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Let $R$ be a *finite* boolean ring. It’s known that there’s a boolean algebra/ring isomorphism $R\cong \mathcal P(\mathsf{Bool}(R,\mathbb Z_2))$.

I’m trying to get a feel for this. The subsets of $\mathsf{Bool}(R,\mathbb Z_2)$ should somehow correspond to elements of $R$. At first I thought of $\mathbb Z_2$ as the usual subobject classifier in the category of sets, though that didn’t get me very far. Then I thought the lattice perspective might help, but the only reasonably canonical map I can think of is $r\mapsto \left\{ \phi\leq \mathrm{eval}_r \right\}$, and I don’t see why this should be an iso.

What’s the intuition behind this isomorphism? How could one have guessed it?

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$\mathsf{Bool}(R,\mathbb{Z}_2)$ has no natural structure of boolean ring; it’s just the set of ring homomorphisms from $R$ to $\mathbb{Z}_2$ (preserving the identity).

Now, what does $f\colon R\to\mathbb{Z}_2$ look like? It’s a ring homomorphism, so its kernel is an ideal; it’s also a maximal ideal, because $\mathbb{Z}_2$ is a field.

Therefore, if $I=\ker f$, the homomorphism is easily described:

$$

f(r)=\begin{cases}

0 & \text{if $r\in I$}\\[6px]

1 & \text{if $r\notin I$}

\end{cases}

$$

This reduces the problem to classifying all maximal ideals in $R$. So, let $I$ be a maximal ideal and set $s=\bigvee_{r\in I}r$. Since $I$ is an ideal, $x,y\in R$ implies $x\vee y=x+y+xy\in I$. Therefore $s\in I$ (because $I$ is finite) and, by maximality, we conclude that “$r\le s$ implies $r\in I$”. Indeed, if $rs=r$ and $r\notin I$, we have $1=rx+y$, where $x\in R$ and $y\in I$; but

$$

s=rsx+ys=rx+ys

$$

so $rx=s+ys\in I$ and then $1\in I$, a contradiction.

In particular $I=\{r\in R: r\le s\}$. Conversely, if $s\in R$, $I_s=\{r\in R:r\le s\}$ is an ideal of $R$, which is maximal if and only if $s$ is maximal in $R\setminus\{1\}$ (a *coatom*).

Now prove that every element of $R$ is in a unique way the meet of pairwise distinct coatoms (the empty set of coatoms, for $1$) and you’ll have the correspondence you wanted.

The isomorphism $i$ sends each element $a\in R$ to the set of those homomorphisms $h:R\to\mathbb Z/2$ that send $a$ to $1$. In other “words”,

$$

h\in i(a)\iff h(a)=1.

$$

As for how one might have guessed something like this, I think the most plausible approach is to begin with an $R$ that you already know to be a power set Boolean algebra, say $R=\mathcal P(X)$ for some finite set $X$, and to ask how you could recover $X$ if you were just given $R$ as an abstract Boolean algebra. The easy way to do the recovery is that the elements of $X$ correspond to the minimal non-zero elements (also called the atoms) of $R$. In other words: The smallest element of $R$ is the empty subset of $X$, and just above this are the singletons $\{x\}$, one for each $x\in X$. Having recovered $X$ as (in canonical bijection with) the set of atoms of $R$, you get an isomorphism $R\cong\mathcal P(\text{Atoms}(R))$.

That argument presupposed that $R$ is already known to be a power set algebra, but once you have this idea, $R\cong\mathcal P(\text{Atoms}(R))$, you could verify that it works for any finite Boolean algebra.

In the question, in place of Atoms$(R)$, you had the set of Boolean homomorphisms $R\to\mathbb Z/2$. For finite Boolean algebras, those homomorphisms are in canonical bijection with the atoms (an atom $a$ corresponds to the homomorphism that sends everything $\geq a$ to $1$ and everything else to $0$). This switch from atoms to homomorphisms is hard to motivate in the case of finite Boolean algebras. The real reason for using homomorphisms is that they still work (with some extra caution) in the case of infinite Boolean algebras; that’s Stone duality.

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