I want to show that a prime ideal in a non-unital Boolean ring $B$ is maximal ideal.
If the ring contains unity then it is easy. As Boolean rings are commutative, for a prime ideal $P$ the ring $B/P$ is both integral domain aswell as Boolean ring. The only non trivial integral domain and Boolean ring is $\mathbb{Z}_2$ which is a field, so $P$ is maximal in $B$.
But there are Boolean rings without unity. For example consider all the finite subsets of real numbers with symmetric difference and intersection; this is a Boolean ring without unity. (This is also the subring of $\mathcal{P}(\mathbb{R})$, the power set of real numbers under the same operations).
Generally for rings with unity, there is a technique to show that an ideal $I$ is maximal ideal. We consider an ideal which strictly contains $I$ and somehow show that it contains unity, so that it is an improper ideal. In the above case this way doesn't work. I'm looking for a technique which can be applied in a wider settings for similar problems such as in non Commutative rings etc..
Best Answer
Note that $P$ being prime is the same as $B/P$ being a ring without nontrivial zero divisors, so multiplication by a fixed nonzero element is injective.
We will show that a nonzero boolean ring $B$ without nontrivial zero divisors has a unit, so it is in fact isomorphic to $\mathbb Z/2\mathbb Z$:
Let $j\in B$ be some nonzero element, then $j$ is the multiplicative identity:
For all $x\in B$, we have $xj=xjj$ therefore $x=xj$ (since multiplication by $j$ is injective), similarly we have $jjx=jx$ so $jx=x$. (This also shows that there is only one nonzero element, since the multiplicative identity is unique)
The point is that in a ring with no nontrivial zero divisors, the only idempotent elements are $0$ and the multiplicative identity.