# Is the tensor product of BAOs a kind of extended BAO?

I've been reading "Boolean algebras with operators. Part I." (Jonsson, Tarski) where, given a subalgebra of a Boolean Algebra, they define its perfect extension. As far as I understand it can be extended to the case of Boolean Algebras with Operators (BAOs) by giving a notion of "extended (n-ary) operators" in the form

$m^{\sigma}=\underset{x\geq y\in C^{n}}{\sum}\underset{y\leq z\in B^{n}}{\prod}m(z)$

where m is an operator in the BAO B, which in its turn is a subalgebra of the extended BAO $B^{\sigma}$ and C is the set of closed elements of B (that is, meets of elements in B).

I do not fully understand that definition of extended operators, the reason to be of that form and if it could be defined in a different way.

I wonder, on the other hand, if it has to do with the definition of tensor product of BAOs, which has a similar form according to the answer given in: https://mathoverflow.net/questions/122800/products-of-boolean-algebras-and-probability-measures-thereon. Since the tensor is a coproduct, could be every BAO (as a factor of the tensor) seen as a subalgebra of the tensor seen as a new BAO (and perhaps treated as a kind of extension of its factors)?

• Well, I can clarify something related to this. I do not think it is a full answer to my own question but could help to understand. – gibarian Sep 6 '16 at 15:53
• The tensor product of BAOs, as the tensor product of algebras over the ring Z/2Z, can not be the perfect extension of its BAOs since they are not subalgebras of the tensor. The reason is that we take quotients to get the tensor and so the supposed inclusion is not injective. – gibarian Sep 6 '16 at 15:59