When is there a natural copower for models of a geometric theory?

55 Views Asked by At

First let's establish the notion of an embedding as it is in universal algebra. Now consider the category of models of a geometric theory over a (nonempty) signature. Now we can make sense of the title question.

Giving more information, when given a finite model of this theory, I want to always be able to construct an infinite (for my interests, countably infinite, but more general answers that imply countably infinite are welcome) model into which the original one embeds. If our category had (infinite) coproducts there would be a natural sense in which we could do this by taking the coproduct of the structure with itself $\omega$-many times. Furthermore, I believe that if the category had a monadic free-forgetful adjunction (as here), we could just take term algebra of the finite model and identify it with the "reduced words" in the term algebra. But we do not always have that- for example, there is no "free field".

So is there any hope to be able to do this uniformly, or does it depend in some complicated way on the theory given/ the category of models of this theory?

EDIT: As pointed out by @EricWofsey, it is not going to be possible in general to hope for the coproduct to be in the same category since the theory may have no infinite models at all, so I would modify the question to asking for existence of a coproduct in some "associated" category.

EDIT2: We should also rule out from consideration theories for which only the empty set is a model.