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Given a family of Boolean algebras $\mathcal{B}=\{B_i\colon i\in I\}$ with respective Stone spaces $S_i$. Recall that the algebra of clopen (both closed and open) subsets of the product space

$$\prod_{i\in I}S_i$$

is said to be the free product of $\mathcal B$. This algebra is typically denoted by

$$\bigotimes_{i\in I}B_i$$

(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 decomposition 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?

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Remark: This tensor product is just the usual tensor product of algebras over $\mathbb{F}_2$. – Martin Brandenburg Jul 21 '11 at 10:26
up vote 7 down vote accepted

Translating this to Boolean spaces, you are looking for a Boolean space X which is not second countable, but cannot be written as a product of two factors of the same type (i.e., not second countable).

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.

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