Let $S$ be an infinite set and consider the ring, $\langle P(S), \triangle, \cap \rangle$.
- Show that the collection, $J$, of finite subsets of $S$ is an ideal of $P(S)$.
- Let $I$ be an ideal of $P(S)$. Show that $I$ is a prime ideal if and only if the following holds: $\forall A \in P(S)$ or $A \subseteq S$, either $A \in I$ or $A^\complement= S \setminus A \in I$ but not both.
It is so much easier to think about that in terms of $P(S) \cong \operatorname{Maps}(S,\mathbb F_2) :=R$ with the identification $$P(S) \ni A \longleftrightarrow \chi_A \in \operatorname{Maps}(S,\mathbb F_2)$$
The first statement then says: Take a map, which vanishes almost anywhere. If you multiply it with any map, it still vanishes almost anywhere.
The second statement says: $I$ is prime iff if either $f$ or $1-f$ is contained in $I$ for any $f \in R$.
Let $I$ be prime. Note That $f(1-f)=0$ and $f+(1-f)=1$. The first shows that one of them is contained in $I$, the latter shows that not both of them can be contained in $I$.
Now let the condition hold: $f \in I$ or $1-f \in I$ tells us, that $f=0$ or $f=1$ in $R/I$. This holds for any $f$, hence $R/I \cong \mathbb F_2$, which shows that $I$ is prime (even maximal).