# Boolean algebra spectrum

The first-order spectrum of a theory, is the set of cardinalities of its finite models. Finite models of Boolean algebras are informally n-dimensional cubes, therefore boolean algebra spectrum is the set of powers of 2. I have trouble proving it.

One direction proof sketch seems easy. Cartesian product of two Boolean algebras is again Boolean algebra, so starting with 2-element boolean algebra we can inductively build the models of higher cardinalities. This is conceptually easy even though the actual proof machinery has to operate in terms of concrete axiom system of Boolean lattice or Boolean ring.

How would one prove that no model with cardinality differing from power of 2 is legitimate? One way may be taking lattice perspective then, defining order relation, proving that it's grading order, go to atoms... This seems too cumbersome, and I'm not sure it would work. Is there an easier approach?

Edit: two suggestions by Brian and Qiaochu (factoring by maximal filter in the lattice, or factoring by ideal in the ring) can be expanded into full proofs. I wonder about one more possibility: leveraging negation operation. Can negation be leveraged to factor Boolean algebra any meaningful way (it certainly partitions sets of elements into equal sets)?

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Show that every finite Boolean ring $B$ is of the form $\mathbb{F}_2 \times B'$ for some Boolean ring $B'$. (Actually the finiteness assumption is not necessary.) –  Qiaochu Yuan Oct 2 '12 at 2:30
@Qiaochu, thanks: boolean ring wikipedia page hints: "The property x ⊕ x = 0 shows that any Boolean ring is an associative algebra over the field $F_2$ with two elements, in just one way. In particular, any finite Boolean ring has as cardinality a power of two." –  Tegiri Nenashi Oct 2 '12 at 16:09

HINT: Let $a\in B$ be an atom, and let $A=\{b\in B:a\le b\}$. Show that $A\cong B\setminus A$; if you think in terms of power set algebras, the isomorphism is pretty obvious. Use this to show that $B\cong 2\times A$. Essentially you’re factoring out a copy of $2$.
A hint for a little different approach: show that every element of $B$ can be uniquely expressed as a sum of minimal nonzero elements.