Boolean rings

# Boolean rings

## Definitions

A ring with unit $R$ is Boolean if the operation of multiplication is idempotent; that is, $x^2 = x$ for every element $x$. Although the terminology would make sense for rings without unit, the common usage assumes a unit.

Boolean rings and the ring homomorphisms between them form a category $Boo Rng$.

## Properties

• $R$ has characteristic $2$ (meaning that $x + x = 0$ for all $x$):
$2 x = 4 x - 2 x = 4 x^2 - 2 x = (2 x)^2 - 2 x = 2 x - 2 x = 0 .$
• $R$ is commutative (meaning that $x y = y x$ for all $x, y$):
$y x = x + y - x - y + y x = (x + y)^2 - x^2 - y^2 + y x = x^2 + x y + y x + y^2 - x^2 - y^2 + y x = x y + 2 y x = x y .$

Define $x \vee y$ to mean $x + x y + y$. Then:

• $\vee$ is commutative (as it would be in any commutative ring) and idempotent:
$x \vee x = x + x^2 + x = 3 x = x .$
• The absorption law ($x \vee x y = x$) also holds:
$x \vee x y = x + x^2 y + x y = x + 2 x y = x .$

We could now prove the other absoprtion law to conclude that $R$ is a lattice using multiplication as meet and $\vee$ as join.

• But in fact, we can skip that step since it follows the distributive law ($x (y \vee z) = x y \vee x z$):
$x (y \vee z) = x (y + y z + z) = x y + x y z + x z = x y + x^2 y z + x z = x y + (x y) (x z) + x z = x y \vee x z .$

Thus $R$ is a distributive lattice.

Next define $\neg{x}$ to be $x + 1$. Then:

• $\neg{x}$ is a pseudocomplement of $x$ (meaning that $x (\neg{x}) = 0$):
$x (\neg{x}) = x (x + 1) = x^2 + x = 2 x = 0 .$

By relativising from $x + 1$ to $x y + x + 1$, we can show that $R$ is a Heyting algebra.

• But don't bother, because $\neg{x}$ is also an op-pseudocomplement of $x$:
$x \vee \neg{x} = x + x (x + 1) + (x + 1) = x^2 + 3 x + 1 = x + x + 1 = 1 .$

Therefore, $\neg{x}$ is a complement of $x$, and $R$ is a Boolean algebra.

Conversely, starting with a Boolean algebra $R$ (with the meet written multiplicatively), let $x + y$ be $x (\neg{y}) \vee (\neg{x}) y$ (which is called exclusive disjunction in $\{\top,\bot\}$ and symmetric difference in $2^X$). Then $R$ is a Boolean ring.

In fact, we have:

Boolean rings and Boolean algebras are equivalent.

This extends to an equivalence of concrete categories; that is, given the underlying set $R$, the set of Boolean ring structures on $R$ is naturally (in $R$) bijective with the set of Boolean algebra structures on $R$.

Here is a very convenient result: although a Boolean ring $R$ is a rig in two different ways (as a ring or as a distributive lattice), these have the same concept of ideal!

## Examples

The most common example is the power set $P(S)$ of any set $S$. It is a Boolean ring with symmetric difference as the addition and the intersection of sets as the multiplication.

## Terminology

Back in the day, the term ‘ring’ meant (more often than now is the case) a possibly nonunital ring; that is a semigroup, rather than a monoid, in Ab. This terminology applied also to Boolean rings, and it changed even more slowly. Thus older books will make a distinction between ‘Boolean ring’ (meaning an idempotent semigroup in $Ab$) and ‘Boolean algebra’ (meaning an idempotent monoid in $Ab$), in addition to (or even instead of) the difference between $+$ and $\vee$ as fundamental operation. This distinction survives most in the terminology of $\sigma$-rings and $\sigma$-algebras.

###### Remark

We pause to note that “idempotent monoid” doesn’t make sense a priori in a general monoidal category: generally speaking the idempotency axiom would be expressed by an equation

$1_M = \left(M \stackrel{\delta_M}{\to} M \otimes M \stackrel{mult}{\to} M \right)$

where $\delta_M$ is an appropriate diagonal map, not generally available for monoidal categories. But in a concrete monoidal category $\mathbf{M}$ where the underlying-set functor $U: \mathbf{M} \to Set$ is lax monoidal, the meaning is that the evident equation holds:

$1_{U(M)} = \left(U(M) \stackrel{\delta}{\to} U(M) \times U(M) \stackrel{\lambda}{\to} U(M \otimes M) \stackrel{U(mult)}{\to} U(M) \right)$

where $\lambda$ is a lax monoidal constraint.

## Analogues

Inasmuch as a semilattice is a commutative idempotent monoid, a Boolean ring may be defined as a semilattice in $Ab$. However, with Boolean rings, we do not need to hypothesize commutativity; it follows. That is, any idempotent monoid in $Ab$ is commutative; indeed, any idempotent magma in $Ab$ is commutative.

Last revised on June 10, 2018 at 00:29:29. See the history of this page for a list of all contributions to it.