Episode 196, October 2026

Torsion-free, yet zero?

A preprint claims a torsion-free group whose group algebra over the two-element field has two nonzero elements with product zero, disproving Kaplansky’s zero-divisor conjecture

Covers result 196 in OpenAI’s catalog: its one paper. See its connections to other results and the work it cites.

Play Episode 196: Torsion-free, yet zero? Three small graphs in which every vertex has odd degree, each with an even number of vertices.

5:48. English captions available. The video loads from YouTube when you press play.

Chapters

The claim, in plain words

Take a group and allow sums of its elements with coefficients 0 or 1, where 1 + 1 = 0; multiply term by term with the group law and keep a term only if it turns up an odd number of times. This is the group algebra 𝔽₂[G]. If g³ = 1, then (1 + g)(1 + g + g²) = 1 + g³ = 0: any nontrivial element of finite order gives two nonzero sums with product zero. In the integers and the integer grid our box searches found no such pair, and the reason is general: in a fixed order, the last term of a product is reached only once.

The question goes back to Graham Higman’s 1940 thesis and was posed by Irving Kaplansky in 1956: if a group has no elements of finite order, is its group algebra free of zero divisors? As the paper tells it, the answer is yes for groups in which every nontrivial finitely generated subgroup maps onto the integers (Higman), for torsion-free elementary amenable groups (Kropholler, Linnell and Moody), and for groups with the unique-product property. Rips and Segev found torsion-free groups without unique products, but over this field a zero product needs every term reached an even number of times. Giles Gardam disproved the related unit conjecture over 𝔽₂ in 2021, and later over ℂ.

Result 196 is one OpenAI preprint, A Torsion-Free Group Algebra with Zero Divisors (26 pages, dated September 23, 2026). Its Theorem 1.1: there are a finitely presented torsion-free group G and nonzero α, β in 𝔽₂[G] with αβ = 0; moreover, G has a finite two-dimensional classifying space. The construction is probabilistic: the paper proves that a suitable choice exists without writing one down. By our arithmetic, the graphs in the Lean proof have more than 10⁸⁶ vertices each.

What it doesn’t claim

Anything over fields of characteristic 0, such as ℚ or ℂ, or of odd characteristic: the cancellation is a parity trick. (By our remark it carries over to every field of characteristic 2, which contains 𝔽₂.) Nothing over the integers, nothing about the idempotent conjecture, and no explicit group. A later paper in the release (result 197) builds on this method.

Where it stands

Posted as a preprint: OpenAI, 26 pages, dated September 23, 2026
Peer reviewed: Not yet
Checked in Lean: OpenAI’s Lean proof of the full statement; not run by us

Last checked October 19, 2026. The preprint has not been revised since it was added to the repository in its initial commit on October 6, 2026. OpenAI’s comparator statement matches Theorem 1.1 in full, including torsion-freeness and the classifying space, and the Lean proof has no sorry, admit, axiom or native_decide by our search: 123 files and about 44,300 lines, excluding Mathlib (source). Like the paper, the Lean proof shows that such a group exists without writing one down. We did not run the check. The paper is not among those OSVerify reviewed, and we have seen no expert review of it.

How the proof goes

An alphabet from a projective plane. The letters are the 16,513 points of the projective plane of order 128 (129 points on each line, any two lines meeting in exactly one point), plus three extra letters.
Two graphs with odd overlaps. Each vertex of graph A or B gets a set of outgoing letters: a line, sometimes with extras (two of the three on A, all three on B). So a vertex of A and a vertex of B always share 1, 3, 129 or 131 letters, an odd number. The edges come from random matchings.
The cancellation. Gluing a cone over each graph makes every closed walk spell the identity, so each vertex in the root’s component has a well-defined group element. Let α be the sum of these over A and β the sum of the inverses over B. Stepping both vertices of a pair along a shared letter leaves the term of αβ unchanged; every pair has an odd number of such steps, so every clump of linked pairs has even size, and αβ = 0. This takes one page (§6).
Two safeguards. No path from a root to a different vertex spells the identity, so α and β are nonzero (Prop. 5.1); the glued complex has no essential spheres, so it is a finite two-dimensional classifying space and the group is torsion-free (Cor. 5.3).
No bad sphere pictures. With high probability the random matchings admit none: small ones by a counting estimate (Prop. 3.1), large ones cut into small pieces by the Lipton–Tarjan planar separator theorem (Prop. 4.5).

The numbers in the video

Box searches (ours): all 16,769,025 pairs of nonzero elements of 𝔽₂[ℤ] supported on t⁰…t¹¹, and all 261,121 pairs in a 3×3 box of 𝔽₂[ℤ²]: no zero products. In {1,x,y}·{1,xy,x²}, x²y is reached twice and cancels; the other seven terms survive.
Our toy on the 7-point plane (98 vertices a side): 9,604 pairs in 1,086 clumps, all of even size; but in the toy’s abelian image both sums vanish, so an easy quotient cannot show they are nonzero.
The paper’s constants (q = 128, 16,513 points, 16,516 letters, shared counts 1, 3, 129, 131) rechecked by us. The size bound n ≥ e²⁰⁰ > 10⁸⁶ and the Lean counts are ours.

Read the source

A Torsion-Free Group Algebra with Zero Divisors (PDF), 26 pages.
The Lean scope note.
The collection’s catalog, where this is result 196.

The paper

OpenAI groups related papers into one result; this one has a single paper.

A Torsion-Free Group Algebra with Zero Divisors

Transcript

Read the full script

0:00The question

Take a group, and allow sums of its elements, with coefficients zero or one, where one plus one is zero. Multiply two such sums term by term using the group law, and keep a term only if it turns up an odd number of times. This is the group algebra over the two-element field. Sometimes two nonzero sums multiply to zero. If an element g has g³ equal to one, then one plus g, times one plus g plus g², is one plus g³, which is zero. Any element of finite order, other than the identity, does this. In the integers, or the integer grid, our box searches found no such pair, and there is a reason: put the terms in order, left to right, and the last term of a product is reached only once, so it never cancels. So the question, going back to Graham Higman’s thesis in 1940, and posed by Irving Kaplansky in 1956: if a group has no elements of finite order, is its group algebra free of zero divisors? As the paper tells it, the answer is yes for several classes: groups in which every nontrivial finitely generated subgroup maps onto the integers, by Higman; torsion-free elementary amenable groups, by Kropholler, Linnell and Moody; and every group with the unique-product property. Rips and Segev found torsion-free groups without unique products, but that alone does not give a zero divisor: over this field, every term of the product must be reached an even number of times. A related conjecture about units was disproved by Giles Gardam in 2021.

1:38The claim

Result 196 is one OpenAI preprint, 26 pages, dated September 23, 2026. It claims: there is a finitely presented group with no elements of finite order, and two nonzero sums in its group algebra over the two-element field whose product is zero. The group even has a finite two-dimensional classifying space. If this holds, Kaplansky’s zero-divisor conjecture is false. But the paper never writes the example down. The construction is random, and the paper proves only that a suitable choice exists, without giving one. By our arithmetic, the graphs in the Lean proof have more than 10⁸⁶ vertices each. Not claimed: anything over fields of characteristic zero, like the rational or complex numbers, or of odd characteristic; the cancellation is a parity trick. By our remark, it carries over to every field of characteristic two, which contains the two-element field. Nothing is claimed over the integers, nor about the idempotent conjecture, nor an explicit group. A later paper in the release builds on this method; that is the next episode.

2:49How the proof goes

The letters come from a finite projective plane of order 128: 16,513 points, with 129 points on each line, and any two lines meeting in exactly one point. Add three extra letters. Now build two graphs, A and B. Each vertex gets a set of outgoing letters: a line of the plane, sometimes with extras, two of the three on side A, all three on side B. So a vertex of A and a vertex of B always share 1, 3, 129 or 131 letters: always an odd number. The edges come from random matchings. Glue a cone over each graph, so that in the resulting group every closed walk spells the identity. Then each vertex x in the root’s part of graph A has a well-defined group element, read along any path from the root. Let alpha be the sum of these elements, and beta the sum of the inverses of those of graph B. Expanding alpha times beta gives one term for each pair of vertices. Now step both vertices along a shared letter at once: the term does not change. Pairs linked by such steps form clumps, and each pair has an odd number of steps. A graph in which every vertex has odd degree has an even number of vertices in each piece, so every clump has even size, and everything cancels. That algebra takes one page. The other 20 pages are two safeguards. First, the sums must not collapse: no path from a root to a different vertex may spell the identity, so each sum keeps the identity exactly once. Second, the group must have no torsion: the glued complex has no essential spheres, which makes it a finite two-dimensional classifying space, and such a group has no elements of finite order. Our toy version, built on the 7-point plane, shows why this needs proof: the cancellation works, but in the toy’s abelian image both sums vanish, so an easy quotient cannot show they are nonzero. The paper shows that, with high probability, the random matchings admit no bad sphere pictures: small ones by a counting estimate, and large ones cut into small pieces by the Lipton–Tarjan planar separator theorem.

5:03Where it stands

The preprint is unrefereed and has not been revised. OpenAI’s Lean statement matches the theorem in full, including torsion-freeness and the classifying space, and its proof is present with no gaps. That is 123 files and about 44,000 lines; we did not run the check. Like the paper, the Lean proof shows that such a group exists without writing one down. The paper is not among those OSVerify reviewed, and we have seen no expert review of it. If it holds, missing torsion is not enough: over the two-element field, some torsion-free group algebra has zero divisors, even though no one has yet written one down.

Spot a mistake?

This episode simplifies on purpose, but it shouldn’t be wrong. If something is, open an issue on GitHub or leave a comment on YouTube. Corrections are noted here with a date.