# Free product of Boolean algebras

Given a family of Boolean algebras $$\mathcal{B}=\{B_i\colon i\in I\}$$ with respective Stone spaces $$S_i$$, the algebra of clopen (both closed and open) subsets of the product space $$\textstyle\prod_{i\in I}S_i$$ is called the free product of $$\mathcal B$$. This algebra is typically denoted by $$\textstyle\bigotimes_{i\in I}B_i$$ (and I will use the standard "tensor" notation for finite free products in the obvious manner).

I am interested in the (possible) Boolean algebras which admit only very particular decompositions in terms of the free product.

Is there an uncountable Boolean algebra $$B$$ such that if $$B$$ is isomorphic to $$A\otimes C$$ then either $$A$$ or $$C$$ is countable?

• Remark: This tensor product is just the usual tensor product of algebras over $\mathbb{F}_2$. – Martin Brandenburg Jul 21 '11 at 10:26
• Wow, a downvote for a question from 2011! – Tomasz Kania Mar 21 '19 at 19:58

Have you considered the compact space $[0,\omega_1]$? It is certainly not the product of two uncountable spaces, as such a product would contain two almost disjoint closed uncountable sets. On the other hand, a countable Boolean space cannot have uncountably many clopen sets.
• It's not even a product of two spaces of cardinal $\ge 2$. The reason is that every point, with a single exception, has a countable neighborhood. – YCor Mar 21 '19 at 9:32
• For completeness, $B$ here is the Boolean subalgebra of subsets of $\mathbf{R}$ generated by the left-closed, right-open intervals. Its Stone space can be viewed as the doubled circle $S^{\pm}$, starting from $\mathbf{R}\cup\{\infty\}$, with each point $x$ replaced by two points $\{x^-,x^+\}$, with the topology of circular order. Theorem 15.14 in the handbook even shows that $B$ doesn't contain the free product of an infinite subalgebra and an uncountable one. In other words, there is no continuous surjection from $S^{\pm}$ onto the product of an infinite with a non-metrizable Stone space. – YCor Mar 21 '19 at 8:15
• Also note that unlike Goldstern's example (whose space is not a product of any two spaces of cardinal $\ge 2$), this space is homeomorphic to its product with any nonempty discrete finite set. – YCor Mar 21 '19 at 9:03