Let $I$ be an index set. For each $i \in I$, let $B_i$ be a Boolean algebra. Let $\prod\limits_{i \in I}B_i$ be the direct product of all the $B_i$'s, which is also a Boolean algebra. (That is, the Cartesian product with meet, join, complement all defined componentwise.)
Let $S(B_i)$ be the Stone space corresponding to $B_i$. (The points are the ultrafilters on $B_i$, and a basic closed set is a set of the form {f: f is an ultrafilter on $B_i$, and b $\in$ f}, where $b$ is some element of $B_i$.)
Similarly, let $S(\prod\limits_{i \in I}B_i)$ be the Stone space corresponding to the Boolean algebra $\prod\limits_{i \in I}B_i$.
Here is my question: what is the relationship between $S(\prod\limits_{i \in I}B_i)$, the stone space corresponding to the product algebra, on the one hand, and the product space of all the $S(B_i)$'s, on the other hand? Is there some kind of homeomorphism between them?
The relationship between Boolean algebras and their Stone spaces is a contravariant equivalence of categories (a.k.a. a duality). Thus, $S(\prod_i B_i)$ is the coproduct of the spaces $S(B_i)$ in the category of Stone spaces. Unfortunately the coproduct of Stone spaces isn't very easy to understand in general (if $I$ is finite it coincides with the disjoint union, but when $I$ is infinite it does not).