2
$\begingroup$

Let $X$ be a set. Then, $\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} )$ is a Boolean ring. In fact, it is a complete atomic Boolean ring and this, together with the the cofunctor from complete atomic Boolean rings to sets $\operatorname{Mor}_{\mathbf{Ring}}({-},\mathbb{Z}/2\mathbb{Z})$ yields a coequivalence of categories between the category of sets and the category of complete atomic Boolean rings---see pg. 26 of Khalkhali's Basic Noncommutative Geometry.

On the other hand, classic Stone Duality says that the category of Boolean rings is coequivalent to the category of Stone spaces, the cofunctors implementing the equivalence being given by $\operatorname{Spec}$ and $\operatorname{Clopen}$ (the functor that takes a topological space to its Boolean algebra of clopen sets)---see Lurie's notes.

I now claim that $\operatorname{Mor}_{\mathbf{Ring}}({-},\mathbb{Z}/2\mathbb{Z})$ is naturally isomorphic to $\operatorname{Spec}({-})$, the natural isomorphism being given by $$ \operatorname{Mor}_{\mathbf{Ring}}(B,\mathbb{Z}/2\mathbb{Z})\ni \phi \mapsto \operatorname{Ker}(\phi )\in \operatorname{Spec}(R). $$ sending a homomorphism $B\rightarrow \mathbb{Z}/2\mathbb{Z}$ to its kernel. (I am regarding them as functors from the category of Boolean rings into $\mathbf{Set}$.) The quotient of a Boolean ring is Boolean ring. Furthermore, the only Boolean ring that is an integral domain is $\mathbb{Z}/2\mathbb{Z}$. Thus, for $\mathfrak{p}\in \operatorname{Spec}(R)$, we must have that $R/\mathfrak{p}\cong \mathbb{Z}/2\mathbb{Z}$. $$ \operatorname{Spec}(B)\ni \mathfrak{p}\mapsto (B\rightarrow B/\mathfrak{p})\in \operatorname{Mor}_{\mathbf{Ring}}(B,\mathbb{Z}/2\mathbb{Z}) $$ then gives us an inverse to $\phi \mapsto \operatorname{Ker}(\phi )$. I will spare you the check of naturality.

Putting these facts together, we have $$ X\cong \operatorname{Mor}_{\mathbf{Ring}}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ),\mathbb{Z}/2\mathbb{Z})\cong \operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} )), $$ $\cong$ here meaning ``naturally isomorphic as endofunctors on $\mathbf{Set}$''. However, Stone Duality says that the right-hand side comes equipped with a canonical Stone (in fact, Stonean) topology (namely the Zariski topology). This suggests that every set has a canonical Stonean topology. I can only think of a couple of topologies one can define on any set (discrete, indiscrete, cofinite) and none of them seem to be Stonean in general.

I imagine there is a flaw in my understanding somewhere, but I'm afraid I am not seeing where. Can someone help point out what I'm missing?

Additionally, after clearing that up, it would be nice to know the following.

Is there a concrete characterization of the topological spaces that are of the form $\operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ))$?

For example, von Neumann algebras also give Boolean rings (their Boolean algebras of projections), and there is a relatively concrete description of the spaces which correspond to these Boolean rings. Can do we do the same for those Boolean rings that arise as the power set of some set?

$\endgroup$
4
$\begingroup$

this, together with the the cofunctor from complete atomic Boolean rings to sets $\operatorname{Mor}_{\mathbf{Ring}}({-},\mathbb{Z}/2\mathbb{Z})$ yields a coequivalence of categories between the category of sets and the category of complete atomic Boolean rings

This is incorrect. To get a coequivalence between the category of sets and the category of complete atomic Boolean rings, your morphisms in the category of complete atomic Boolean rings must be complete Boolean homomorphisms (i.e., they must preserve infinite joins and meets as well). These are not the same as ring-homomorphisms (they are only certain very special ring-homomorphisms).

As a result, you don't have a natural isomorphism $$X\cong \operatorname{Mor}_{\mathbf{Ring}}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ),\mathbb{Z}/2\mathbb{Z}).$$ You just have a natural isomorphism $$X\cong \operatorname{Mor}_{\mathbf{CompleteBool}}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ),\mathbb{Z}/2\mathbb{Z})$$ and a natural injection $$\operatorname{Mor}_{\mathbf{CompleteBool}}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ),\mathbb{Z}/2\mathbb{Z})\to \operatorname{Mor}_{\mathbf{Ring}}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ),\mathbb{Z}/2\mathbb{Z}).$$ Putting this together, you get a natural injection from $X$ to the Stone space $\operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} )).$ This induces a canonical topology on $X$, but this topology need not be a Stone topology because the image of this injection need not be closed. Indeed, it turns out that the canonical topology on $X$ is just the discrete topology, and this canonical injection can be naturally identified with the canonical inclusion $X\to \beta X$, where $\beta X$ is the Stone-Cech compactification of $X$ with the discrete topology. That is, the spaces of this form are exactly the free compact Hausdorff spaces.

For an explicit description, $\operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ))$ is the space of ultrafilters on the set $X$. Indeed, $\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\}$ is just the power set $\mathcal{P}(X)$ with its usual Boolean structure, and a map $\mathcal{P}(X)\to \mathbb{Z}/2\mathbb{Z}$ is a Boolean homomorphism iff it is the characteristic function of an ultrafilter on $X$. So $\operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ))$ is the set of (characteristic functions of) ultrafilters on the set $X$. A bit of definition chasing gives that the inclusion $X\to \operatorname{Spec}(\operatorname{Mor}_{\mathbf{Set}}(X,\{ 0,1\} ))$ just sends a point $x\in X$ to the principal ultrafilter at $x$.

For another characterization of these spaces, you can just translate "complete atomic Boolean ring" via Stone duality. Complete Boolean rings correspond to Stonean spaces, and a Boolean ring is atomic iff isolated points are dense in its Stone space. So these spaces are exactly the Stonean spaces in which isolated points are dense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.