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A (first-order) theory is $\kappa$-categorical if all its models of cardinality $\kappa$ are isomorphic. Now is there a similar notion which doesn't require isomorphism, but only elementary equivalence? (Let's say "elementary categorical".) Some questions about:

1) Are there any theories with this property (for some $\kappa$ for which they have a model of cardinality $\kappa$) which are not $\kappa$-categorical? And are there any which are not complete?

2) Are there any theories which satisfy the property for every $\kappa$, and are not $\kappa$ categorical for any $\kappa$? And are there any which are not complete?

Sorry for the lack of preciseness in the question, but being a somehow "unheard" problem for me I didn't really know how to state it.

Thank you in advance.

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    $\begingroup$ The theory of algebraically closed field of characteristic 0 is not $\aleph_0$-categorical, but every two model of the theory is elementarily equivalent. The theory of dense linear orders is also good example. It is not $\kappa$-categorical for all uncountable $\kappa$ but the $\aleph_0$-categoricity implies elementary equivalence. $\endgroup$
    – Hanul Jeon
    Jul 3, 2017 at 15:56

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Assuming the language is countable "elementary $\kappa$-categoricity" of $T$ as you define is equivalent to $$T \cup \{\exists x_1, ..., x_n \bigwedge_{i < j} x_i \neq x_j : n < \omega\}$$ being complete. (A contradictory theory is asumed complete here.) In particular this does not depend on $\kappa$.

An example where $T$ is not complete is the theory of vector spaces over a finite field. An example of where $T$ is not categorical is your favourite complete theory that is not categocial, e.g. $DCF_0$.

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Your elementary categoricity implies every model of the theory is elementarily equivalent, if the theory has no finite model. The simplest proof of the Łoś–Vaught test would work. Therefore, every elementarily categorical theory is complete.

That is, the notion of complete theory and elementarily categorical theory are same for theories with no finite model.

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  • $\begingroup$ Nice! Just how do you eliminate the hypotesis of not having finite models? $\endgroup$
    – W. Rether
    Jul 3, 2017 at 16:12
  • $\begingroup$ @W.Rether I have forgot a theory can have a finite model :( $\endgroup$
    – Hanul Jeon
    Jul 3, 2017 at 16:16

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