Show that a finite Boolean algebra is made of its atoms

Question

Let $\mathcal{B}$ a finite boolean algebra.

Only using the calculus rules about logic operators: and ($\land$), or ($\lor$) and not ($\neg$), how do I prove that every element $p \in \mathcal{B}$ is equal to a disjonction of atoms ?

Define an atom as an element $p_a \in \mathcal{B}$ such that for every other element $p \in \mathcal{B}$, either $p_a \land p = p_a$ or $p_a \land p = 0$. Denote $\mathcal{A}$ the set of atoms: $$\mathcal{A} = \{p_a \in \mathcal{B} \mid \forall p \in \mathcal{B}, p_a \land p = p_a \text{ or } 0\} $$ For an element $p \in \mathcal{B}$, I want to show that it is a combination of atoms, more precisely: $$\forall p \in \mathcal{B}, \quad p = \bigvee_{p_a \in E} p_a$$ where: $$E = \{p_a \in \mathcal{A} \mid p_a \land p = p_a\} $$ However, I'm not sure how to show that these two elements in $\mathcal{B}$ are equal only using $\land, \lor, \neg$.

Answer 1

Hints:

1. Boolean algebras can be ordered by $x\le y\ \overset{def}\iff x\lor y=y\ (\iff\ x\land y=x) $.
2. Atoms are exactly the minimal nonzero elements, i.e. $a$ is an atom iff $0<a$ and $0<x\le a\implies x=a$.
3. In a finite Boolean algebra, for every $x$ we can form $x':=\bigvee\{a\text{ atom} :a\le x\}$ and we have $x'\le x$.
4. By finiteness, if $z:=x\land\lnot x'\ne 0$, we can find an atom $a$ below $z$.

Answer 2

I'm tempted to show this from the perspective of correspondence between Boolean algebras and Boolean rings.

Recall that since in a Boolean ring $x^2=x$, so $x^n=x$ for all $n \geq 1$. This implies that a Boolean ring has no nilpotents, for if $x^n=0$ then $x=0$.

Now, by Chinese remainder theorem, any finite ring without nilpotents is isomorphic to a product of fields. Since a Boolean ring is also a $\mathbb F_2$-algebra, these fields necessarily have $\operatorname{char}=2$. Since the equation $x^2=x$ cannot have more than $2$ solutions in a field, we conclude that all these fields are in fact $\mathbb F_2$ and the Boolean ring is isomorphic to $\mathbb F_2^n$.

Now, it is easy to see that the atoms are tuples of the form $(0,0,\dots,0,1,0,\dots,0)$, and any element of the ring is just the sum of these.