A counterexample to Kaplansky’s zero-divisor conjecture

The explanation

Take a clock with three positions, 0, 1 and 2. Let g mean "move one step forward". Three steps bring you back, so g³ = 1, where 1 means "stay put". Now do arithmetic where only odd and even matter, so 1 + 1 = 0. Multiply (1 + g) by (1 + g + g²). The six terms give 1, g and g² twice each. Every pair cancels, and the product is 0.

So two things that are not zero multiply to zero. Such things are called zero divisors. Ordinary numbers never do this. The clock did it because g has finite order: some power of g brings you back to the start. The paper states this trick in general. If g has order m above 1, then (1 − g)(1 + g + … + g^(m−1)) = 0.

The setting is a group algebra. Start with a group, a set of moves that can be combined and undone. A group algebra element is a finite sum of moves, here with each move counted odd or even times. Products expand term by term, as with ordinary brackets. A group is torsion-free if no move except "stay put" ever returns to the start. Kaplansky's zero-divisor conjecture says that for a torsion-free group, the group algebra over any field has no zero divisors.

The question is old. The paper traces it to Graham Higman's 1940 thesis. Irving Kaplansky posed it in a 1956 conference contribution, published in 1957. Many cases were settled in its favour. Higman proved it for groups in which every nontrivial subgroup maps onto the whole numbers. In 1988, Kropholler, Linnell and Moody proved it for torsion-free elementary amenable groups. In 2021, Giles Gardam disproved a related claim, the unit conjecture. A unit, though, is never a zero divisor.

Cancellation alone is not enough. In 1987, Rips and Segev built torsion-free groups without the unique-product property. That property means some product term always appears exactly once, so nothing can cancel it. Losing it is not enough. Over odd-and-even arithmetic, every term must appear an even number of times, not just more than onceSource 1.

This paper proves the conjecture false. It builds a torsion-free group G, described by finitely many moves and rules. It gives two nonzero sums whose product is 0, counting odd or evenSource 2. The paper builds parity into the design. It builds two random graphs whose edges carry letters, and the letters become the group's basic moves. The letters are the 16513 points of a finite geometry, plus three extra lettersSource 3,Source 4. At each position, the letters on the outgoing edges form one line of that geometry, plus possibly some extrasSource 3. Two line parts meet in 1 point, or in 129 points when they are the same lineSource 5. Both counts are odd. The extra parts on the two sides overlap in 0 or 2 letters, both evenSource 6. Odd plus even stays odd. So every relevant overlap is odd, and the paper turns that into even counts that cancel.

The paper calls the cancellation itself simpleSource 7. The difficulty lies in two safeguards. Both sums must stay nonzero, and the group must stay torsion-freeSource 8. The paper gets both from a shape argument. It argues that the space built from its graphs has no essential spheresSource 9. An essential sphere is a sphere-shaped surface in the space that cannot be shrunk to a point. That makes it a finite two-dimensional classifying space, which rules out finite-order movesSource 10.

There is a limit worth knowing. The construction is probabilisticSource 11. The paper argues that suitable random choices exist, but it does not write down a specific group. So nobody can yet read off the two sums and multiply them by hand. The clock example works because g has finite order. The new group has no such move, yet zero divisors still appear.

How sure is this? A computer proof checker called Lean has checked a formal statement of the counterexample. The Lean scope note describes that statement. It gives a finitely presented torsion-free group G. It also gives two nonzero sums in its group algebra, counted odd or even, whose product is 0. The note also mentions the classifying space. No specialists have published a review yet.

What it doesn't show. This does not write down an explicit group or explicit zero divisors, because the construction is probabilistic.