Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Related MO discussion. At least a partial answer is given in the accepted answer. – Asaf Karagila Jun 1 '12 at 20:56
Thanks. That does give some interesting information. – tomasz Jun 1 '12 at 21:06
up vote 2 down vote accepted

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.

share|cite|improve this answer
I think any algebraically closed field is homogeneous, so it's really a special case of the example I've supplied, but the thesis you linked looks quite comprehensive, if not exactly simple. :) Thanks. – tomasz Jun 2 '12 at 10:29

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 $\omega+\Bbb Z\cdot\big(\Bbb Q\cap[0,1)\big)$.

share|cite|improve this answer
@André: Which two? – Brian M. Scott Jun 1 '12 at 22:50
@André: True, but each has an elementary (indeed isomorphic) embedding into the other. – Brian M. Scott Jun 1 '12 at 23:04
@André: Aargh. Yes, you’re right. My model theory is much rustier than my set theory. – Brian M. Scott Jun 2 '12 at 3:33
It is easy to see there's no isomorphism, but how to simply show that the obvious embeddings are elementary? – tomasz Jun 2 '12 at 10:46
@tomasz: I’m sorry, but it’s been about 40 years since I did any model theory, and while I might even have done something very similar to this back then, at this point I’m not at all sure that I could produce a rigorous demonstration in any reasonable period. Still, I’m pretty sure that in both cases any existential statement with parameters from the submodel that’s satisfied by something in the large model is also satisfied by something in the submodel: there just isn’t that much that you can say in this language. – Brian M. Scott Jun 2 '12 at 11:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.