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 $Boo Rng$.

- $R$ 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!

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.

Back in the day, the term ‘ring’ meant (more often than now is the case) a possibly *non*unital 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.

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.

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 04:29:29. See the history of this page for a list of all contributions to it.