A negative answer to a Bahturin-Regev conjecture about regular algebras in positive characteristic

TL;DR

This paper disproves a conjecture by Bahturin and Regev in positive characteristic fields, showing that the invertibility of a certain matrix does not necessarily imply minimality of a regular algebra decomposition.

Contribution

The paper provides a counterexample in positive characteristic fields, demonstrating that the Bahturin-Regev conjecture does not hold beyond characteristic zero.

Findings

Counterexamples in positive characteristic fields

The matrix associated with regular decompositions can be singular

Minimal regular decompositions can have non-invertible matrices

Abstract

Let $A=A_1\oplus\cdots\oplus A_r$ be a decomposition of the algebra $A$ as a direct sum of vector subspaces. If for every choice of the indices $1\le i_j\le r$ there exist $a_{i_j}\in A_{i_j}$ such that the product $a_{i_1}\cdots a_{i_n}\ne 0$, and for every $1\le i,j\le r$ there is a constant $\beta(i,j)\ne 0$ with $a_ia_j=\beta(i,j) a_ja_i$ for $a_i\in A_i$, $a_j\in A_j$, the above decomposition is regular. Bahturin and Regev raised the following conjecture: suppose the regular decomposition comes from a group grading on $A$, and form the $r\times r$ matrix whose $(i,j)$th entry equals $\beta(i,j)$. Then this matrix is invertible if and only if the decomposition is minimal (that is one cannot get a regular decomposition of $A$ by coarsening the decomposition). Aljadeff and David proved that the conjecture is true in the case the base field is of characteristic 0. We prove that the conjecture does not hold for algebras over fields of positive characteristic, by constructing algebras with minimal regular decompositions such that the associated matrix is singular.