This exercise K (f) of Chapter 2 from General Topology by Kelley.
I proved that there is exactly one maximal ideal in the Boolean ring, but am stuck with showing that there is one homomorphism to $\Bbb Z_2$.
Preliminaries
The $\underline{natural\;ordering}$ of a Boolean ring is defined by agreeing that $r \geqslant s\; $ iff $r\cdot s=s$;
The $S$ is an ideal in a Boolean ring $(R,+,\cdot)$ iff $S$ is an additive subgroup and $r\cdot s \in S$ whenever $r\in R$ and $s\in S$;
The ideal $S$ is maximal iff $R\neq S$ and no ideal other than $R$ properly contains $S$.
Problem
There is a one-to-one correspondence between maximal ideals in $R$ and homomorphisms into $\Bbb Z_2$ which are not identically zero. $\;\;\;\;$(The kernel of such a homomorphism is a maximal ideal.).
Proof: Let $U,S$ be distinct maximal ideals in $R$, then their elements must be related by the natural ordering.
Suppose $\max(U) \leqslant \max(S)$, then $\max(U)\cdot \max(S)= \max(U)\in S$. So $U\subset S$. Hence there is only one maximal ideal.
On the other hand, suppose that $\varphi,\gamma$ are homomorphisms from Bool. ring $R$ on $\Bbb Z_2$. And let $s$ be element of max. ideal $S$ and $r\notin S$.
Then $\varphi(r\cdot s)=\varphi(s)=\varphi(r)\cdot\varphi(s) \implies\varphi(r)=1$.
Hence there is $s\in S$ s.t. $\varphi(s)=0$.
Likewise define $\gamma$ s.t. $\gamma(r)=1,\; \gamma(s)=0$ and $\gamma(\max(S) :=q)=1$ for $r\geqslant q\geqslant s, \quad s\in S, r\notin S$
(I think that somehow following reasoning is in the direction of disproving $\gamma$)
Then $\gamma(r\cdot q)=\gamma(r)\cdot\gamma(q)=1\cdot\gamma(q)=1=\gamma(r)$.