Boolean ideal
Let $A$ be a Boolean algebra^{} and $B$ a subset of $A$. The following are equivalent^{}:

1.
If $A$ is interpreted as a Boolean ring^{}, $B$ is a ring ideal.

2.
If $A$ is interpreted as a Boolean lattice, $B$ is a lattice ideal.
Before proving this equivalence, we want to mention that a Boolean ring is equivalent to a Boolean lattice, and a more commonly used terminology is a Boolean algebra, which is also valid, as it is an algebra over the ring of integers^{}. The standard way of characterizing the ring structure^{} from the lattice^{} structure is by defining $a+b:=({a}^{\prime}\wedge b)\vee (a\wedge {b}^{\prime})$ (called the symmetric difference^{}) and $a\cdot b=a\wedge b$. From this, we can “solve” for $\vee $ in terms of $+$ and $\cdot $: $a\vee b=a+b+a\cdot b$.
Proof.
First, suppose $B$ is an ideal of the “ring” $A$. If $a,b\in B$, then $a\vee b=a+b+a\cdot b\in B$. Suppose now that $a\in B$ and $c\in A$ with $c\le a$. Then $c=c\wedge a=c\cdot a\in B$ as well. So $B$ is a lattice ideal of $A$.
Next, suppose $B$ is an ideal of the “lattice” $A$. If $a,b\in B$, then both ${a}^{\prime}\wedge b$ and $a\wedge {b}^{\prime}$ are in $B$ since the first one is less than or equal to $b$ and the second less than or equal to $a$, so their join is in $B$ as well, this means that $ab=a+b=({a}^{\prime}\wedge b)\vee (a\wedge {b}^{\prime})\in B$. Furthermore, if $a\in B$ and $c\in A$, then $c\cdot a=c\wedge a\le a\in B$ as well. As a result, $B$ is a ring ideal of $A$. ∎
A subset of a Boolean algebra satisfying the two equivalent conditions above is called a Boolean ideal, or an ideal for short. A prime Boolean ideal is a prime lattice ideal, and a maximal Boolean ideal is a maximal lattice ideal. Again, these notions and their ring theoretic counterparts match exactly. In fact, one can say more about these ideals in the case of a Boolean algebra: prime ideals^{} are precisely the maximal ideals^{}. If $A$ is a Boolean ring, and $M$ is a maximal ideal of $A$, then $A/M$ is isomorphic to $\{0,1\}$.
Remark. The dual notion of a Boolean ideal is a Boolean filter, or a filter for short. A Boolean filter is just a lattice filter of the Boolean algebra when considered as a lattice. To see the connection between a Boolean ideal and a Boolean lattice, let us define, for any subset $S$ of a Boolean algebra $A$, the set ${S}^{\prime}:=\{{a}^{\prime}\mid a\in S\}$. It is easy to see that ${S}^{\prime \prime}=S$. Now, if $I$ is an ideal, then ${I}^{\prime}$ is a filter. Conversely, if $F$ is a filter, ${F}^{\prime}$ is a ideal. In fact, given any Boolean algebra, there is a Galois connection
$$I\mapsto {I}^{\prime},F\mapsto {F}^{\prime}$$ 
between the set of Boolean ideals and the set of Boolean filters in $A$. In addition^{}, $I$ is prime iff ${I}^{\prime}$ is. As a result, a filter is prime iff it is an ultrafilter (maximal filter).
Title  Boolean ideal 

Canonical name  BooleanIdeal 
Date of creation  20130322 17:01:59 
Last modified on  20130322 17:01:59 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03G05 
Classification  msc 03G10 
Related topic  BooleanRing 
Defines  Boolean filter 
Defines  prime Boolean ideal 
Defines  maximal Boolean ideal 