TITLE: ω-categorical, ω-stable structure with trivial geometry not definable in the pure set
QUESTION [13 upvotes]: Briefly, my question is the following.

does every countable ω-categorical, ω-stable structure with 
disintegrated strongly minimal sets interpret in the countable pure set?

By countable pure set I mean a structure with countable universe and equality relation only.

This is a repetition of this question A totally categorical structure with trivial geometry which is not interpretable in the trivial structure. However, I do not understand why the answer provided there is marked correct. (I agree with Dima Sustretov in the comments that the structure does interpret in the pure set).
The background to my question is the following.
It is shown in the paper of Cherlin, Harrington and Lachlan that
every ω-categorical, ω-stable structure is coordinatized by a collection of projective spaces, affine spaces and pure sets (which appear as strictly minimal sets in the expansion of the original structure by imaginaries). I'm interested in those structures in which only the pure sets appear. These were studied in the paper of Lachlan titled "Structures coordinatized by indiscernible sets". In this paper, it is shown that every such structure interpret in an arbitrary countable linear order,
and also, that such structures correspond precisely to reducts of totally categorical structures with trivial geometry (of the strongly minimal sets).
It is easy to see that every structure which interprets in the pure set is 
ω-categorical and ω-stable, and it follows from the paper of Lachlan that it is coordinatized by indiscernible sets (this even shown for structures which interpret in a dense linear order). 
Therefore, we have the following implications:

interprets in countable pure set → ω-categorical, ω-stable, with disintegrated strongly minimal sets → interprets in countable dense linear order.

The second implication cannot be reversed (the dense linear order itself is not ω-stable). My question is whether the first implication can be reversed. In other words, this is the same question as asking about the existence of A totally categorical structure with trivial geometry which is not interpretable in the trivial structure.

REPLY [4 votes]: This is a follow-up to Szymon Toruńczyk's answer in which I will prove the claims in it.
Proposition.
$\mathrm{Th}(M)$ is totally categorical with trivial geometry.
Proof. First, note that $M = \mathrm{acl}(D)$. This implies that $D$ is stably embedded in $M$. Furthermore, we have that every permutation of $D$ extends (non-uniquely) to an automorphism of $M$. This implies that the induced structure on $D$ is that of an infinite pure set, so, in particular, $D$ is strongly minimal.
For any $N \equiv M$, we will also have that $N = \mathrm{acl}(D^N)$, so we get a unique model of each infinite cardinality. $\square$
Proposition. $\mathrm{Th}(M)$ is not interpretable in the theory of an infinite pure set.
Proof. Let $N$ be a model of $\mathrm{Th}(M)$, and let $X$ be an infinite pure set. Suppose that $N$ is interpretable in $X$. So in particular, we have imaginary sorts $D'$ and $C'$ as well as a definable binary relation $E' \subseteq C'\times C'$ and a definable function $f':C' \to D'$ such that $(D',C',E',f')$ is isomorphic to $N$.
Since $\mathrm{Th}(X)$ is uncountably categorical, there is a strongly minimal set $S\subseteq D'$. Furthermore, there must be a definable finite-to-finite correspondence between $X$ and $S$. Since $X$ is a pure set, this must actually be a bijection on some cofinite set.
Let $Y\succ X$ be an elementary extension such that $|Y \setminus X| = 2$. Let the two new elements be $a_0$ and $a_1$. Let $S^Y$ and $C'^Y$ be the corresponding sets in the elementary extension, and let $b_0$ and $b_1$ be the new elements of $S^Y$ corresponding to $a_0$ and $a_1$, respectively. Finally let $c_0,$ $c_1,$ $c_2,$ and $c_3$ be elements of $C'^Y$ satisfying that $f'(c_0) =f'(c_2)=b_0$, $f'(c_1)=f'(c_3)=b_1$, and $c_0E'c_1E'c_2E'c_3E'c_0$.
Let $\sigma$ be the automorphism of $Y$ that fixes $X$ and has $\sigma a_0=a_1$ and $\sigma a_1 =a_0$. We necessarily have that $\sigma b_0 = b_1$ and $\sigma b_1 = b_0$.
There are only two possibilities for the value of $\sigma c_0$. Either $\sigma c_0 = c_1$ or $\sigma c_0 = c_3$. In both cases $\sigma^2 c_0 = c_2$, but $\sigma^2$ is the identity automorphism, so this is absurd.
Therefore no such interpretation can exist. $\square$