5
$\begingroup$

A semilattice $(S,\cdot)$ is a commutative idempotent semigroup.

A congruence on a semilattice is an equivalence relation that preserves multiplication, i.e. $x_1\mathrel{\theta} y_1$ and $x_2\mathrel{\theta} y_2$ imply that $x_1\cdot x_2\mathrel{\theta} y_1\cdot y_2$.

It turns out that the set of all congruences on a semilattice $\mathbf{S}=(S,\cdot)$ forms a lattice under inclusion, called the congruence lattice, denoted $\mathrm{Con}\,\mathbf{S}$.

A lattice $(L,\vee,\wedge)$ is meet-semi-distributive if for every $x,y,z\in L$, we have $x\wedge y=x\wedge (y\vee z)$ whenever $x\wedge y=x\wedge z$.

How can I show that $\mathrm{Con}\,\mathbf{S}$ is meet-semi-distributive for semilattice $\mathbf{S}$?

Here's what I have so far: Let $\theta,\phi,\psi\in\mathrm{Con}\,\mathbf{S}$. Assume $\theta\cap\phi=\theta\cap\psi$. The inequality $\theta\cap\phi\leq\theta\cap(\phi\vee\psi)$ holds in every lattice, so we only need to show $\theta\cap(\phi\vee\psi)\leq\theta\cap\phi$. Suppose $x\mathrel{(\theta\cap(\phi\vee\psi))}y$. So $x\mathrel{\theta}y$ and $x\mathrel{(\phi\vee\psi)}y$.

Since semilattices are a generalization of lattices and semi-distributivity is a generalization of distributivity, I believe the proof will be a generalization of this proof. Unfortunately, I do not know an alternate characterization of the join of two semilattice congruences.

$\endgroup$

1 Answer 1

3
$\begingroup$

I found a proof in this paper:

Papert, Dona. Congruence relations in semi-lattices. J. London Math. Soc. 39 1964 723–729.

The trick is to notice that, for $\theta\subseteq \phi$, we can define a relative pseudo-complement $\phi\ast\theta$, i.e. $\phi\cap(\phi\ast\theta)=\theta$ and if $\phi\cap\psi=\theta$, then $\psi\subseteq\phi\ast\theta$.

Then, if we let $\zeta$ denote $\phi\cap\theta=\psi\cap\theta$, we can define $\theta\ast\zeta$. Since $\phi\cap\theta=\psi\cap\theta=\zeta$, we see that $\phi,\psi\subseteq\theta\ast\zeta$. Hence $\phi\vee\psi\subseteq \theta\ast\zeta$, so $(\phi\vee\psi)\cap\theta\subseteq(\theta\ast\zeta)\cap\theta=\zeta=\phi\cap\theta$, as desired.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .