Scott rank of Boolean algebras (Q811719)

Elementary theories of Boolean algebras were intensively studied by many authors. It is known that the first-order language does not provide sufficient tools for the description of Boolean algebras. In the present article, the Boolean algebras are studied by means of an infinitary logic, i.e., the logic admitting conjunctions and disjunctions of an infinite set of formulas. The Scott rank is a measure of complexity of an algebraic system formulated in terms of this logic. The author considers numerous problems concerned with the Scott rank of Boolean algebras. The Boolean algebras with a finite Scott rank are completely described. It turns out that a Boolean algebra is of finite rank if and only if its elementary characteristic is \((0,k,0)\) or \((0,k,1)\) for some natural \(k\). For every such algebra \(\mathcal B\) its Scott rank \(sr({\mathcal B})\) is computed. A formula reducing the Scott rank of a Boolean algebra to the rank of its quotient w.r.t. an iterated Frechet ideal is obtained. If \(\alpha\) is an ordinal, \(\mathcal A\) an \(\alpha\)-atomic Boolean algebra, \(F_\alpha ({\mathcal A})\) an \(\alpha\)th Frechet ideal, and \({\mathcal A}/F_\alpha ({\mathcal A})\) is not trivial, then \(sr({\mathcal A})=\omega\cdot\alpha+sr({\mathcal A}/F_\alpha ({\mathcal A}))\). The Scott rank is computed exactly for superatomic Boolean algebras, for arbitrary Boolean algebras a lower bound is indicated. Namely, the following theorem is proved. Theorem. Let \(\mathcal A\) be a Boolean algebra of nonzero ordinal rank \(\alpha\). The following assertions hold: (1) if \(\alpha\) is limit, then \(sr({\mathcal A})\geq\omega\cdot\alpha,\) (2) if \(\alpha=\alpha_0+1\) and \({\mathcal A}/F_{\alpha_0}({\mathcal A})\) has infinitely many atoms, then \(sr({\mathcal A})\geq\omega\cdot\alpha,\) (3) if \(\alpha=\alpha_0+1\) and \({\mathcal A}/F_{\alpha_0}({\mathcal A})\) has exactly \(k\) atoms, \(k\in\omega\), then \(sr({\mathcal A})\geq\omega\cdot\alpha_0 + sr({\mathcal B}_k),\) (4) if \(\mathcal A\) is superatomic, \(\alpha=\alpha_0+1\), and \({\mathcal A}/F_{\alpha_0}({\mathcal A})\) has exactly \(k\) atoms, \(k\in\omega\), then \(sr({\mathcal A})=\omega\cdot\alpha_0 + sr({\mathcal B}_k)\). Here \({\mathcal B}_k\) denotes a finite Boolean algebra with \(k\) atoms whose Scott rank equals 0 if \(k=1\), and \([\log_2(k-1)]\) if \(k\geq 2\).