Does Bernstein theorem hold for models with elementary functions?

Question

Bernstein theorem is a general pattern that occurs in many areas of mathematics (see the Wikipedia article for some examples). Does it hold for arbitrary models with elementary embeddings?

To be more precise: if there are two models $M,N$ with an elementary embedding of $M$ in $N$ and vice versa, does it imply that they are isomorphic?

If that is so, is there a simple proof? If not, are there any simple counterexamples, and what are some interesting subclasses of the class of all models for which it might hold?

I'm pretty sure it holds for homogeneous models, which can be shown in the following way:

1. Existence of elementary embeddings implies that the models have the same signature and are elementarily equivalent, and are of the same cardinality (by usual Bernstein theorem, incidentally).
2. Elementarily equivalent homogeneous models of the same cardinality are isomorphic iff they realize the same types, but elementary embeddings preserve types of elements, so the two models are isomorphic.

Andre Nicolas' answer suggests another somewhat wide class of examples: strongly minimal models (which includes all algebraically closed fields, infinite vector spaces over fields, and iirc some objects in algebraic geometry, but I don't know much about it). They have all the nice properties of algebraic closure (unique acl-dimension, uniqueness of a model of a given dimension etc.).

However, I'm pretty sure any strongly minimal model is homogeneous, so while motivating, it is only really a subclass of the class I specified earlier.

Answer 1

The thesis of John Godrick has a good summary of results, including a collection of counterexamples, and a number of positive results. One example where we do get isomorphism is the theory of algebraically closed fields of characteristic $0$. There the only possible counterexamples are countable. However, if there are elementary embeddings in both directions, the cardinality of transcendence bases is the same, so we get isomorphism.

Answer 2

It fails for linear orders, even countable ones. You can even throw in a successor function and a $0$, requiring every element to have a unique successor and every non-$0$ element to have a unique predecessor: take $\omega+\Bbb Z\cdot\big(\Bbb Q\cap[0,1]\big)$ and $\omega+\Bbb Z\cdot\big(\Bbb Q\cap[0,1)\big)$.