nLab
Boolean ring

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 $\mathrm{Boo}\mathrm{Rng}$.

Properties

• $R$ has characteristic $2$ (meaning that $x+x=0$ for all $x$):

  $$2x=4x-2x=4x^2-2x=(2x{)}^2-2x=2x-2x=0.$$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 $xy=yx$ for all $x,y$):

  $$yx=x+y-x-y+yx=(x+y{)}^2-x^2-y^2+yx=x^2+xy+yx+y^2-x^2-y^2+yx=xy+2yx=xy.$$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 $x+xy+y$. Then:

• $\vee$ is commutative (as it would be in any commutative ring) and idempotent:

  $$x\vee x=x+x^2+x=3x=x.$$x \vee x = x + x^2 + x = 3 x = x .

• The absorption law ($x\vee xy=x$) also holds:

  $$x\vee xy=x+x^{2}y+xy=x+2xy=x.$$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 ($x(y\vee z)=xy\vee xz$):

  $$x(y\vee z)=x(y+yz+z)=xy+xyz+xz=xy+x^{2}yz+xz=xy+(xy)(xz)+xz=xy\vee xz.$$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 $x(\neg x)=0$):

  $$x(\neg x)=x(x+1)=x^2+x=2x=0.$$x (\neg{x}) = x (x + 1) = x^2 + x = 2 x = 0 .

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

• But don't bother, because $\neg x$ is also an op-pseudocomplement of $x$:

  $$x\vee \neg x=x+x(x+1)+(x+1)=x^2+3x+1=x+x+1=1.$$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 $\{\top ,\perp \}$ 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!

Terminology

Back in the day, the term ‘ring’ meant 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 $\mathrm{Ab}$) and ‘Boolean algebra’ (meaning an idempotent monoid in $\mathrm{Ab}$), 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.

Analogues

Inasmuch as a semilattice is a commutative idempotent monoid, a Boolean ring may be defined as a semilattice in $\mathrm{Ab}$. However, with Boolean rings, we do not need to hypothesise commutativity; it follows.