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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?

Arrow
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  • What does bool(R,\Bbb Z_2)$$ denote exactly? – rschwieb Nov 12 '15 at 01:15
  • @rschwieb the boolean ring of boolean ring homomorphisms $R\rightarrow \mathbb Z _2$. – Arrow Nov 12 '15 at 08:16
  • Don't you mean Stone's theorem? A possible formulation for it would give that every boolean algebra is isomorphic to the field of sets consisting of certain sets of $2$-valued boolean morphisms. – Tarc Nov 12 '15 at 15:40
  • @Arrow thanks for clarifying. Apparently you're not supplied with the proof that constructs the isomorphism? – rschwieb Nov 12 '15 at 16:11
  • @rschwieb no, I was just told such a thing exists. Tarc, I haven't studied the proof of Stone's theorem in any generality. I'm trying to get a feel for this before deciding whether to dive in. – Arrow Nov 12 '15 at 20:05
  • @Arrow, most likely you already know, but observe that not every boolean algebra is isomorphic to some power set, just the complete atomic ones. Also there's another formulation of the representation theorem that uses ultrafilters instead of $2$-valued morphisms, maybe you'll find it more intuitive. See Theorem 3.4 from http://www.math.harvard.edu/~waffle/boolean.pdf – Tarc Nov 13 '15 at 14:51
  • @Tarc actually, I did not know this, but it reminds me that I have forgotten a very important word throughout the question - finite. Thanks for the link. I don't have time to study this topic in depth, so I'll correct my question for now. – Arrow Nov 14 '15 at 11:14

2 Answers2

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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.

Andreas Blass
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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.

egreg
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