# Boolean ideal

1. 1.

If $A$ is interpreted as a Boolean ring  , $B$ is a ring ideal.

2. 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 $c\leq a$. 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 $a-b=a+b=(a^{\prime}\wedge b)\vee(a\wedge b^{\prime})\in B$. Furthermore, if $a\in B$ and $c\in A$, then $c\cdot a=c\wedge a\leq a\in B$ 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

 $I\mapsto I^{\prime},\qquad F\mapsto F^{\prime}$

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 BooleanIdeal 2013-03-22 17:01:59 2013-03-22 17:01:59 CWoo (3771) CWoo (3771) 10 CWoo (3771) Definition msc 03G05 msc 03G10 BooleanRing Boolean filter prime Boolean ideal maximal Boolean ideal