Elementary embedding and infinite atomic Boolean algebras Let $A$ and $B$ be two infinite atomic Boolean algebras with a monomorphism $h: A \rightarrow B$ that embeds $A$ into $B$. Also, let it be such that for any atom $a \in A$, $h(a)$ is an atom in $B$.
Let $\mathscr{L}$ be the language of Boolean algebra and $T$ be the theory of infinite atomic Boolean algebra in $\mathscr{L}$. $A$ and $B$, as models of $\mathscr{L}$, are both models of $T$ (and hence they are elementarily equivalent). Now my question is: is $h: A \rightarrow B$ an elementary embedding, in the sense that for any $\phi(x_1, ..., x_n)$ of $\mathscr{L}$, and $a_1, ..., a_n \in A$,
$$A \models \phi(x_1, ..., x_n)[a_1, ..., a_n] \, \, \text{iff}\, \, B \models \phi(x_1, ..., x_n)[h(a_1), ..., h(a_n)]$$
I can see that the statement definitely holds on the atomic formulas, but I struggle about the quantified formulas.
 A: Noah's answer is correct: an embedding of atomic Boolean algebras which preserves the atoms is an elementary embedding. Let me give a few more details.
The theory of atomic Boolean algebras, when extended by unary predicates $P_m(x)$ expressing "$x$ has at least $m$ atoms below it" has quantifier elimination (as discussed in your recent question).
Let $h\colon A\to B$ be an atom-preserving embedding of atomic Boolean algebras. I claim that for $a\in A$, $a$ has at least $m$ atoms below it if and only if $h(a)$ does.  If $a_1,\dots,a_m$ are distinct atoms below $a$, then $h(a_1),\dots,h(a_m)$ are distinct atoms below $h(a)$ (since $h$ is injective, preserves atoms, and preserves $\leq$). Conversely, if $a$ has fewer than $m$ atoms below it, say $a_1,\dots,a_k$ with $k<m$, then since $A$ is atomic, $a = a_1\vee \dots \vee a_k$, and $h(a) = h(a_1)\vee \dots \vee h(a_k)$ (since $h$ preserves $\vee$). Now if $c\leq h(a)$ is an atom, we have \begin{align*}
c &= c\wedge h(a)\\
&= c\wedge (h(a_1)\vee \dots \vee h(a_k))\\
&= (c\wedge h(a_1))\vee \dots \vee (c\wedge h(a_k)).
\end{align*}
Each $h(a_i)$ is an atom, so $c\wedge h(a_i) = \bot\text{ or }h(a_i)$. And since the $h(a_i)$ are distinct and $c$ is an atom, exactly one of the terms $c\wedge h(a_i)$ is not $\bot$. Thus we have $c = h(a_i)$ for some $1\leq i \leq k$, which shows that there are exactly $k$ atoms below $h(a)$.
What we have shown is that if we expand $A$ and $B$ by the predicates $P_m(x)$, then $h$ is an embedding in the expanded language. By QE, $h$ is an elementary embedding in the expanded language, so it is an elementary embedding in the language of Boolean algebras.

Another perspective on what we've shown here is that the theory of atomic Boolean algebras is model complete in the language where we include a predicate $\mathrm{At}(x)$ naming the atoms. Recall that a theory is model complete if every embedding between models is an elementary embedding. And it's not hard to see that an embedding between atomic Boolean algebras maps atoms to atoms if and only if it is an embedding in the language expanded by $\mathrm{At}(x)$ (this is not immediate: we also need to check that if $a$ is not an atom, then $h(a)$ is not an atom).
A well-known characterization of model complete theories is that $T$ is model complete if and only if every formula is equivalent to both an existential and a universal formula. Now if $T$ has an expansion to a theory with quantifier elimination, and every new atomic formula in the expansion is equivalent to both an existential and a universal formula in the original language, then $T$ is model complete (since any Boolean combination of formulas, each of which is both existential and universal, is both existential and universal).
So we could rephrase the argument above as follows: Show that the theory of atomic Boolean algebras with $\mathrm{At}(x)$ is model complete by showing that all the predicates $P_n(x)$ are both existential and universal in this language.
And indeed, we have that $P_n(x)$ is equivalent to both $$\exists z_1\dots z_n\,\left(\bigwedge_{i\neq j}z_i\neq z_j\land \bigwedge_{i=1}^n (\mathrm{At}(z_i)\land z_i\leq x)\right)$$
and $$x\neq \bot \land \forall z_1\dots z_{n-1}\lnot \left( (z_1\vee \dots \vee z_{n-1} = x)\land \bigwedge_{i=1}^{n-1} \mathrm{At}(z_i)\right).$$
A: A useful idea for this sort of problem is to look for quantifier elimination, if not in the original language then at least in some "not-too-great" enlargement of it. And indeed this works here.
Augment the usual language of Boolean algebras $\{\wedge,\vee,'\}$ with predicates $A_m$ for $m\in\omega$. Tarski (if memory serves) showed that the theory of infinite atomic Boolean algebras in this language with $A_m$ interpreted as "is above at least $m$ atoms" is complete and admits quantifier elimination. This means that embeddings in the sense of this enriched language are automatically elementary. See Theorem $5.3.4$ here.
Since an atomicity-preserving monomorphism between infinite atomic Boolean algebras is such an embedding (once we enrich the algebras appropriately), we get an affirmative answer to your question.
