The following result holds for relational structures:
If $A$ is a countable structure, with an $\omega$−categorical theory $Th(A)$, that admits quantifier elimination, then $A$ is ultrahomogeneous.
I am looking for an algebraic counterexample. So: an algebraic structure with an $\omega-$categorical theory, that admits QE , but is not ultrahomogenous.
(Maybe some abelian group??)
Thank you for any help!
Think about the proof of this fact that you know for relational structures. Does it use the fact that the language is relational? If so, is this use essential?
When thinking about Fraïssé limits and ultrahomogeneous structures, the benefit to working in a finite relational language is that there are only finitely many possible quantifier-free types in $n$ variables for each $n$. When you move to a language with function symbols, this is no longer true, since a structure generated by $n$ elements can be arbitrarily large or even infinite. This is the main difference: there can be infinitely many quantifier-free types in $n$ variables for a fixed $n$. For this reason, working with a language with function symbols is very much like working with an infinite relational language.
To regain the benefits of a finite relational language, one often assumes uniform local finiteness: a finite bound on the size of substructures generated by $n$ elements will ensure only finitely many quantifier-free types in $n$ variables, as long as your language is finite.
However, if you assume $\aleph_0$-categoricity of the countable structure at the outset, then there are only finitely many complete types, hence only finitely many quantifier-free types! In this case, whether the language is finite or infinite or has function symbols or not is fairly irrelevant.