nLab Boolean ring

Boolean rings

Boolean rings


A ring with unit RR is Boolean if the operation of multiplication is idempotent; that is, x 2=xx^2 = x for every element xx. 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 BooRngBoo Rng.


  • RR has characteristic 22 (meaning that x+x=0x + x = 0 for all xx):
    2x=4x2x=4x 22x=(2x) 22x=2x2x=0. 2 x = 4 x - 2 x = 4 x^2 - 2 x = (2 x)^2 - 2 x = 2 x - 2 x = 0 .
  • RR is commutative (meaning that xy=yxx y = y x for all x,yx, y):
    yx=x+yxy+yx=(x+y) 2x 2y 2+yx=x 2+xy+yx+y 2x 2y 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 xyx \vee y to mean x+xy+yx + x y + y. Then:

  • \vee is commutative (as it would be in any commutative ring) and idempotent:
    xx=x+x 2+x=3x=x. x \vee x = x + x^2 + x = 3 x = x .
  • The absorption law (xxy=xx \vee x y = x) also holds:
    xxy=x+x 2y+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 RR 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(yz)=xyxzx (y \vee z) = x y \vee x z):
    x(yz)=x(y+yz+z)=xy+xyz+xz=xy+x 2yz+xz=xy+(xy)(xz)+xz=xyxz. 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 RR is a distributive lattice.

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

  • ¬x\neg{x} is a pseudocomplement of xx (meaning that x(¬x)=0x (\neg{x}) = 0):
    x(¬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+1x + 1 to xy+x+1x y + x + 1, we can show that RR is a Heyting algebra.

  • But don't bother, because ¬x\neg{x} is also an op-pseudocomplement of xx:
    x¬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, ¬x\neg{x} is a complement of xx, and RR is a Boolean algebra.

Conversely, starting with a Boolean algebra RR (with the meet written multiplicatively), let x+yx + y be x(¬y)(¬x)yx (\neg{y}) \vee (\neg{x}) y (which is called exclusive disjunction in {,}\{\top,\bot\} and symmetric difference in 2 X2^X). Then RR 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 RR, the set of Boolean ring structures on RR is naturally (in RR) bijective with the set of Boolean algebra structures on RR.

Here is a very convenient result: although a Boolean ring RR 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)P(S) of any set SS. 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 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 AbAb) and ‘Boolean algebra’ (meaning an idempotent monoid in AbAb), in addition to (or even instead of) the difference between ++ and \vee as fundamental operation. This distinction survives most in the terminology of σ\sigma-rings and σ\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=(Mδ MMMmultM)1_M = \left(M \stackrel{\delta_M}{\to} M \otimes M \stackrel{mult}{\to} M \right)

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

1 U(M)=(U(M)δU(M)×U(M)λU(MM)U(mult)U(M))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 λ\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 AbAb. However, with Boolean rings, we do not need to hypothesize commutativity; it follows. That is, any idempotent monoid in AbAb is commutative; indeed, any idempotent magma in AbAb 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.