Congruence pairs of principal MS-algebras and perfect extensions

TL;DR

This paper introduces a simplified notion of congruence pairs for principal MS-algebras, explores their relation to substructures, and addresses a problem of characterizing congruence lattices, including special cases involving perfect extensions.

Contribution

It presents a simpler approach to congruence pairs in principal MS-algebras and offers insights into their structure and congruence characterization, including special cases with perfect extensions.

Findings

Congruences correspond to MS-congruence pairs on substructures.

A simple, elegant solution to a Grätzer-like problem for principal MS-algebras.

Characterization of perfect extensions via de Morgan and Boolean subalgebras.

Abstract

The notion of a congruence pair for principal MS-algebras, simpler than the one given by Beazer for $K_2$-algebras \cite{6}, is introduced. It is proved that the congruences of the principal MS-algebras $L$ correspond to the MS-congruence pairs on simpler substructures $L^{\circ\circ}$ and $D(L)$ of $L$ that were associated to~$L$ in \cite{4}. An analogy of a well-known Gr\"atzer's problem \cite[Problem 57]{11} formulated for distributive p-algebras, which asks for a characterization of the congruence lattices in terms of the congruence pairs, is presented here for the principal MS-algebras (Problem 1). Unlike a recent solution to such a problem for the principal p-algebras in \cite{2}, it is demonstrated here on the class of principal MS-algebras, that a possible solution to the problem, though not very descriptive, can be simple and elegant. As a step to a more descriptive solution of Problem 1, a special case is then considered when a principal MS-algebra $L$ is a perfect extension of its greatest Stone subalgebra $L_{S}$. It is shown that this is exactly when de Morgan subalgebra $L^{\circ\circ}$ of $L$ is a perfect extension of the Boolean algebra $B(L)$. Two examples illustrating when this special case happens and when it does not are presented.