# On a theorem concerning free Boolean algebras

In Sikorski's book "Boolean Algebras" (3rd edition), p. 42, one finds the following theorem:

In order that $\mathfrak{A}$ be a free Boolean algebra with $n$ free generators, it is necessary and sufficient that $\mathfrak{A}$ be the Boolean product of an indexed set of $n$ four-element Boolean algebras.

Suppose that $\mathfrak{A}$ has 1 generator, hence 4 elements.Then $\mathfrak{A}$ is trivially the Boolean product of 1 four-element Boolean algebra.

Suppose next that $\mathfrak{A}$ has 2 generators, hence 16 elements. Then $\mathfrak{A}$ is the Boolean product of 2 four-element Boolean algebras (which indeed has 16 elements).

Consider now the case that $\mathfrak{A}$ has 3 free generators, hence, 256 elements. Then the Boolean product of 3 four-element Boolean algebras has 64 elements (not 256).

What am I missing here?

• What is the definition of the "Boolean product"? I';m not familiar with the term. – bof Jan 14 '16 at 12:05
• Boolean product and Cartesian product are synonymous. Note that the definition of the Cartesian product of Boolean algebras is not that straightforward, though... – puzzled Jan 14 '16 at 14:49
• The Boolean product of algebras is not the same as the direct product. If I am not mistaken the former is the particular case of the general construction of the free product of algebras, while the latter is just the usual Cartesian product with component-wise operations (and indeed in this case the cardinality of the cartesian product of $n$ four-element Boolean algebras is $4^n$). – Random Jack Jan 16 '16 at 9:57