# Isomorphism type of strongly minimal sets in an uncountably categorical theory

Let $T$ be an uncountably categorical theory in a countable language. If $\mathcal{M}$ is a model of $T$ then there is a strongly minimal definable subset $D$ of $\mathcal{M}$ such that the dimension of $\mathcal{M}$ over $D$ uniquely determines it up to isomorphism.

The set $D$ need not be unique up to finitely many elements, and it may be that there is no strongly minimal set which is definable without parameters: see here.

However, if $D_1$ and $D_2$ are two strongly minimal sets definable in $\mathcal{M}$, must they be isomorphic? Here I am leaving isomorphic vague intentionally, but it could mean the following. Enrich the original structure with a predicate $R_\phi$ for every formula $\phi$ in the obvious way and give $D_1$ and $D_2$ the structure induced by $\mathcal{M}$. Then must they be isomorphic in the enriched language after possibly removing finitely many elements from both?

## 1 Answer

No. Consider the following structure: Two infinite sets, $A$, and $B$, and a function $f:A\rightarrow B$ such that for every $b\in B$, the preimage of $b$ in $A$ consists of exactly two elements.

If you're comfortable with multisorted structures, you can make $A$ and $B$ sorts and $f$ a function between them. If not, make $A$ and $B$ unary relation symbols in the language, define $f$ in the way described above on $A$, and define $f$ to be the identity on $B$ (so that it's total).

The theory of this structure is categorical in every infinite cardinal: a model $M$ is determined up to isomorphism by $|B| = |A| = |M|$. Both $A$ and $B$ are strongly minimal sets, but they are not isomorphic: the induced structure on $B$ is just that of an infinite set, but the induced structure on $A$ is that of an infinite set together with the equivalence relation $a \sim a'$ iff $f(a) = f(a')$, which has infinitely many $2$-element classes. No such equivalence relation can be defined on $B$.