I need to prove the following statement:
In a Boolean algebra (a lattice with complement operation) an ideal is maximal if and only if it is prime.
For every distributive lattice we have the $\implies$. For the reverse direction I could have only figured out that the ideal generated by $x$ and $x'$ are comaximal. I know that Boolean algebras are equivalent to Boolean rings so I would find interesting also a translated answer. Thanks in advance.
Suppose $I$ is a prime ideal in a Boolean algebra $B$. Note that for each $x\in B$, $x\wedge x'=0\in I$, so either $x$ or $x'$ in $I$. Now suppose $I$ is strictly contained in an ideal $J$; say $x\in J\setminus I$. Then since $x\not\in I$, $x'\in I$. But then both $x$ and $x'$ are in $J$, so $x\vee x'=1\in J$. Thus $J$ is not a proper ideal, which proves $I$ is maximal.