Kaplansky’s idempotent conjecture for integral quandle rings
For every finite Latin quandle , let be the free abelian group with basis , equipped with the bilinear multiplication extending the quandle operation on basis elements. Then every idempotent , meaning , is trivial: or for some .
References
Primary source
Additional references
Progress summary
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 , the integral quandle ring has only trivial idempotents. The retrieved sources do not identify its proposer or date.
Known results
- The conjecture was previously known for and .
- 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 , , and .
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
- arxiv.org
- arxiv.org
- export.arxiv.org
- ar5iv.labs.arxiv.org
- arxiv.org
- maths.ox.ac.uk
- en.wikipedia.org
- mathoverflow.net
- heldermann-verlag.de
- gilkalai.wordpress.com
- quantamagazine.org
- quantamagazine.org
- arxiv.org
- export.arxiv.org
- mathstodon.xyz
- mathstodon.xyz
- mathstodon.xyz
- mathstodon.xyz
- mathstodon.xyz
- quantamagazine.org
- quantamagazine.org
- quantamagazine.org
- scientificamerican.com
Solutions 0
No solutions have been posted yet.