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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?

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    $\begingroup$ Remark: This tensor product is just the usual tensor product of algebras over $\mathbb{F}_2$. $\endgroup$
    – Martin Brandenburg
    Commented Jul 21, 2011 at 10:26
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    $\begingroup$ Wow, a downvote for a question from 2011! $\endgroup$
    – Tomasz Kania
    Commented Mar 21, 2019 at 19:58

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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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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 9:32
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By Theorem 15.14 of the Boolean algebra handbook, the interval algebra of the real numbers is such an algebra B.

Reference: S. Koppelberg. Handbook of Boolean algebras. Vol. 1. Edited by J. D. Monk and R. Bonnet. North-Holland Publishing Co., Amsterdam, 1989.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 8:15
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 9:03

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