Boolean algebras, Stone theorem and being isomorphic to a field of sets

Question

I'm a little bit confused about duality between boolean algebras and topological spaces or sets. I know the following theorem (which is due to Stone, as far as I know):

Every boolean algebra $B$ is isomorphic to the algebra of clopen sets of its Stone space (denote this by $B \cong CO(S(B))$.

Q1. Is it also true that given some zero dimensional, compact, Hausdorff space $X$ one have a homeomorphism $X \cong S(CO(X))$?

Q2 Is it true that this theorem establishes an equivalence of catgeories of Boolean algebras and zero dimensional, compact Hausdorff spaces?

From the other hand, there is also a following theorem:

Every Boolean algebra is isomorphic to a field of sets.

Q3 In some notes I found this theorem but with additional assumption that the algebra has to be complete and atomic. I wonder what is the difference: my guess is that the above theorem states that given a boolean algebra $B$ one can find a set $X$ and an embedding of $B$ into $2^X$. And there is a result which states that this embedding is in fact isomorphism provided that $B$ is complete and atomic. Please correct me if I'm wrong or confirm if I'm right.

Q4Do we also have an equivalence of categories of complete atomic Boolean algebras and sets?

And finally I would like to know some historical context and general framework: as far as I know, the theorem of Stone is rather deep result. How about the other cited results, are they also deep or rather simple observations, are they also due to Stone and what is the chronology?

Forgive me such an elaborate question but I would like to have a clear picture: maybe also somebody will find this discussion as a good occasion for the overview of dualities involving boolean algebras.

Answer 1

Q1. Yes, it is true.

Q2. Yes, but it is a duality or a contravariant equivalence since the functors between the categories are contravariant.

Q3. The embedding is $B \cong CO(S(B)) ⊆ \mathcal{P}(S(B))$, but if $B$ is infinite, then there are free ultrafilters on $B$ and a singleton containing a free ultrafilter is never in image of $B$ via this embedding. (if $p$ is an ultrafilter on $B$ and there is $b ∈ B$ such that $p$ is the only ultrafilter containing $b$, then $b$ is an atom and $p$ is principal and generated by $b$).

The result about an isomorphism between $B$ and $\mathcal{P}(X)$ for some $X$ considers a different homomorphism. Namely, $h: B \to \mathcal{P}(At(B))$ assigning to each $b ∈ B$ the set of atoms under $b$. This homomorphism is injective iff $B$ is atomic, and if $B$ is complete, then it is surjective.

Q4. Yes, but it is again a duality / contravariant equivalence. And the morphisms on complete Boolean algebras are complete. Consider the contravariant powerset functor $\mathcal{P}(f): \mathcal{P}(Y) \to \mathcal{P}(X)$ defined as $\mathcal{P}(f)(B) = f^{-1}[B]$ for $f: X \to Y$, and a contravariant functor $At(h): At(B) \to At(A)$ defined as $At(h)(b) = \bigwedge h^{-1}[\uparrow b]$ for $h: A \to B$ a complete homomorphism of complete atomic Boolean algebras.