Equivalence of simplicity conditions for Bell–Rogalski algebras

Determine whether, for a Bell–Rogalski algebra B = R_{\boldsymbol{p}}(\boldsymbol{\sigma},\mathbf{H},\mathbf{J}) of rank n, the family of conditions BI_i^{(k)} t_i^{k} B = B for all i in {1,…,n} and all k in N_{>0} implies B x B = B for all x in the Ore set S = ( { a t^{\alpha} | a in I^{(\alpha)}, \alpha in Z^n with \alpha > 0 } ∪ {1} ) \setminus {0}, where I_i^{(k)} is defined by I_i^{(k)} = J_i \sigma_i(J_i) … \sigma_i^{k-1}(J_i) for k > 0, I_i^{(0)} = R, and I_i^{(k)} = \sigma_i^{-1}(H_i) … \sigma_i^{k}(H_i) for k < 0. Ascertain whether these conditions are actually equivalent, i.e., whether the reverse implication holds.

Background

Theorem 6.7 provides a simplicity criterion for a Bell–Rogalski (BR) algebra B, which includes the condition that BxB = B for all x in a specific Ore set S generated by positive-degree homogeneous elements of the form a t^{\alpha} with a in the canonical ideals I^{(\alpha)}.

To make this criterion more checkable, the authors introduce an alternative family of conditions parameterized by i ∈ [n] and k > 0, namely BI_i^{(k)} t_i^{k} B = B, where I_i^{(k)} are the canonical products of the ideals J_i and H_i under the automorphism \sigma_i. They show various equivalences for fixed i, and it is clear that BxB = B implies all BI_i^{(k)} t_i^{k} B = B.

However, it is not established whether the converse holds, i.e., whether verifying BI_i^{(k)} t_i^{k} B = B for all i and k suffices to guarantee BxB = B for all x in S. Resolving this would yield a more practical simplicity test for higher-rank BR algebras.