplansky conjecture over $\mathbb{Z}/p$ implies the Kaplansky conjecture over $\mathbb{Z}$. –  Greg Kuperberg Apr 21 '11 at 20:30
    
You are right, I got the logic wrong. I made corrections so that it makes more sense now. –  Andreas Thom Apr 21 '11 at 20:42
    
Maybe a better way to say the second half is that if the Kaplansky conjecture is true for $\overline{\mathbb{F}_p}$ for some prime $p$, then it is true for $\mathbb{C}$. –  Greg Kuperberg Apr 21 '11 at 20:55
    
You are right that all finite fields of characterstic $p$ is a good choice, but many can apparently also mean a set of finite fields, so that $\mathbb C$ embeds into their ultraproduct. It is not clear whether there is a good way of saying what the right condition on the set of finite fields is. –  Andreas Thom Apr 23 '11 at 9:40

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