# partitions of Boolean algebras

A partition of a Boolean algebra is a collection of pairwise disjoint nonzero elements with supremum 1. For any infinite Boolean algebra $A$ let $a(A)$ be the least size of an infinite partition of $A$. Is $a(A+B)=\min\{a(A),a(B)\}$, where $A+B$ is the free product of $A$ and $B$?

• In forcing terms, is $A+B$ just the lottery sum of the two Boolean algerbas? – Asaf Karagila Nov 20 '17 at 16:34
• I think he means the analogue of product forcing, but with Boolean algebras that are not necessarily complete (since then there is always a countable partition). Don, could you tell us a little more about what you mean by free product? – Joel David Hamkins Nov 20 '17 at 16:39
• If he means the lottery sum (direct sum), then the answer is easily yes, since a partition of $1$ in $A\oplus B$ gives a partition of $1_A$ in $A$ and similarly in $B$ of the same size. The more interesting question is with products, since there are Ramsey-like issues concerning on which coordinate the incompatibility arises. – Joel David Hamkins Nov 20 '17 at 16:54
• @Joel: With product, I suspect that a product of a Suslin tree with itself will provide a counterexample. – Asaf Karagila Nov 20 '17 at 17:11
• The free product is the coproduct. Given disjoint presentations of Boolean algebras by generators and relations, $\langle G_1|R_1\rangle$ and $\langle G_2|R_2\rangle$, the free product is the BA with presentation $\langle G_1\cup G_2|R_2\cup R_2\rangle$. Alternatively, the free product of $B_1$ and $B_2$ is the BA with Stone space that is the product of the Stone space of $B_1$ and the Stone space of $B_2$. – Keith Kearnes Nov 20 '17 at 17:29

1. $a(A+B)=\min\{a(A),a(B)\}$ for every $A,B$
2. $a(A+A)=a(A)$ for every $A$.
Is there a partial order $\mathbb{P}$ such that $a(\mathbb{P})<a(\mathbb{P}\times \mathbb{P})$?
There are several partial results which imply that the answer to the OP´s question is yes in at least two cases: When the boolean algebras are atomic (see Theorem 6.1) and when the boolean algebras have size at most $\aleph_1$ (see Corollary 4.2).
Kurilić also mentions that it is not even clear what happens with the algebra $A:=\mathcal{P}(\omega)/\mathrm{Fin}$. It is known by results of Spinas that $a(A+A)=a(A)$ in models of $\mathfrak{a}=\mathfrak{b}$ (so it is true under $CH$ or $MA$). However, it is not known whether $a(A+A)<a(A)$ is consistent.