1
vote
1answer
50 views

Showing finite lattices are isomorphic to their sets of ideals.

I'm working out a problem from Gratzer '71. We are to show that any finite lattice $L$ $\simeq$ $I(L)$. My attempt was as follow: prove the contrapositive. So assume $L$ is not isomorphic to $I(L)$. ...
1
vote
1answer
53 views

Factor congruences of non trivial Lattices

A pair of congruences $\theta$ and $\theta^*$ are called factor congruences if $\theta \vee \theta^*$ = full congruence. $\nabla$ $\theta \wedge \theta^*$ = trivial congruence. $\triangle$ I need ...
1
vote
1answer
22 views

Does the closure of the set of all irreducible elements always equal the whole set?

Let $X$ denote a set and $\mathrm{cl}$ denote a finitary closure operator on its powerset. Call $x \in X$ irreducible iff for all $A \subseteq X$ we have that if $x \in \mathrm{cl}(A)$, then $x \in ...
1
vote
2answers
56 views

Notation for “incommensurate” elements?

Say that $x\wedge y\not=x,y$, that is neither $x\leq y$ nor $y\leq x$. Is there a way to denote this? I've been saying $x<>y$ but that's completely made up.
3
votes
1answer
151 views

Do filters on a Boolean algebra also make a Boolean algebra?

Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...
4
votes
2answers
164 views

Stone duality for ideals and filters (exercise)

In A Course in Universal Algebra (Burris, Sankapannavar), the exercise 4.4.7-8, p.158, says: Let $A$ be a Boolean algebra. Denote $A^\ast:=\{\text{ultrafilters of }A\}$, and give $A^\ast$ the ...
3
votes
1answer
138 views

$M_3$ is a simple lattice

I'd like to prove (exercise 9.5 in Roman's Lattices and Ordered Sets, p.203) that the lattice $M_3$ is simple, meaning that the only congruences on $M_3$ are the trivial ones (the 'equality' ...
2
votes
1answer
118 views

ideals of a ring form a modular lattice

We know that if $M$ is a left $R$-module, then $(\{\text{submodules of }M\},\subseteq)$ is a modular lattice. Taking $M\!=\!R$, we deduce that $(\{\text{ideals of }R\},\subseteq)$ is a modular ...