# Stable examples from Algebra such that the model theoretic algebraic closure of a substructre is no model

Let $T$ be a stable theory. Let $A$ be a subset or substructure of a model $M$ of $T$. Now in some theories the (model theoretic) algebraic closure of $A$ is already a (sub)model of $T$. For example, in the theory of algebraic closed fields this holds even for the empty set. Now of course there are many model theoretic examples were this fails, but these are rather constructed. My questions is:

Which are the (from an algebraic point of view) interesting examples were this property fails?

A motivation for this is the failure of higher amalgamation as described in Hrushovski's Groupoids, imaginaries and internal covers http://arxiv.org/abs/math/0603413. Since by results of Tristram De Piro, Byunghan Kim, Jessica Millar in Constructing the hyperdefinable group from the group configuration http://arxiv.org/abs/math/0508583 higher amalgamation over models holds in any theory. Hence any theory which will potentially fail higher amalgamation, has to fail the property described above.

• A silly example: the theory of an infinite structureless set. (Or any theory with trivial ACL whose models are all infinite.) – Noah Schweber Mar 19 '14 at 18:47
• ... and almost all counter examples to failure higher amalgamation seem to be extensions of such silly examples. – TimZ Mar 19 '14 at 21:03

This happens in differentially closed fields. The model theoretic algebraic closure of a $A$ is just the field theoretic algebraic closure of the differential field generated by $A$. This will rarely be a differentially closed field. For example, if $A$ is contained in the constants, then its algebraic closure will be as well.