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.



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> 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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Free product of Boolean algebras
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.


