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Every Boolean algebra $A$ is isomorphic to a field of set. In particular, if $A$ is finite, then $A$ is isomorphic to the power set of its atoms.

Now, suppose that $A$ is free Boolean algebra with 2 free generators (or atoms). Because $A$ is a finite Boolean algebra, it is isomorphic to the power set of its atoms. However, $A$ has $16$ elements, whereas the power set of its atoms has only 4 elements.

Could anybody explain what I seem to miss.

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    $\begingroup$ Actually two generators x and y give four atoms (x and y, x and not y, not x and y, not x and not y). $\endgroup$
    – David C. Ullrich
    Sep 11, 2015 at 14:20
  • $\begingroup$ @DavidC.Ullrich I think your comment should be an answer. $\endgroup$
    – Alex Kruckman
    Sep 14, 2015 at 23:49

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"Free generators" and "atoms" are not the same thing. An atom in a Boolean algebra is a minimal nonzero element, while a set of free generators is a set such that any map from the set to another Boolean algebra extends uniquely to a homomorphism of Boolean algebras. There is no reason to expect these to coincide.

In particular, if $B$ is freely generated by $\{x,y\}$, it turns out that $B$ has four atoms, namely $x\wedge y$, $\neg x\wedge y$, $x\wedge \neg y$, and $\neg x\wedge\neg y$. More generally, the free Boolean algebra on a finite set with $n$ elements has $2^n$ atoms, namely all the ways to form meets of the generators or their complements.

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