There is a well-known result that every boolean algebra $(R,0,1,\land,\lor,\neg)$ can be made a boolean ring by defining $$x\cdot y=x\land y\qquad\text{and}\qquad x+y=(x\land\neg y)\lor(y\land\neg x)\,.$$ However the proof requires a lot of computation. For $R=\mathfrak{P}(X)$, there is a simpler argument. Namely, one can show that the bijection $$\mathfrak{P}(X)\longrightarrow(\mathbb{Z}/2)^X\,,\qquad A\longmapsto\chi_A$$ sending a subset on the characteristic funtion defined by that set respects addition and multiplication. It follows that $\mathfrak{P}(X)$ is a ring and the above bijection is a ring-homomorphism.

My question is: can one deduce the general statement about boolean algebras from the special case above? This would work out, for example, if either

Every boolean algebra embeds into the boolean algebra defined by a power set.


Every boolean algebra is a quotient of the boolean algebra defined by a power set.

The first of these conjectures would be in the same spirit as Cayley's theorem (or Yoneda's Lemma), the second would be in the same spirit as a polynomial ring (or free objects in general). Both techniques can be used to find identities in algebraic structures: To check that $(xy)^{-1}=y^{-1}x^{-1}$ holds in groups, it would suffice to check this in symmetric groups or in free groups.

However I know very little boolean algebras and I'm not able to check, if there is a way like this to avoid the computation, and if it is, which of my conjectures holds.

Edit: The answer by rschwieb suggests that my first conjecture is true, using some non-trivial ring theory and the equivalence of boolean algebras and boolean rings. So my question becomes: Is there a simple way to write down an embedding $(R,0,1,\land,\lor,\neg)\longrightarrow(\mathfrak{P}(X),\emptyset,X,\cap,\cup,X\setminus-)$ of boolean algebras? Perhaps $X=R$ would be a good choice, but I'm not sure.

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    $\begingroup$ The natural embedding is given by taking $X$ to be the space of Boolean algebra homomorphisms from $R$ into the Boolean algebra of order $2$, equipped with what is essentially the Zariski topology. This is part of the proof of the Stone representation theorem: en.wikipedia.org/wiki/… $\endgroup$ Jan 2, 2015 at 12:14
  • $\begingroup$ Thank you, Qiaochu. Put in elementary terms, this is $$r\longmapsto\left\{f\colon R\to\{0,1\}\mid f(r)=1\right\}\,,$$correct? I will try to show this is a boolean algebra homomorphism. $\endgroup$
    – user114885
    Jan 2, 2015 at 12:29

1 Answer 1


I think the theorem that you might be looking for is this:

Theorem: A (commutative) von Neumann regular ring is a subdirect product of fields.


Corollary: A boolean ring is a subdirect product of some number of copies of the field of two elements $\Bbb F_2$.

Since you can naturally interpret an element of $\prod_{i\in I} \Bbb F_2$ as a member of $\mathcal{P}(I)$, I think this is a proof of your first conjecture.

  • $\begingroup$ Ok, using the equivalence of boolean algebras and boolean rings, this shows that every boolean algebra embeds into a boolean algebra of the form $\mathfrak{P}(X)$. But since this equivalence is, what I want to show, I still have to work out this fact without using any ring theory. I will see what I can do. Thank you so far. $\endgroup$
    – user114885
    Jan 2, 2015 at 11:44

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