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A course I am taking has started to introduce Boolean rings: rings where every element is idempotent. It was proved that every finite Boolean ring $R$ is isomorphic to a power set ring $\wp (S)$ for some set $S$ (with the usual operations $+=\triangle,\;\cdot =\bigcap$).

The proof was to consider $S=\{\text{homomorphisms}:R\to\mathbb{F}_2\}$ and take the isomorphism: $$\Omega:\;\;R\to\wp(S)\,;\quad r\mapsto\big\{\phi\in S\;\big|\;r\notin \text{ker}\,\phi\big\}$$ Bringing $\mathbb{F}_2$ into this is natural, and this question helped shed light on how one would arrive at the isomorphism $\Omega$.

I understand the proofs, but I feel like I am missing why we would expect these rings to admit this isomorphism. In other words, is there something about the condition that every element is idempotent which suggests that the ring has a power-set structure?

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Geometry

Start by proving that a Boolean ring is commutative. Next, show that if $R$ is a commutative ring, idempotent elements $e \in R$ are in natural bijection with decompositions

$$R \cong Re \times R(1 - e)$$

of $R$ into a direct product of two rings. Geometrically this says that idempotents in $R$ correspond to ways to disconnect the affine scheme $\text{Spec } R$ into two components: think of an idempotent as a function $\text{Spec } R \to \{ 0, 1 \}$ to see the topological intuition behind this.

So a Boolean ring is a very special sort of ring: it's a commutative ring where every element corresponds to a way to decompose $\text{Spec } R$ into two pieces. The proof that finite Boolean rings are power sets can be done by inducting on this observation, repeatedly disconnecting $\text{Spec } R$ into pieces until you get down to pieces that can't be disconnected any further. This means Boolean rings where the only idempotents are $0$ and $1$ (rings with this property are called connected, as you might expect), and of course $\mathbb{F}_2$ is the unique such ring.

So geometrically a better way to state the classification is that every finite Boolean ring is the ring $X \to \mathbb{F}_2$ of $\mathbb{F}_2$-valued functions on a finite set $X$. In fact something even better is true: the category of finite Boolean rings is equivalent to the opposite of the category of finite sets.

Logic

Boolean rings are the same thing, in a very strong sense, as Boolean algebras (more precisely, they are equivalent as concrete categories). Boolean algebras are supposed to describe how Boolean logic works, and power sets are the most natural models of Boolean logic for the following reason. In the language of Boolean logic, you can interpret a morphism $B \to \mathbb{F}_2$ from a Boolean ring $B$ to $\mathbb{F}_2$ as a consistent assignment of truth values to each element of $B$, thought of as logical propositions. If $B$ is a collection of questions about the state of a system, then such morphisms are possible answers, generated by possible states of the system.

So conversely it's very natural to associate to every $b \in B$ the set of all states in which $b$ is true. If there are "enough" states, then this gives a power set description of $B$.

Commutative algebra

Another approach is the following. Associated to every commutative ring $R$ is a canonical map

$$R \to \prod_P R/P$$

from $R$ to the product of the quotients $R/P$ of $R$ by every prime ideal. (In algebraic geometry terms, we are evaluating elements of $R$ at every point of the prime spectrum.) The kernel of this map is the intersection of the prime ideals of $R$, which is known to be the nilradical. Now, show that

  1. A Boolean ring has no nontrivial nilpotent elements, and
  2. If $B$ is a Boolean ring, then so is every quotient of $B$, and if $B$ is also an integral domain, then $B = \mathbb{F}_2$.

From here it follows that every Boolean ring admits a canonical injection

$$B \to \prod_P B/P \cong \prod_P \mathbb{F}_2$$

into a product of copies of $\mathbb{F}_2$ (and in particular it follows that homomorphisms $B \to \mathbb{F}_2$ are in natural bijection with prime ideals of $B$). From here you need some extra argument to show that the above map is also surjective when $B$ is finite. I think you can use the Chinese remainder theorem.

The benefit of this approach is that it generalizes to some related situations. For example, it follows that every commutative ring in which every element satisfies $x^n = x$ for some fixed $n$ (Jacobson famously showed that you can drop "commutative" here) canonically injects into a product of finite fields of order $q$ where $q - 1$ divides $n - 1$, and for finite rings this is a bijection.

Further

To get some context, you can also learn about what infinite Boolean rings are like here. "Finite set" gets replaced by "profinite set."

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