# On the link between elementary equivalence and isomorphism in first-order logic

My question is the following.

Assume that, for a theory $$T$$, there exist $$\mathcal{M,N} \vDash T$$ and constants $$a,b$$ (included in the theory's language) such that $$(\mathcal{M},a) \equiv (\mathcal{N},b)$$ (they satisfy the same formulas in their respective models).

Can we conclude that there exist $$\mathcal{M}',\mathcal{N}' \vDash T$$ and an isomorphism $$\mathcal{M}' \rightarrow \mathcal{N}'$$ mapping $$a$$ to $$b$$?

I am trying to put together the facts that

• $$\mathcal{M}$$ and $$\mathcal{N}$$ have $$\kappa$$-saturated elementary extensions for any $$\kappa$$
• $$\mathcal{M}$$ and $$\mathcal{N}$$ have elementary extensions of arbitrary large cardinality
• For saturated structures of same cardinality, elementary equivalence implies isomorphism.

I suspect tunnel vision made me miss something obvious here. Thank you for your help.

• I suppose you mean that $\mathcal{M}'$ and $\mathcal{N}'$ are elementary extensions of $\mathcal{M}$ and $\mathcal{N}$ respectively? – Mark Kamsma Sep 1 '19 at 10:55
• Not necessarily. That's only how I am trying to obtain them. – user453758 Sep 1 '19 at 11:07

Yes, there are such models $$\mathcal{M'}$$ and $$\mathcal{N'}$$, and we may take them to be extensions of $$\mathcal{M}$$ and $$\mathcal{N}$$ respectively. Note that $$a$$ and $$b$$ being constants of the language is not really a requirement. If they were just elements of $$\mathcal{M}$$ and $$\mathcal{N}$$ with the same type, then we could add constants to the language and interpret them as $$a$$ and $$b$$.

In general, to build saturated models, we need certain set-theoretic assumptions, or certain assumptions on our theory (if the theory is stable). Indeed, we have the following proposition:

Proposition. Let $$\mathcal{M}$$ be some $$L$$-structure, then there is a $$\kappa^+$$-saturated elementary extension $$\mathcal{N}$$ with $$|N| \leq |M|^\kappa$$.

This we can do without any additional assumptions. It is now an easy corollary that we can build saturated models under set-theoretic assumptions.

Corollary. If $$2^\kappa = \kappa^+$$, then there is a saturated model for our theory $$T$$ of size $$\kappa^+$$. So under assumption of the Generalised Continuum Hypothesis, there are saturated models of size $$\kappa^+$$ for all $$\kappa$$.

Proofs of these statements can be found as Theorem 4.3.12 and Corollary 4.3.13 in Model Theory: An Introduction by David Marker. In fact, the corollary after that (4.3.14) discusses how we can also build a very big saturated model under the assumption of a large cardinal (a strongly inaccessible cardinal in this case). I will come back to that at the end of this answer.

So if you are willing to make certain set-theoretic assumptions, then we already have what you want. Because indeed, elementary equivalent saturated structures of the same cardinality are isomorphic (as you already stated yourself). However, we can do without additional set-theoretic assumptions. This is done by using the concept of special structures. This time, I will refer to A Course in Model Theory by Katrin Tent and Martin Ziegler. You can find all of this in chapter 6.1 (right at the start).

Definition. A structure $$\mathcal{M}$$ of cardinality $$\kappa$$ is special if $$\mathcal{M}$$ is the union of a chain of elementary extensions $$\mathcal{M}_\lambda$$, where $$\lambda$$ runs over all cardinals less then $$\kappa$$ and each $$\mathcal{M}_\lambda$$ is $$\lambda^+$$ saturated.

Using the earlier proposition and some clever choice of cardinals (which we can do without additional set-theoretic assumptions), we can always find an elementary extension to a special structure of a certain cardinality. Why is this relevant? Well, the result you mentioned for saturated structures generalises to special structures. This is in fact a generalisation, because every saturated structure is special (just take the chain that is constantly that structure). That is, we have:

Theorem. Two elementary equivalent special structures of the same cardinality are isomorphic.

All of this is actually part of a bigger story: that of monster models. A monster model is a very big saturated model, that contains every smaller model as an elementary substructure. So you can think of it as the union of all smaller models. There are various ways to make this precise. A few are as follows.

1. Assume the existence of some large cardinal $$\kappa$$. Then smaller models are the ones that have cardinality $$<\kappa$$, and your monster model has cardinality $$\kappa$$.
2. Work in Bernays-Gödel set theory with global choice (BGC), then your monster model becomes a proper class (which makes sense in that setting).
3. Work in a stable theory and just pick a big enough cardinal such that every model you will ever want to consider is smaller than that.

Again, I would refer you to the books I mentioned. David Marker discusses the monster model in chapter 6.2, and Tent and Ziegler do it in 6.1.

The connection to your question is that once you are comfortable with monster models, the answer would just be: "yes, $$\mathcal{M}$$ and $$\mathcal{N}$$ are elementary substructures of the monster model, and then since $$a$$ and $$b$$ have the same type, there must be an automorphism of the monster model taking $$a$$ to $$b$$".