# The Unapologetic Mathematician

## Boolean Rings

A “Boolean ring” is a commutative ring $R$ with the additional property that each and every element is idempotent. That is, for any $r\in R$ we have $r^2=r$. An immediate consequence of this axiom is that $r+r=0$, since we can calculate

\displaystyle\begin{aligned}0&=(r+r)-(r+r)\\&=(r+r)^2-(r+r)\\&=r^2+r^2+r^2+r^2-(r+r)\\&=r+r+r+r-(r+r)\\&=r+r\end{aligned}

The typical example we care about in the measure-theoretic context is a ring of subsets of some set $X$, with the operation $E\Delta F$ for addition and $E\cap F$ for multiplication. You should check that these operations satisfy the axioms of a Boolean ring. Since this is our main motivation, we will just consistently use $\Delta$ and $\cap$ to denote addition and multiplication in Boolean rings, whether they arise from a measure theoretic context or not. From here it looks a lot like set theory, but keep in mind that the objects we’re looking at may have nothing to do with sets.

We can use these operations to define the other common set-theoretic operations. Indeed

$\displaystyle E\cup F=(E\Delta F)\Delta(E\cap F)$

and

$\displaystyle E\setminus F=E\Delta(E\cap F)$

and we can then define orders in the usual manner: $E\subseteq F\Leftrightarrow E\cap F=E$.

As usual, the union of two elements is the “smallest” (with respect to this order) element above both of them, and the intersection of two elements is the “largest” element below both of them. The same goes for any finite number of elements, but if we try to move to an infinite number of elements there is no guarantee that there is any element above or below all of them, much less that such an element is unique. A “Boolean $\sigma$-ring” is a Boolean ring so that every countably infinite set of elements has a union. In this case, it is immediately true that any countably infinite set of elements has an intersection as well. The typical example, of course, is a $\sigma$-ring of subsets of a set $X$.

A “Boolean algebra” is a Boolean ring for which there is some element $X\neq0$ so that $E\subseteq X$ for all elements $E$. A “Boolean $\sigma$-algebra” is both a Boolean $\sigma$-ring and a Boolean algebra.

In the obvious way we have a full subcategory $\mathcal{B}oolean$ of the category of rings. It contains full subcategories of Boolean $\sigma$-rings, Boolean algebras, and Boolean $\sigma$-algebras.

August 4, 2010 - Posted by | Algebra, Ring theory

## 11 Comments »

1. […] on Boolean Rings and Measure Rings We’re not just interested in Boolean rings as algebraic structures, but we’re also interested in real-valued functions on them. Given a […]

Pingback by Functions on Boolean Rings and Measure Rings « The Unapologetic Mathematician | August 5, 2010 | Reply

2. […] of Boolean Rings We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite […]

Pingback by Completeness of Boolean Rings « The Unapologetic Mathematician | August 10, 2010 | Reply

3. […] Boolean Rings It will be useful to have other ways of showing that a given function between boolean rings is a morphism. The definition, of course, is that and , since and here denote our addition and […]

Pingback by Specifying Morphisms Between Boolean Rings « The Unapologetic Mathematician | August 11, 2010 | Reply

4. […] Theorem I Today we start in on the representation theorem proved by Marshall Stone: every boolean ring is isomorphic (as a ring) to a ring of subsets of some set . That is, no matter what looks like, […]

Pingback by Stone’s Representation Theorem I « The Unapologetic Mathematician | August 18, 2010 | Reply

5. where is $\Delta$ defined?

Comment by isomorphismes | February 23, 2014 | Reply

6. I guess Δ is the https://en.wikipedia.org/wiki/Symmetric_difference ?

Comment by isomorphismes | January 2, 2015 | Reply

7. Yes, sorry I missed that comment before; that is correct.

Comment by John Armstrong | January 2, 2015 | Reply

• No problem. Thanks! This XOR is much crazier than I realised … thanks for walking me through it.

Comment by isomorphismes | January 3, 2015 | Reply

8. Why does existence of a unique intersection follow from existence of a unique union?

Comment by isomorphismes | January 3, 2015 | Reply

9. DeMorgan’s Laws hold for infinite unions and intersections as well. The intersection of any collection is the complement of the union of their complements.

Comment by John Armstrong | January 3, 2015 | Reply