9
$\begingroup$

The Kaplansky conjecture says that: for any field $F$ and any torsion free group $G$, the group ring $F[G]$ does not have nontrivial idempotent elements.

Questions
Do we assume that $F$ has any characteristic ?

Does the conjecture imply that any finitely generated projective $F[G]$-module is free ? or eventually stably free?

For which field $F$ and which class of groups $G$, the conjecture is known to be true?

$\endgroup$
3
  • 2
    $\begingroup$ For starters: mathoverflow.net/questions/79559/… $\endgroup$ Commented May 23, 2017 at 18:36
  • $\begingroup$ @T.Amdeberhan I read that link before asking my question. $\endgroup$ Commented May 23, 2017 at 18:44
  • 2
    $\begingroup$ The answer to your first question is already in your question, since you're stating the conjecture correctly: there's no characteristic assumption, otherwise it would be explicit. $\endgroup$
    – YCor
    Commented May 24, 2017 at 20:36

1 Answer 1

11
$\begingroup$

Kaplansky's zero divisor conjecture: Let $\mathbb{F}$ be a field and $G$ be a torsion-free group. Then $\mathbb{F}[G]$ does not contain a zero divisor.

The existence of a nontrivial idempotent $a$ in a ring $R$ implies the existence of a zero divisor in $R$, because $a(a-1)=0$. So, the zero divisor conjecture implies the idempotent conjecture.

The idempotent conjecture has been confirmed in special cases. For example, Formanek (1973) showed that if $G$ is a torsion-free group satisfying the ascending chain condition on cyclic subgroups and $\mathbb{F}$ is a field of characteristic $0$, then $\mathbb{F}[G]$ has no nontrivial idempotents. Also, Bass (1976) proved that if $G$ is a torsion-free linear group, then $\mathbb{C}[G]$ has no nontrivial idempotents.

These conjectures have not been confirmed for any fixed field and it seems that confirming the conjecture even for the finite field $\mathbb{F}_2$ is still out of reach.

For the zero divisor conjecture case, our work Zero divisors and units with small supports in group algebras of torsion-free groups may be helpful. Also for more details about these two conjectures you can see Zero-divisors and idempotents in group rings.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your answer and references! Is there any result about the classification of finitely generated projective $F[G]$-modules under the assumption that the Kaplansky conjecture is true ? $\endgroup$ Commented May 25, 2017 at 2:21
  • $\begingroup$ I am not aware about results related to this question. $\endgroup$ Commented May 25, 2017 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.