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To be more precise, suppose that $M$ is a model of ZF, for simplicity (or tactility) a set in some larger model $V$ of ZFC+Con(ZF), and suppose that $M\vDash``\alpha$ is an ordinal$"$. Must there be a model $N$ of ZFC (appearing in $V$) such that $\alpha\in N$ and an isomorphism of the partial structures given by the transitive closures of $\alpha$ in $M$ and $N$?

If yes, then can we also assume that $M$ and $N$ agree on cardinality for ordinals ${\leq}\alpha$?

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    $\begingroup$ Note that we can always get choice by shrinking: if $M\models\mathsf{ZF}$ then $L^M\models\mathsf{ZFC}$ and $M$ and $L^M$ "agree on the ordinals" (although of course they may disagree about cardinalities). In the other direction, many but not all models of $\mathsf{ZF}$ have tame class forcing extensions with the same ordinals to models of choice. Gitik's model, in which all uncountable ordinals are singular, is an example where this can't be done. $\endgroup$ Commented Mar 22, 2023 at 17:53
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    $\begingroup$ I mean, we covered the Feferman–Levy model in class... $\endgroup$
    – Asaf Karagila
    Commented Mar 23, 2023 at 23:09
  • $\begingroup$ @AsafKaragila does FefLev necessarily not admit a model of ZFC agreeing on cardinalities of ordinals? We'd disagree on cofinality, but nothing immediately strikes me as contradictory. $\endgroup$ Commented Mar 24, 2023 at 10:08
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    $\begingroup$ No, because if $M$ is the FL model, and $N$ is a model of $\sf ZFC$ with the same ordinals, then either $\omega_1^M$ is countable, or $\omega_1^M=\omega_\omega^N$. $\endgroup$
    – Asaf Karagila
    Commented Mar 24, 2023 at 10:27

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As Noah mentioned in the comments, the answer to the first question is "yes".

The answer to the second question is consistently "no", assuming a little consistency: if there is a transitive model of ZF in $V$, then there will be transitive models $M$ of ZF in $V$ and ordinals $\alpha$ of $M$ for which the extra requirement fails. For let $P$ be a countable transitive model of ZF of minimal height. Let $Q=L^P$, so $Q$ models ZFC and has the same ordinals as does $P$ (and $Q$ is also countable). Let $\lambda=\aleph_\omega^Q$. So $Q\models$ "$\lambda$ is singular". Let $G$ be $(Q,\mathrm{Coll}(\omega,{<\lambda}))$-generic. Let $\mathbb{R}^*$ be the resulting symmetric set of reals, i.e. $\mathbb{R}^*=\bigcup_{\alpha<\lambda}\mathbb{R}\cap Q[G\upharpoonright\alpha]$. Let $M=(L(\mathbb{R}^*))^{Q[G]}$. Then $M\models$ ZF and $\mathbb{R}^*=\mathbb{R}\cap M$ and $M\models$ "$\lambda=\omega_1$", so $M\models$ "$\omega_1$ is singular", since $Q=L^M\subseteq M$.

Now suppose there is a model $N$ of ZFC such that the ordinals of $N$ which are ${\leq\lambda}$ are (externally) isomorphic to those of $M$ which are ${\leq\lambda}$, and $M,N$ have the same cardinals $\leq\lambda$. Thus, $\lambda=\omega_1^N$. But $L_\lambda^N=L_\lambda^M$, and the sequence $\left<\aleph_n^{L^M}\right>_{n<\omega}=\left<\aleph_n^{L_\lambda^M}\right>_{n<\omega}$ is definable over $L_\lambda^M$, so it is in $N$, so $N\models$ "$\lambda$ is singular", a contradiction.

(Without the assumption that $\lambda+1$ is wellfounded, it doesn't seem clear that we must get $L_\lambda^N=L_\lambda^M$, so the last paragraph seems to become problematic.)

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