Correspondence between ideals and congruences of Boolean algebras I'm trying to prove that 
$con(B) \simeq I(B)$ where $B$ is a Boolean algebra and $con(B)$ is the lattice of congruences on $B$ and $I(B)$ is the lattice of ideals on $B$
I found a map s.t. for a congruence $\theta$, $\theta \mapsto \boldsymbol 0/\theta$
I'm having difficulty concretely showing it though.
I found a problem asking to show that every congruence on a Boolean algebra is of the form $a \theta b $ iff $(\exists c \in J:\text{Ideal of B})(a \vee c = b \vee c)$
 A: We need to show that there is a bijective homomorphism between these two lattices. You have defined two maps:


*

*$f \colon con(B) \to I(B), \theta \mapsto f(\theta) = 0/\theta$

*$g \colon I(B) \to con(B), I \mapsto g(I) = \theta$, where $a \theta b$ iff
$\exists c \in I\colon a \vee c = b \vee c$.


We will show that $fg$ and $gf$ are the identity maps and hence $f$ and $g$ are bijections between these lattices.
At first consider $fg$. Fix some ideal $I \in I(B)$ and let $\theta = g(I)$. We need to show that $f(g(I)) = f(\theta) = 0/\theta = I$. If $x \in I$ then $0 \vee x = x \vee x$, so $0\theta x$ and so $x \in 0/\theta$. Coversely, if $x \in 0/\theta$ then there is $c \in I$ with $0 \vee c = x \vee c \Leftrightarrow c = x \vee c \Leftrightarrow x \leqslant c$ and hence $x \in I$, since $I$ is a down-set w.r.t. $\leqslant$.
Now consider $gf$. Fix some congruence $\theta \in con(B)$ and let $I = f(\theta)$. We need to show that $g(f(\theta)) = g(I) = \theta$. If $a g(I) b$ then there is $c \in I = 0/\theta$ with $a \vee c = b \vee c$, but since $\theta$ is a congruence and $0\theta c$ we have $a = (a \vee 0) \theta (a \vee c) = (b \vee c) \theta (b \vee 0) = b$, that is $a \theta b$. If $a\theta b$ then consider $c = a'b \vee ab'$. Note that $c\theta(a'a\vee aa') = 0$ and so $c \in I$. Also $$a \vee w = a \vee a'b \vee ab' = a \vee a'b = a \vee b = b \vee ab' = b \vee ab' \vee a'b = b \vee w,$$
hence $a g(I) b$.
It is left to show that one of these maps is a lattice homomorphism. Since these lattices are complete it is enough to check that $f$ preserves only the arbitrary intersections, which is not hard to do by definition, and I'll leave it up to you. If you have any questions, feel free to ask.
