Stone’s Representation 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, we can find some space and a ring of subsets of so that as rings.
We start by defining the “Stone space” of a Boolean ring . This is a representable functor, and the representing object is the two-point Boolean ring . That is, , the set of (boolean) ring homomorphisms from to . To be clear, consists of the two points , with the operations for addition and for multiplication, and the obvious definitions of these operations. This is a contravariant functor — if we have a homomorphism of Boolean rings we get a function between the Stone spaces , which takes a function to the function .
The Stone space isn’t just a set, though; it’s a topological space. We define the topology by giving a base of open sets. That is, we’ll give a collection of sets — closed under intersections — which we declare to be open, and we define the collection of all open sets to be given by unions of these sets. For each element , we define a basic set like so:
To see that this collection of sets is closed under intersection, consider two such sets and . I say that the intersection of these sets is the set . Indeed, if and , then
Conversely, if , then . Thus
and so , and . Similarly, . Thus we see that .
In fact, this map from to the basic sets is exactly the mapping we’re looking for! We’ve already seen that our base is closed under intersection, and that the map preserves intersections. I say that we also have . If but , then and . Then
and similarly if but . Thus . Conversely, if , then
and so either and or vice versa. Thus .
So we know that is a homomorphism of (boolean) rings. However, we don’t know yet that it’s an isomorphism. Indeed, it’s possible that has a nontrivial kernel — could be for some . We must show that given any there is some so that .
For a finite boolean ring this is easy: we pick some minimal element and define if and only if . Such a exists because there’s at least one element below — itself is one — and there can only be finitely many so we can just take their intersection. Clearly by definition, and it’s straightforward to verify that is a homomorphism of boolean rings using the fact that is an atom of .
For an infinite boolean ring, things are trickier. We define the set of all functions , not just the ring homomorphisms. This is the product of one copy of for every element of . Since each copy of is a compact Hausdorff space, Tychonoff’s theorem tells us that is a compact Hausdorff space. If is any finite subring of containing , let be the collection of those functions which are homomorphisms when restricted to and for which .
I say that the class of sets of the form has the finite intersection property. That is, if we have some finite collection of finite subrings and the finite subring they generate, then we have the relation
Indeed, is clearly contained in the generated ring . Further, if is a homomorphism on then it’s a homomorphism on each subring .
Okay, so since is a finite boolean ring, the proof given above for the finite case shows that is nonempty. Thus the intersection of any finite collection of sets is nonempty. And thus, since is compact, the intersection of all of the is nonempty.
That is, there is some function which is a homomorphism of boolean rings on any finite boolean subring containing , and with . Given any other two points and there is some finite boolean subring containing , , and , and so we must have and within this subring, and thus within the whole ring. Thus is a homomorphism of boolean rings sending to , which shows that .
Therefore, the map is a homomorphism sending the boolean ring isomorphically onto the identified base of the Stone space .