Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? Let $(B, +, \cdot)$ be a ring (not necessarily unital!) with the property that every $x \in B$ satisfies $x \cdot x = x$.
How does one show that the kernel of any non-zero homomorphism of rings $h:B\rightarrow \mathbb{Z}_2$ is a maximal ideal of $(B, +, \cdot)$?
I'm looking for an elementary proof, not requiring anything more than the definitions of a ring, of an ideal, and of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.
 A: Let $f:B\to\mathbb{Z}_2$ be a non-zero homomorphism, and let $I=\ker(f)$. Suppose $I\subsetneq J\subseteq B$. There exists an element $x\in J\setminus I$ (because $I\subsetneq J$), and we must have $f(x)=1$ (because there are only two elements of $\mathbb{Z}_2$ to go to), so that $$f(x-1_B)=f(x)-f(1_B)=1-1=0,$$ and hence $x-1_B\in \ker(f)=I\subset J$. Thus 
$$1_B=x-(x-1_B)\in J,$$
and therefore $J=B$. Thus $I$ is maximal.

The standard proof:
Any non-zero homomorphism of rings to $\mathbb{Z}_2$ is surjective; apply the first isomorphism theorem. Then use that an ideal $I$ of a ring $R$ is maximal $\iff$ $R/I$ is a field.
(None of this depended on any properties of $B$.)
A: Since $h:B \to \mathbb{Z}_2$ is a non-zero homomorphism, $\ker h$ is a proper ideal of $h.$ If $xy\in \ker h$ then $h(xy)=h(x)h(y)=0$ in $\mathbb{Z}_2$ so either $x$ or $y$ is in $\ker h,$ hence $\ker h$ is a prime ideal of $B.$ 
Now suppose $I$ is an ideal of $B$ that strictly contains $\ker h.$ Pick $x\in I\setminus \ker h.$ For any $y\in B$ we have $x(y-xy)=0\in \ker h$ and since $\ker h$ is a prime ideal we have $y-xy \in \ker h.$ Thus $y = (y-xy) +xy \in I$ and we conclude that $\ker h$ is maximal.
