Kaplansky’s idempotent conjecture for integral quandle rings

For every finite Latin quandle QQ, let Z[Q]\mathbb{Z}[Q] be the free abelian group with basis QQ, equipped with the bilinear multiplication extending the quandle operation x∗yx*y on basis elements. Then every idempotent e∈Z[Q]e\in\mathbb{Z}[Q], meaning e2=ee^2=e, is trivial: e=0e=0 or e=qe=q for some q∈Qq\in Q.

References

Primary source

arXiv

Progress summary

Refreshed
Claimed progress

A new paper settles the question for two broad families, but no general proof for all integral quandle rings has appeared.

The conjecture asks whether, for a finite Latin quandle QQ, the integral quandle ring Z[Q]\mathbb{Z}[Q] has only trivial idempotents. The retrieved sources do not identify its proposer or date.

Known results

  • The conjecture was previously known for R3R_3 and R5R_5.
  • Integral quandle rings of free quandles have only trivial idempotents.
  • Certain finite-type coverings and knot quandles provide infinitely many or nontrivial idempotents, so the broad formulation requires hypotheses.
  • Earlier work computed idempotents for several specific quandle rings, including Z[R3]\mathbb{Z}[R_3], Z[R4]\mathbb{Z}[R_4], and Z[Cs(4)]\mathbb{Z}[\mathrm{Cs}(4)].

September 2026 special-case resolution

A September 2026 paper claims the conjecture for Takasaki quandles, including dihedral quandles, and for medial commutative quandles. It introduces Fourier analysis for Alexander quandles and proves necessity of a previously known sufficient condition for counterexamples. This is a substantial advance, but not a resolution for all integral quandle rings.

Current status (as of September 2026): The conjecture is claimed for Takasaki and medial commutative quandles, while the general case remains open and the new claims are unverified.

Sources

Solutions 0

No solutions have been posted yet.