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.



Is there a model of ZFC that can define a "longer" model of ZFC to which it is isomorphic?
The answer is yes.


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.




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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:


* *Gitman, Victoria; Hamkins, Joel David*, [**A natural model of the multiverse axioms**](http://jdh.hamkins.org/naturalmodelofmultiverseaxioms/), Notre Dame J. Formal Logic 51, No. 4, 475-484 (2010). [DOI:10.1215/00294527-2010-030](http://dx.doi.org/10.1215/00294527-2010-030), [ZBL1214.03035](https://zbmath.org/?q=an:1214.03035).


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. $\Box$


