What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Question

Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group.

A couple of good resources for this problem that gives some historical overview are:

• Passman, Donald S. The algebraic structure of group rings. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.

• Passman, Donald S. Group rings, crossed products and Galois theory. CBMS Regional Conference Series in Mathematics, 64. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1986.

The conjecture has been proven affirmative, when $G$ belongs to special classes of groups. I tried to write down some of the history:

• Ordered groups (A.I. Malcev 1948 and B.H. Neumann 1949)
• Supersolvable groups (E. Formanek 1973)
• Polycyclic-by-finite groups (K.A. Brown 1976, D.R. Farkas & R.L. Snider 1976)
• Unique product groups (J.M. Cohen, 1974)

Here are my questions:

1. Was Irving Kaplansky the first one to state this conjecture? Can someone provide me with a reference to a paper or book that claims this?
2. Since the publications of Passman's expository note (above) in 1986, has there been any major developments on the problem? Are there any new classes of groups that will yield a positive answer to the conjecture? Can someone help me to extend my list above?

The zero-divisor conjecture (let's denote it by "(Z)") is related to the following two conjectures:

(I): If $G$ is torsion-free, then $K[G]$ has no non-trivial idempotents.

(U): If $G$ is torsion-free, then $K[G]$ has no non-trivial units.

Now, if $G$ is torsion-free, then one can show that:

(U) $\Rightarrow$ (Z) $\Rightarrow$ (I).

Has there been any developments, since 1986, to any partial answers on conjecture (U)? Passman claims that "this is not even known for supersolvable groups". Is this still the case?

I want to point out that this post is related to another old MO-post.

Answer 1

Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $K=\mathbb{F}_2$ and virtually abelian $G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \rangle$ (as mentioned in David Craven's answer) in arXiv:2102.11818 .

Addendum (2024-01-05): the unit conjecture is also false over $\mathbb{C}$, see arXiv:2312.05240 (having previously been disproved in all positive characteristics by Murray in arXiv:2106.02147).

Answer 2

I suggest you have a look at Chapter 10 of W. Lueck's book, "$L^2$-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002. There he discusses the Atiyah conjecture (conjecture 10.3): if $K$ is a subfield of $\mathbb{C}$, a group $G$ satisfies the Atiyah conjecture with coefficients in $K$ if for any $m\times n$ matrix $A$ with coefficients in $K[G]$, the von Neumann dimension of the kernel of the operator $r_A:\ell^2(G)^m\rightarrow \ell^2(G)^n:x\mapsto Ax$, is an integer.

Lueck then proves (lemma 10.15) that if $G$ is torsion-free and satisfies the Atiyah conjecture with coefficients in $K$, then it satisfies Kaplansky's conjecture (Z). For $K=\mathbb{C}$ and $G$ amenable, the converse is true (lemma 10.16).

Lueck goes on to prove (Theorem 10.19) a remarkable result by P. A. Linnell (Division rings and group von Neumann algebras. Forum Math., 5(6):561-576,1993): Let $\cal{C}$ be the smallest class of groups containing free groups and closed under directed unions and extensions with elementary amenable quotients; if $G$ is in $\cal{C}$ and has finite subgroups of bounded order, then the Atiyah conjecture with coefficients in $\mathbb{C}$ holds.

For further reading and more recent results, try typing "Atiyah conjecture" on Google or in the ArXiV.

Answer 3

Since there's been a request for an update on the zero divisor conjecture, let me give one. The zero divisor conjecture is open. I'll update this answer if I hear otherwise.

Alain already mentioned the positive result coming from the Atiyah conjecture, so I'll focus on positive results that aren't to do with Atiyah, or the unit conjecture. (However I feel I should mention that some big classes of groups recently shown to satisfy the Atiyah conjecture are locally indicable groups by Jaikin-Zapirain–López-Álvarez and virtually compact special groups by Schreve.)

A theorem of Kropholler–Linnell–Moody is that the group ring (with coefficients in any skew field) of a torsion-free elementary amenable group embeds into a skew field and thus the zero divisor conjecture is true for groups which are residually (torsion-free elementary amenable).

Recent work of Fisher–Sánchez-Peralta proves among other things that the group ring (with coefficients in any skew field) of a torsion-free group which is either virtually compact special or a 3-manifold group embeds into a skew field (see arxiv doi).

Finally, Lewin proved the class of groups whose integral group rings have no zero divisors is closed under amalgamation by amenable groups. (That 1972 paper formulates it in terms of the Ore condition but it's equivalent by Kielak's appendix to a paper of Bartholdi. The conjecture in that paper is thus false because there exist non-amenable groups without free subgroups.)

Jaikin-Zapirain, Andrei; López-Álvarez, Diego, The strong Atiyah and Lück approximation conjectures for one-relator groups, Math. Ann. 376, No. 3-4, 1741-1793 (2020). ZBL1481.20093.

Schreve, Kevin, The strong Atiyah conjecture for virtually cocompact special groups, Math. Ann. 359, No. 3-4, 629-636 (2014). ZBL1353.20027.

Kropholler, P. H.; Linnell, P. A.; Moody, J. A., Applications of a new K-theoretic theorem on soluble group rings, Proc. Am. Math. Soc. 104, No. 3, 675-684 (1988). ZBL0691.16013.

Lewin, Jacques, A note on zero divisors in group-rings, Proc. Am. Math. Soc. 31, 357-359 (1972). ZBL0242.16006.

Bartholdi, Laurent, Amenability of groups is characterized by Myhill’s theorem, J. Eur. Math. Soc. (JEMS) 21, No. 10, 3191-3197 (2019). ZBL1458.37017.

Answer 4

It is still the case that (U) is not known for supersoluble groups. In fact, it is not known for the 'easiest' supersoluble group that isn't abelian-by-cyclic (for which (U) is known). The group is an extension of $\mathbb{Z}^3$ by the Klein four group.

Peter Pappas and I, in a paper in the early 2010s, gave some general statements about (U) for supersoluble groups, and a general strategy for how to prove (U) for this particular group, but it hasn't been done, even for this one group, in the intervening decade or so.