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.

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    $\begingroup$ Possible duplicate of A totally categorical structure with trivial geometry which is not interpretable in the trivial structure $\endgroup$
    – verret
    Feb 21, 2016 at 23:19
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    $\begingroup$ Posting a comment on John Baldwin's answer to the other question and tagging Dima Sustretov might draw their attention to the question again, so they can reevaluate or explain what's missing. $\endgroup$ Feb 22, 2016 at 0:07
  • $\begingroup$ Unfortunately I cannot comment there because I have no reputation :(. As soon as I earn it I will post a comment there. $\endgroup$ Feb 22, 2016 at 9:08
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    $\begingroup$ Dear Szymon, I cannot remember now exactly why, but I accepted John Baldwin's anwer by mistake. I think one can construct the counterexample from my question as a non-split cover of the theory of $(M,=)$. This can be done for example using groupoids, as in this article of Hrushovski $\endgroup$ Feb 23, 2016 at 14:58
  • $\begingroup$ Dear Dima, thanks a lot for this. I will try to look into this article and exhibit an example; if I succeed, I will post it here. $\endgroup$ Feb 23, 2016 at 16:20

2 Answers 2


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. $\square$

  • $\begingroup$ Cool. If I recall correctly, there is a theorem that if $M$ and $N$ are $\omega$-categorical structures then $M$ interprets $N$ iff there is a surjective morphism $\mathrm{Aut}(M) \to \mathrm{Aut}(N)$. I think your argument basically rules out such a morphism directly. $\endgroup$ Apr 10 at 3:15
  • $\begingroup$ Now of course I want to know if this thing is trace definable in the trivial structure. I think that it should be, the main thing would be to figure out what the definable sets in this structure actually are. $\endgroup$ Apr 10 at 3:33
  • $\begingroup$ Let $f_1 =f$ ($f$ as before) and $f_2$ be the function $C\to D$ where $f_2(a)=b$ if there is $c$ such that $aEc$ and $f_1(c)=b$. For all $(i,j) \in \{1,2\}^2$ let $R_{ij}$ be the binary relation on $C$ where $R_{ij}(a,a')$ iff $f_i(a) = f_j(a')$. So $(C,R_{ij})$ is bi-interpretable with $M$. I'm pretty sure that $(C,R_{ij})$ is homogeneous. $\endgroup$ Apr 12 at 0:58
  • $\begingroup$ Assuming that claim, it's not hard to show that $(C,R_{ij})$ is trace definable in the trivial structure. But I don't see how to generalize this to the general Hrushovski construction that $M$ is a special case. $\endgroup$ Apr 12 at 1:01
  • $\begingroup$ @ErikWalsberg Where is the general construction defined? $\endgroup$ Apr 12 at 2:16

The answer to my question was given to me by Ehud Hrushovski, and is as follows (the following formulation is my own).

Let $D$ be the pure set. For each pair $\{a,b\}$ of distinct elements of $D$ let $C_{\{a,b\}}$ be a directed square, i.e., a directed cycle of length four, and let $f_{\{a,b\}}$ map the two elements of one of the diagonals of $C_{\{a,b\}}$ to $a$ and the remaining two elements to $b$. Let $C$ be the disjoint union, over all pairs $\{a,b\}$, of the cycles $C_{\{a,b\}}$; $C$ is equipped with the directed edge relation $E$. Let $f:C\to D$ be the disjoint union of all the maps $f_{\{a,b\}}$.

Now, the two-sorted structure $M=(D,C,E,f)$ is $\omega$-categorical, $\omega$-stable, has disintegrated strongly minimal sets, but does not interpret in the pure set. However, $M$ does interpret in $(\mathbb Q,\le)$ (this also proves $\omega$-categoricity of $M$). It is also a finite cover of a structure which interprets in the pure set, and a special case of a construction studied in Section 4 of the paper 'Totally categorical structures' by Hrushovski.

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    $\begingroup$ This looks interesting, but can you enlighten us how to prove all the claims? $\endgroup$ Apr 2, 2019 at 7:00
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    $\begingroup$ I would also like to know how to show that this structure does not interpret in the pure set. $\endgroup$ May 18, 2019 at 17:33
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    $\begingroup$ Like Emil and Alex, I would also be interested in a seeing a proof of this. $\endgroup$ Oct 3, 2020 at 23:54
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    $\begingroup$ @EmilJeřábek I have posted a second answer containing proofs of the relevant claims. $\endgroup$ Apr 7 at 19:51
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    $\begingroup$ @ErikWalsberg I have posted a second answer containing proofs of the relevant claims. (Apparently you cannot @ multiple people in a single comment.) Also, I have already shown this to Alex Kruckman via email. $\endgroup$ Apr 7 at 19:52

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