# Is there a model of ZFC that can define a "longer" model of ZFC to which it is isomorphic?

Suppose $$ZFC$$ is consistent. Is there a model $$\mathcal{M}$$ of $$ZFC$$ and formulae $$\varphi_D(x)$$ and $$\varphi_\in(x,y)$$ that define (in $$\mathcal{M}$$) the domain and membership relation of a model $$\mathcal{N}$$ of $$ZFC$$ such that we have $$\mathcal{N}\cong\mathcal{M}$$ externally (i.e., in $$V$$) and yet $$\mathcal{M}$$ thinks $$\mathcal{N}$$ is well-founded but not set-like?

I think it's clear that if this were true $$\mathcal{M}$$ would need to be ill-founded.

I suspect the Ehrenfeucht-Mostowski theorem, which gives automorphisms for such models, can be adapted to obtain an embedding from some $$\mathcal{N}$$ onto an initial segment of itself, say $$\mathcal{M}$$, but where the respective spines of indiscernibles used to generate these models are (externally) isomorphic. If that works, I think we'd have $$\mathcal{N}\cong\mathcal{M}$$ but - among other things - I can't see that (an isomorphic copy of) $$\mathcal{N}$$ would be definable in $$\mathcal{M}$$.

If the issue were that assuming $$ZFC$$ was consistent was not sufficient, I'd still be interested in the assumption that did suffice.

Let $$M$$ be a countable computably saturated model of ZFC with a measurable cardinal $$\kappa$$. Let $$N$$ be the Ord-length iterated ultrapower of a measure on $$\kappa$$. The model $$M$$ thinks $$N$$ is a definable well-founded class structure, which is strictly taller than $$M$$. But $$M$$ and $$N$$ are two countable models of ZFC with the same theory, same standard system, and computably saturated, and this ensures that they are isomorphic (externally). The model $$N$$ is computably saturated, since any definable class in a computably saturated structure is also computably saturated.

Here is my previous answer to your question. Here is an example of the dual situation, where a model of set theory can define a model that it thinks is strictly shorter, but to which it is actually isomorphic. I am posting this because I find it interesting and related, even though it doesn't answer your question.

Theorem. Every countable computably saturated model of ZFC is isomorphic to a rank-initial segment of itself.

Proof. Suppose that $$M$$ is a countable computably saturated model of ZFC. It follows that the theory of $$M$$ is in the standard system of $$M$$. That is, there is a natural number $$t$$ in $$M$$ that codes a sequence of natural numbers, such that the standard part of that sequence is exactly the Gödel codes of the sentences true in $$M$$. By reflection, increasing large standard fragments of this theory are true in the rank-initial segments $$V_\alpha^M$$, and so by overspill, there must be some $$N=(V_\alpha)^M$$ that satisfies a nonstandard fragment of the theory coded by $$t$$. Thus, $$N$$ and $$M$$ have the same theory, and the same standard system. From this, it follows that they are isomorphic. $$\Box$$

The model $$N$$ is definable in $$M$$ from parameters, since $$M$$ thinks $$N$$ is just $$V_\alpha$$.

You can see various versions of this argument in my paper:

Here is another variation on the theme:

Theorem. There is a countable model $$M$$ of ZFC that is isomorphic to a forcing extension $$M[c]$$ of itself.

This is a little closer to what you asked for, since $$M[c]$$ can define $$M$$, and it is isomorphic to $$M$$. But this model is still set-like, so ultimately it doesn't fulfill your requirement.

Proof. Let $$M_0$$ be a countable computably saturated model of ZFC, and let $$M=M_0[d]$$ be the model obtained by forcing to add a Cohen real $$d$$. Let $$c$$ be $$M$$-generic, and consider $$M[c]$$. Since both are obtained by forcing over $$M_0$$ to add a Cohen real, they have the same theory, and they are both computably saturated and have the same standard system, hence isomorphic. $$\Box$$

• Could something like this work? Let M be a computably saturated model of ZFC with a measurable cardinal (an extra assumption, I know). Let U be a normal ultrafilter witnessing this. Take the iterated ultrapower of M via U and its images of length Ord as defined in M (but don't try to collapse it). Call this N. I think N and M would have the same standard system, but is N computably saturated? Oct 30 '20 at 18:41
• Yes, I think this works, and I have posted an edit to my answer. Oct 30 '20 at 19:03
• But is this really longer? In what sense does $\Bbb R$ longer than $(0,1)$? It's just a scaling issue. On the other hand, $\omega_1$ is certainly longer than $\omega$. Oct 30 '20 at 20:39
• It's definitely longer. M has a definable map from it's ordinals to an initial segment of the ordinals of N. Namely, the iteration takes $\kappa$ to Ord itself, and the ordinals above $\kappa$ to "ordinals" above Ord. There is no definable converse map, even if there is one externally. Oct 30 '20 at 20:50
• Joel, it might be worth adding that if we take just a single ultrapower and collapse it in our computably saturated M, then we get an inner model which is isomorphic to M. Then shorter, wider, taller and thinner are all covered. Oct 30 '20 at 23:51