Could someone explain what an atom in Boolean algebra means? I am acquainted with ring theory and group theory but not Boolean algebra. As far as I can tell from browsing around, it is something like a generator, but then again not exactly... Your help will be very much appreciated. Thanks in advance.

Added: It would be particularly helpful if it could be placed in the context of rings or groups. I think there are such entities as Boolean rings...


You can think of Boolean algebras as something which looks a little bit like a power set of some set $X$. This is naturally equipped with complement, union and intersection operations which corresponds to $-,\lor,\land$ operations in the Boolean algebra.

An atom is something which cannot be decomposed into two proper subsets, this is much like a singleton which cannot be written as a union of two strictly smaller subsets.

Indeed, when we return to the world of Boolean algebras, $a\in B$ is an atom if whenever $b\lor c=a$ either $b=a$ or $b=0$.


An atom is a minimal non-zero element.

$x$ is an atom iff for every $y$, either $y\wedge x = x$ or $y\wedge x = 0$.

In some Boolean algebras, there are no atoms. These are called atomless Boolean algebras. An example is the countably infinite Boolean algebra freely generated by $a_1,a_2,a_3,\ldots$. Every element can be reached by taking meets, joins, and complements of finitely many of these. No element $x$ can be an atom since if $x$ can be formed by taking meets, joins, and complements of $a_1,\ldots,a_n$, then $x\wedge a_{n+1}$ is neither $x$ nor $0$.

In some Boolean algebras, for every element $a$, there is some atom $x\le a$. These are called atomic Boolean algebras. For example the set of all subsets of any set is a Boolean algebra in which every singleton set is an atom.

In some Boolean algebras, there are atoms and there are some elements with no atoms below them. For example the set of all clopen subsets of $C\cup\{2\}$, where $C$ is the Cantor set.

Some people use the term "non-atomic" instead of "atomless". I think that's a bad idea, because "non-atomic" is easily confused with "not atomic", which is not the same thing. My third example above is neither atomless nor atomic. I would rather not have to explain that it's neither atomic nor non-atomic.


If $B$ is a Boolean algebra then $x \in B$ is an atom if, for all $y \in B$, either $x \wedge y = x$ or $x \wedge y = 0$. Intuitively, it's a sort of minimal element: the way to think is "if $x$ is an atom and $0 \le y \le x$ then either $y=0$ or $y=x$".

In a Boolean ring $R$, we can think of $\wedge$ as multiplication, so we have that if $x$ is an atom then $xy=x$ or $xy=0$ for all $y \in R$.

For example, let $X$ be a set, let $R = \mathcal{P}X$ be its power set and define set multiplication by intersection and addition by symmetric difference. Then the zero element is $\varnothing$, and the atoms are the elements $A \subseteq X$ such that whenever $Y \subseteq X$ either $A \cap Y = A$ or $A \cap Y = \varnothing$. That is, the atoms of $\mathcal{P}X$ are precisely the singletons $\{ x \}$, and $\varnothing$.

  • $\begingroup$ Would you mind putting it in the context of a Boolean ring? In other words, what are the atoms of a Boolean ring? $\endgroup$ – harvey Feb 16 '12 at 15:33
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    $\begingroup$ It's the same. $x \wedge y$ is the "product" in the ring, "$0$" is the zero of the ring. Now read Clive's first sentence. $\endgroup$ – GEdgar Feb 16 '12 at 15:39
  • $\begingroup$ @harvey: I've added a concrete example. $\endgroup$ – Clive Newstead Feb 16 '12 at 15:47
  • $\begingroup$ @CliveNewstead: Thank you very much! $\endgroup$ – harvey Feb 16 '12 at 17:18

If $(B,\cup,\cap,\lnot,0,1)$ is a Boolean algebra, the Boolean ring is just the same algebra with $\cap,0,$ and $1$ corresponding to $\times,0,$ and $1$, respectively, in the ring, and $x+y = (x\cap \lnot y)\cup(\lnot x\cap y)$. From this view, the more useful thing to look at is the "co-atoms" - the elements of the Boolean algebra whose complements are atoms.

That's because, when we consider a Boolean algebra as a ring, the ideals of the ring turn out to be sets $I$ such that, if $x,y\in I$ then $x\cup y\in I$, and, if $x\in I$, $y$ any element, $x\cap y\in I$. That such sets are ideals is easy to show. To prove that all ideals of the Boolean ring satisfy this property, you need only prove the following formula: $$x + y + (x\cap y) = x\cup y$$

If $x$ is any element, the ring ideal, $\left<x\right>$, generated by $x$, is the set of $y$ such that $x\cap y = y$.

Now, it turns out that $x$ is a co-atom if and only if $\left<x\right>$ is a maximal ideal of the ring.

So $a$ is an atom of your Boolean algebra if and only if $\left<\lnot a\right>$ is a maximal ideal of the Boolean ring. Then $B/\left<\lnot a\right>\cong \mathbb Z/\left<2\right>$.

Not all maximal ideals in a boolean algebra necessarily correspond to atoms, however.


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