I want to prove the following:
Let $A$ be a Boolean ring and let $\mathfrak{a}\neq A$ be an irreducible ideal. Then $\mathfrak{a}$ is maximal.
I already know that the prime ideals of $A$ are maximal and that $\mathfrak{a}$ is the intersection of the maximal ideals that contain $\mathfrak{a}$.
Suppose $\mathfrak{a}$ is not maximal and let $M$ be the set of maximal ideals containing $\mathfrak{a}$. I would like to find two subsets $R,S\subset M$ such that $\bigcap_{\mathfrak{m}\in R}\mathfrak{m}\neq\mathfrak{a}$, $\bigcap_{\mathfrak{m}\in S}\mathfrak{m}\neq\mathfrak{a}$ and $\bigcap_{\mathfrak{m}\in R\cup S}\mathfrak{m}=\mathfrak{a}$. But I'm not sure if this is possible, especially if $M$ is very large.
Can someone give me a hint?