English

The finite basis problem for matrix semirings $\mathbf{M}_n(S_7)$

Rings and Algebras 2026-06-12 v1

Abstract

We first prove an embedding theorem for matrix semirings Mn(S)\mathbf{M}_n(S) over an additively idempotent semiring SS: for all n2n \geq 2, Mn(S)\mathbf{M}_n(S) embeds into Mn+1(S)\mathbf{M}_{n+1}(S). This yields an ascending chain of varieties V(M2(S))V(M3(S))\mathsf{V}(\mathbf{M}_2(S)) \leq \mathsf{V}(\mathbf{M}_3(S)) \leq \cdots, which is strictly ascending when SS is the two-element distributive lattice. We then show that every variety in the interval [V(Sc(abc)),V(Mn(S7))][\mathsf{V}(S_c(abc)), \mathsf{V}(\mathbf{M}_n(S_7))] is nonfinitely based (i.e., has no finite basis for its identities), where Sc(abc)S_c(abc) is an eight-element flat semiring and S7S_7 is the unique nonfinitely based three-element additively idempotent semiring. Consequently, Mn(S7)\mathbf{M}_n(S_7) is nonfinitely based, yielding an ascending chain V(M2(S7))V(M3(S7))\mathsf{V}(\mathbf{M}_2(S_7)) \leq \mathsf{V}(\mathbf{M}_3(S_7)) \leq \cdots; moreover, every variety in [V(S7),V(Mn(S7))][\mathsf{V}(S_7), \mathsf{V}(\mathbf{M}_n(S_7))] is also nonfinitely based, and this interval contains at least countably infinitely many distinct varieties. Although we do not know whether V(Mn(S7))=V(Mn+1(S7))\mathsf{V}(\mathbf{M}_n(S_7)) = \mathsf{V}(\mathbf{M}_{n+1}(S_7)) holds, we show that the multiplicative reduct of Mn(S7)\mathbf{M}_n(S_7) without the constant matrix [1]n[1]_n is 55-nilpotent, which strongly suggests that the equality may indeed hold for all n2n \geq 2.

Keywords

Cite

@article{arxiv.2607.09677,
  title  = {The finite basis problem for matrix semirings $\mathbf{M}_n(S_7)$},
  author = {Jun Jiao and Miaomiao Ren},
  journal= {arXiv preprint arXiv:2607.09677},
  year   = {2026}
}