Is the full AE theory of a structure companionable? Let $M$ be a structure, and $T$ be the full $\forall\exists$ theory of $M$, i.e., the set of $\forall\exists$ sentences true of $M$.  Is $T$ companionable?
What few non-companionable theories I know are not of this form, and in fact it seems that companionability is usually asked of a theory defined in terms of axioms as opposed to this form.
 A: Not necessarily. One of the simplest examples of a non-companionable theory is the theory of acyclic graphs. The language is $\{E\}$, and the theory consists of universal axioms saying that $E$ is symmetric and antireflexive, and for each $n\geq 3$, there are no $E$-cycles of length $n$. Call this theory $\mathsf{AG}$.
$\mathsf{AG}$ is not companionable because an existentially closed model is connected (if there is no path from $a$ to $b$ in a graph $G$, we can embed $G$ into a larger graph $G'$ with a new vertex $c$ connected to $a$ and $b$) but the complete theory of an existentially closed model admits disconnected models by compactness. So the class of existentially closed models is not axiomatizable.
As you probably know, a theory $T$ is companionable if and only if the theory $T_\forall$ (consisting of the universal consequences of $T$) is companionable. It follows that for any structure $M$, $\mathrm{Th}(M)$ is companionable if and only if $\mathrm{Th}_{\forall\exists}(M)$ is companionable if and only if $\mathrm{Th}_\forall(M)$ is companionable, since all three of these theories have the same universal consequences (namely $\mathrm{Th}_\forall(M)$).
So all we need is to find a graph $M$ such that $\mathrm{Th}_\forall(M) \equiv \mathsf{AG}$. Let $M$ be the countable infinitely branching tree (i.e. the unique countable connected acyclic graph in which every vertex has countably many neighbors). It is easy to see that every finite acyclic graph embeds in $M$, so the finite substructures of $M$ are exactly the models of $\mathsf{AG}$, and hence $\mathrm{Th}_\forall(M) \equiv \mathsf{AG}$. So $\mathrm{Th}_{\forall\exists}(M)$ is not companionable.
