Take the 2-minute tour ×  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?  share|improve this question Remark:
This tensor product is just the usual tensor product of algebras
over $\mathbb{F}_2$. –  Martin Brandenburg Jul 21 '11 at 10:26
1 Answer 1  up vote 7 down vote accepted  Translating this to
Boolean spaces, you are looking for a Boolean space X which is
not second countable, b