Let $S$ be a set, we can give a commutative ring structure on $P(S)$ by setting addition as xor and multiplication as intersection.
Can you please help me with some questions about the underlying topological space of $Spec(P(S))$
($a, b, …$ denote subsets of $P(S)$, $x, y, z, …$ denote elements of $s$)
1) is this list of criteria : contains $\emptyset$, every subset, every union, does not contain all of $S$ and maximal with respect to these properties, sufficient for a subset of $P(S)$ to be a prime ideal?
2) is there an easier way to describe the underlying set?
3) Is the topological space totally disconnected?
The below is my attempt at solving these problems:
Writing out the definition of a prime ideal: a prime ideal $p$ must contain $\emptyset$, the xor of any two sets in $p$, and any subset of any set in $p$. Furthermore, if $a \cap b \in p$ then $a \in p$ or $b \in p$.
It must also contain the union of two elements in $p$, since $(b \backslash a) \text{ xor } a = a \cup b$.
Suppose $p \subseteq q$ is an inclusion of prime ideals. Then, let $a \in q$. Then, $a \cap a^c = \emptyset \in p$, so either $a \in p$ or $a^c \in p$. If $a^c \in p$, then since $p \subseteq q$, $a \cup a^c = S \in q$, a contradiction. Thus, all prime ideals are maximal.
The conditions are necessary for a subset of $P(S)$ to be a prime ideal, but I don't know if they're sufficient.
The underlying topological space is compact (since it's the Spec of something) and Hausdorff (since all prime ideals are maximal)
If $p$ and $q$ are prime ideals, then since they're maximal then there is $a \in p \backslash q, b \in q \backslash p$. However, I don't know if $D(a) \cap D(b)$ = $D(a \cap b)$ is empty
$R=\mathscr{P}(S)$ is indeed a Boolean ring in the operations you describe, and in this case a complete Boolean algebra as well. In $R$ a prime ideal is maximal and $X:=\text{Spec}(R)$ is essentially just the Stone space of $R$ (there is an easy duality between ultrafilters in the BA sense and maximal ideals ring sense) and this is a compact zero-dimensional (which implies totally disconnected) Hausdorff space. In this case, as $R$ is complete, $X$ is extremally disconnected (for any open $O$, $\overline{O}$ is also open).
An ideal $I \subseteq R$ is indeed maximal iff for all $a \in R$, either $a \in I$ or $a^c \in I$.