Can there be only one (uncountable transitive model of ZFC)?

Question

It is an immediate consequence of Cohen's forcing that if there is one countable transitive model of $\sf ZFC$, then there are many of them. Even if all of these models are of the same height, there are still many.

But the proof of existence of generic filters don't carry over for uncountable models of set theory. Not without further assumptions (like Martin's axiom) anyway.

Is it possible to have an uncountable transitive model which is the only uncountable transitive model? Since that model is going to have to be $L_\alpha$ for some $\alpha>\omega_1$, is it also consistent with $V\neq L^1$, that a unique uncountable model exists?

Can this be pushed arbitrarily high (perhaps with, or without the assistance of $\sf MA$ and the Tennenbaum-Solovay theorem) so there are many transitive models of size $<\kappa$, but only one of size $\kappa$?

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Footnotes:

1. Clearly one can just add a Cohen real to the universe without damaging any uncountable sets in a mind-shattering way. And on the other hand, if $V\neq L$ in a strong enough way, i.e. $0^\#$ existing in the universe, then there are arbitrarily large uncountable transitive models of $\sf ZFC$. Simply $L_\kappa$ for $\kappa$ a cardinal, or a Silver indiscernible.

   So my purposely vague question about $V\neq L$ is some middle-ground between the very uninteresting $L[r]$ for some "simple" real number $r$, and the existence of large cardinals (which automatically implies there are many uncountable transitive models).

Answer 1

No, this is not possible.

As Mohammad argued in the comments, such a model $M$ must contain all countable ordinals.

We may assume that $M = L^M$, otherwise $L^M$ is another transitive model of ZFC which is uncountable since $\omega_1 \subseteq L^M$. If $M = L^M$, then $M = L_\alpha$ for some $\alpha \geq \omega_1$. The cardinal $\omega_1$ must be an uncountable regular cardinal in $L$ and I will now write $\kappa$ instead of $\omega_1$ to avoid confusion with $\omega_1^L$. We may assume that $\alpha \lt (\kappa^+)^L$ (i.e. $L \vDash |M| = \kappa$). Otherwise we could apply Löwenheim-Skolem in $L$ to get $N \prec M$ with $\kappa \subseteq N$ and $|N| = \kappa$; the transitive collapse of $N$ would then be an uncountable model of ZFC different from $M$.

There are now two cases:

• If $\alpha \gt \kappa$, then the poset $(2^{\lt\kappa})^L$ is in $M$. Working in $L$, using the fact that $2^{\lt\kappa}$ is $(\lt\kappa)$-closed and that $|M| \leq \kappa$, we can easily construct a $M$-generic $G$ for $2^{\lt\kappa}$. Then, the generic extension $M[G]$ is an uncountable transitive model of ZFC different from $M$.

• If $\alpha = \kappa$ then every set in $M$ is countable in $V$. Therefore, every set forcing in $M$ has an $M$-generic set in $V$, which leads to a plethora of different uncountable transitive models of ZFC.