Two-sidedness conjecture for null ideals of finite rings

For every finite associative unital ring RR, let N(R):={f(x)∈R[x]∣f(a)=0 for every a∈R}N(R):=\{f(x)\in R[x]\mid f(a)=0\text{ for every }a\in R\} be the null ideal of RR. The conjecture asserts that N(R)N(R) is a two-sided ideal of R[x]R[x].

References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A preprint reports that the conjecture is false, with failures at every higher nilpotency level and a positive result for the lowest levels.

Werner conjectured that, for every finite ring RR, the null polynomials form a two-sided ideal of R[x]R[x]. The general assertion was already contradicted by an explicit 4×44\times4 upper-triangular example.

Known results

  • Werner: the property holds for local rings, semisimple rings, matrix rings over commutative rings, and rings of odd order.
  • Frisch: it holds for upper-triangular and structural matrix rings.

August 26, 2026: counterexamples at all higher levels

A new paper reports counterexamples for every nilpotency level n≥5n\ge5 and proves the property when radical nilpotency is at most 33. A separate preprint gives an explicit finite-ring counterexample and states that GPT-5.6 Sol generated the counterexample and initial proof; the author reports independent verification and simplification.

Current status (as of August 2026): The general conjecture has a claimed counterexample, while the all-level family and the positive range through radical nilpotency 33 remain unverified in this record.

Sources

Solutions 0

No solutions have been posted yet.