A ring $R$ is called Boolean if $x^2 = x$ for all $x \in R$. It follows that Boolean rings have characteristic $2$ and are commutative.
Let $S$ be a non-empty set, then $P(S)$ with $A + B = (A - B) \cup (B - A)$ and $AB = A \cap B$ is a Boolean ring.
The ideal generated by $T$ in $P(S)$ is $(T) = P(S) \cdot T = P(T)$, and by $A \cup B = A + B + AB$ we have that any finitely generated ideal is principal.
If $S$ is infinite then $\widetilde{I} = \{X \in P(S)\, . X \text{ is finite}\}$ is an ideal of $P(S)$ that is not principal, since it cannot be generated by a finite set(it has an infinite number of elements) and does not contain any infinite sets hence it cannot be generated by an infinite set either.
Let $I$ be an ideal of a Boolean ring $B$, $I$ is a prime ideal iff $B / I$ is a domain (that is also a Boolean ring) hence iff $B / I = \mathbb{Z}_2$ iff $I$ is a maximal ideal.
Going back to $P(S)$ and $\widetilde{I}$, it is clear that the ideal is neither prime or maximal: two infinite sets with finite intersection contraddict primality, and if $T$ is any infinite set with infinite complement it follows that $\widetilde{I} \subsetneq \widetilde{I} + (T) \subsetneq P(S)$.
My questions:
Can I say anything interesting on $P(S) / \widetilde{I}$?
Is it possible to give an explicit description of maximal(eq. prime) ideals in $P(S)$?
For any $s \in S$, the set $\{A \in P(S) | s \notin A\}$ is a maximal ideal (and a hyperplane in the $\mathbf{F}_2$-algebra $P(S)$). But as soon as $S$ is infinite, there are maximal ideals which are not of this form (as @MooS said, $\widetilde{I}$ is not contained in such an ideal).