Let $\mathsf{Stone}$ denote the category of Stone spaces (compact, totally disconnected Hausdorff spaces) and continuous maps. The forgetful functor $U : \mathsf{Stone} \to \mathsf{Set}$ has a left adjoint $F : \mathsf{Set} \to \mathsf{Stone}$ which maps a set $X$ to the space $F(X)$ of ultrafilters on $X$ (i.e. the Stone-Čech compactification of the discrete space $X$).

Question. Is $F$ comonadic?

I have tried to use Beck's monadicity criterion in its dualized form: $F$ is comonadic iff $F$ is a left adjoint ($\checkmark$), $F$ is conservative ($\checkmark$) and $\mathsf{Set}$ has ($\checkmark$) and $F$ preserves equalizers of $F$-split pairs. Only the last point is unclear to me. I hope that it is true, and that actually $F$ preserves much more equalizers.

Here is some background: In his (fantastic!) paper on codensity, Tom Leinster mentions an interesting monad on the category of commutative rings $\mathsf{CRing}$ defined by $T(A) = \prod_{\mathfrak{p} \in \mathrm{Spec}(A)} Q(A/\mathfrak{p})$ and asks for a description for its category of algebras. Each $T(A)$ is von Neumann regular, so I thought it would be good to first look at the restriction to the category of Boolean rings and determine the algebras for the induced monad. By Stone duality, this is monad is dual to the comonad $FU : \mathsf{Stone} \to \mathsf{Stone}$ from above, and comonadicity of $F$ would imply that the boolean $T$-algebras are just products of copies of $\mathbb{F}_2$.


2 Answers 2


I believe the answer is yes: via the ultrafilter functor, $Set$ is comonadic over $Stone$ (and also over $CpctHaus$). Instead of the plain comonadicity theorem, let's use the crude comonadicity theorem, which says the following:

Let $F: C \to D$ be a conservative left adjoint such that $C$ has and $F$ preserves equalizers of coreflexive pairs. Then $F$ is comonadic.

So consider a coreflexive pair in $Set$, i.e. we have $f,g: A \rightrightarrows B$ which are both sections of a map $p: B \to A$. Let $E = \{a \in A \mid f(a) = g(a)\}$ be the equalizer. Let $\mathcal F$ be an ultrafilter on $A$ such that $f_\ast(\mathcal F) = g_\ast(\mathcal F)$, i.e.

$(\ast)$ For all $V \subseteq B$ we have $f^{-1}(V) \in \mathcal F \Leftrightarrow g^{-1}(V) \in \mathcal F$.

It will suffice to show that $E \in \mathcal F$. Suppose not. Then $(A \setminus E) \in \mathcal F$. Let $V \subseteq B$ be the set $V = \{f(a) \mid a \in (A \setminus E)\}$. Then we have $f^{-1}(V) = (A \setminus E) \in \mathcal F$, but $g^{-1}(V) = \emptyset \not \in \mathcal F$, contradicting $(\ast)$. Thus we must have $E \in \mathcal F$ after all, and so the ultrafilter functor preserves this coreflexive equalizer.

Note that the same proof shows that the ultrafilter functor $\beta: Set \to \mathcal C$ preserves coreflexive equalizers for any of $\mathcal C \in \{ Stone, CpctHaus, Set\}$. In particular, the ultrafilter functor exhibits $Set$ as comonadic over the category of compact Hausdorff spaces as well as over the category of Stone spaces.

  • $\begingroup$ Thanks! The proof for $g^{-1}(V)=\emptyset$ uses $p$. Why is it sufficient to prove $E \in \mathcal{F}$? Why need an ultrafilter $\mathcal{F}'$ on $E$ with $\mathcal{F}= i_* \mathcal{F}'$, where $i : E \to A$ is the inclusion. Do you take $\mathcal{F}' = \{T \cap E : T \in \mathcal{F}\}$? $\endgroup$ Feb 7, 2020 at 21:57
  • 1
    $\begingroup$ @MartinBrandenburg Exactly. This hold in general -- if $E \to A$ is any injection of sets, then $\beta E \to \beta A$ is also an injection, whose image consists of precisely those ultrafilters containing the image of $E$. And yes, the use of $p$ is crucial -- I think of $p$ as corresponding to a decomposition $B = \amalg_{a \in A} p^{-1}(a)$, and $f,g$ as choice functions selecting $f(a),g(a) \in p^{-1}(a)$ for each $a$. Then $E$ is the set where the choice functions agree, and $A \setminus E$ is the set where they disagree. $\endgroup$
    – Tim Campion
    Feb 7, 2020 at 21:59

Just a couple of comments in relation to the $T$-algebras mentioned.

Any product $P=\prod_{\beta\in B}k_{\beta}$ of fields $k_{\beta}$ is a $T$-algebra. Indeed, writing $ult(B)$ for the set of ultrafilters on $B$, one has $T(P)=\prod_{g\in ult(B)}P/g$, the product of all ultraproducts of the $k_{\beta}$. The structure map $T(P)\rightarrow P$ is given by the projection $\prod_{g\in ult(B)}P/g\rightarrow\prod_{\beta\in B}P/g_{\beta}$, where $g_{\beta}$ is the principal ultrafilter corresponding to $\beta$. Here, of course, $P/g_{\beta}\cong k_{\beta}$. And $T(T(P))=\prod_{G\in ult(ult(B))}T(P)/G$ has a natural projection onto $\prod_{g\in ult(B)}T(P)/G_{g}$, where $G_{g}$ is the principal ultrafilter on $ult(B)$ corresponding to $g\in ult(B)$. And the ultraproduct $T(P)/G_{g}$ equals the ultraproduct $P/g$. This gives $\mu_{P}:T(T(P))\rightarrow T(P)$.

Conversely, if $A$ is a $T$-algebra, put $P:=T(A)=\prod_{\mathfrak{p}\in Spec(A)}Q(A/\mathfrak{p})$, a product of fields. $P$ is VNR (von Neumann regular $\iff$ absolutely flat $\iff$ zero dimensional and reduced $\iff$ zero dimensional subring of a product of fields). We have maps $\eta_{A}:A\rightarrow P$ and $h:P\rightarrow A$ with composition $1_{A}$, so that $A$ is a quotient ring of $P$ via $h$. Hence $A$ is zero dimensional. Therefore, in fact $Q(A/\mathfrak{p})=A/\mathfrak{p}$ for all prime ideals $\mathfrak{p}$ of $A$. And as $\eta_{A}$ is injective, the ring $A$ is also VNR.

Assume that $A$ is not a product of fields. Put $B:=Spec(A)$ and $k_{\beta}:=A/\mathfrak{p}$ for $\beta=\mathfrak{p}\in B$, so that we have $P=\prod_{\beta\in B}k_{\beta}$ as above. $A$ is a quotient of $P$, hence $A=P/f$, a so called reduced product, for a suitable filter $f$ on the set $B$. That is, $A$ equals $P$ modulo the equivalence relation $(x_{\beta})_{\beta}\equiv (y_{\beta})_{\beta}$ iff $\{\beta\in B\mid x_{\beta}=y_{\beta}\}\in f$. This results from the correspondence between ideals of $P$ and filters on $B$. Now $f$ cannot be a principal filter: if $f=\{E\subseteq B\mid D\subseteq E\}$ for some subset $D\subseteq B$, then clearly we have $A=P/f\cong\prod_{\beta\in D}k_{\beta}$, a product of fields after all. In particular, it follows that the set $B$ is not finite.

Using the definition of $T$, we find $P=T(A)=\prod_{g\in ult_{f}(B)}P/g$, where $ult_{f}(B)=\{g\in ult(B)\mid f\subseteq g\}$. This product must be isomorphic to $\prod_{\beta\in B}k_{\beta}$. I do not know what can be concluded from two products of sets of fields being isomorphic, other than that each field in one set must be an ultraproduct of the fields in the other set, and conversely.

There does not appear to be a "natural" structure map $\prod_{g\in ult_{f}(B)}P/g\rightarrow P/f$ for the $T$-algebra $A$, even when all fields $k_{\beta}$ are identical. But that is not to say that such $T$-algebras cannot exist, of course...

  • $\begingroup$ As mentioned in my comment at the ncafé, the $T$-algebra structure on a product of fields is trivial using that $T$-algebras have products. No ultrafilters are needed. $\endgroup$ Feb 8, 2020 at 19:30

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