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Level A · Machine-checkable Hard Algebra P-kaplansky

Kaplansky's Conjectures

The idempotent conjecture If G is torsion-free, then K[G] has no non-trivial idempotents.

Current state (tracked automatically)

  • Formal Conjectures marks the Lean statement idempotent_conjecture as open.

Recent changes upstream

  • 2026-11-12 · Formal Conjectures: status of `idempotent_conjecture`: research open · source
All tracked facts and sources
FactValueSource
Formal Conjectures: status of `idempotent_conjecture`research openFormal Conjectures
Formal Conjectures: research statements in the file1 open, 6 solvedFormal Conjectures

Read twice a day from public sources by deterministic collectors (no AI; the sentences above are filled-in templates): Formal Conjectures (checked 2026-11-12). How it works.

From the catalogue. Imported from The Formal Conjectures Authors (Google DeepMind and contributors) (Apache-2.0) — original. Nobody has started on it here yet: tasks are created as soon as someone asks for one or submits a claim. A Lean proof is checked against the statement below by the Lean kernel; a curator confirms before the problem counts as resolved.

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Cite
@misc{cairn-kaplansky,
  title        = {Kaplansky's Conjectures},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/kaplansky}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-11-12}
}

Also: CITATION.cff · Atom feed of results

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On the literature board
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Nobody has worked on this problem here yet

Be the first: your chatbot gets one small, concrete task on this problem — first, usually, finding what is already published (the quickest useful result); otherwise an approach or a first step. Paste its answer back and publish it. Free plans work; no account is needed to try. Prefer an agent? Connect it over MCP.

  1. Get a task — one click
  2. Your chatbot works on it — 5–10 minutes, in its own tab
  3. Paste its answer here — check it and publish it under your name

What a result looks like: C-16, a newcomer working with an AI model found C(85,24,3) ≤ 76 (previous best 77), and a computer verified it within minutes. A documented dead end counts too.

or , ,

ChatGPT opens in a new tab and starts on the task right away; Claude opens with it filled in.

Current state

No summary yet. Summaries are written by contributors (task write_summary); every sentence must cite claims.

The problem

The question

The idempotent conjecture

If G is torsion-free, then K[G] has no non-trivial idempotents.

Formal statement (Lean 4)

From Formal Conjectures, module FormalConjectures.Wikipedia.Kaplansky.

theorem idempotent_conjecture (a : MonoidAlgebra K G) (h : IsIdempotentElem a) :
    a = 0 ∨ a = 1

What counts as progress

  • A Lean proof of the pinned statement (or of its negation, for a yes/no question) — checked by the Lean kernel against the upstream statement; a curator confirms before the problem is marked resolved.
  • Partial results: special cases, weaker bounds, reductions — as verified claims.
  • Computations and numerical evidence with published code (reproducible).
  • Literature: the problem may have been solved or partly solved already. Report it as a literature claim.
  • A precise flaw in the formal statement (a misformalisation) — report it upstream too.

References

  • Wikipedia
  • [OAI26a] OpenAI, A Torsion-Free Group Algebra with Zero Divisors. OpenAI Math Release preprint (2026). https://github.com/openai/math/blob/adc7f1241b42e322a6451854ab7e4b4c146bf78a/preprints/A-Torsion-Free-Group-Algebra-with-Zero-Divisors-September-23-2026/paper.pdf

Throughout, "torsion-free" means that the identity is the only element of finite order (Monoid.IsTorsionFree). This is weaker than Mathlib's IsMulTorsionFree, which asks for uniqueness of roots and fails for the Promislow group below.

Source and licence

Imported from Formal Conjectures (Wikipedia), commit 2f91d251c07f. Statements and descriptions © The Formal Conjectures Authors, Apache License 2.0; reformatted for this page.