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I was reading this answer which states that the free Boolean algebra on countably many generators is atomless.

Which Boolean algebra do we get if we insist that $a \land b = 0$ for two of the generators but impose no additional relations?

If this Boolean algebra is atomless, then it's isomorphic to the free Boolean algebra by a $\omega$-categoricity argument, but I'm not sure whether it ends up being atomless.

I was thinking for a bit and wondering why this free Boolean algebra is atomless, and why our generators themselves weren't the atoms, but then it made sense.

Let the generation of an item be the minimum number of function symbols and generators needed to write it.

If $a$ and $b$ are arbitrary generators, then we cannot conclude that $a \land b = \alpha$ for any $\alpha$ of generation 1 or 2, thus $a \land b$ has generation 3. Therefore none of the generators are atoms.

Similarly, let $\alpha$ be an expression of generation $n$ and let $\beta$ be an expression obtained by replacing each generator in $\alpha$ with another generator in such a way that the generators in $\alpha$ and $\beta$ are disjoint. $\alpha \land \beta$ has generation greater than $n$ and is less than $\alpha$ and is less than $\beta$.

Also, intuitively, if they were atoms, then we could argue by symmetry they'd have to be co-atoms too, which would drop us into the 4-element Boolean alegbra case, I think.

If we take our free Boolean algebra on the countable set $\{a, b, \cdots\}$ and impose just the relation $a \land b = 0$ for the particular generators $a$ and $b$, what Boolean algebra do we get? If we further impose $b = 0$ or $b = \lnot a$, then we get back an algebra isomorphic to our original free Boolean algebra, but I'm not sure what we get with just $a \land b = 0$.

Greg Nisbet
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    The title says "minimum" (which doesn't seem to make sense) but the body says $a\land b$ the meet of $a$ and $b$. – bof Dec 31 '22 at 00:09
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    @bof I get "meet" and "join" confused and can't remember which one is supposed to be the small one, so I just call them minimum and maximum. I'll update the question. – Greg Nisbet Dec 31 '22 at 00:13
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    I would call them "greatest lower bound" and "least upper bound", the "maximum" and "minumum" only exist if $a$ and $b$ are comparable. – bof Dec 31 '22 at 00:19

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I find this easier to think about topologically via Stone duality. If you take the free Boolean algebra on $a$ and $b$ modulo the relation $a\wedge b=0$, its Stone space is a discrete space with $3$ points. If you then adjoin countably infinitely many more free generators, the Stone space becomes a product $\{0,1,2\}\times\{0,1\}^\mathbb{N}$. It is clear that this space has no isolated points, so the Boolean algebra is atomless and thus isomorphic to the free Boolean algebra on countably infinitely many generators.

Or in algebraic terms, note that any element $x$ of your Boolean algebra is in a subalgebra generated by finitely many of the generators, which can be assumed to include $a$ and $b$. So, the entire Boolean algebra is obtained by then adjoining infinitely many more generators freely. For any one of these generators $c$, both $x\wedge c$ and $x\wedge \neg c$ will be nonzero, as long as $x$ is nonzero. (This corresponds to the topological argument that given any basic open set in the product topology $\{0,1,2\}\times\{0,1\}^\mathbb{N}$, you can always make it smaller by restricting it on one more coordinate.)

Eric Wofsey
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