Tagged Questions
2
votes
1answer
98 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 $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...
3
votes
2answers
109 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 ...
2
votes
1answer
78 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
78 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 ...
