Q:

Uniqueness of countable version of $L[U]$?

Suppose $\alpha$ is a countable ordinal and $U_0,U_1,\kappa$ are such that $L_\alpha[U_i] \models \mathrm{ZFC} + U_i$ is a normal ultrafilter on $\kappa$.  Does $U_0 = U_1$?
The argument for uniqueness of $L[U]$ due to Kunen uses an iteration of length a $V$-regular cardinal above $\kappa$.  This is not available for countable models.  I am not assuming the models are fully iterable from the outside.

A:

Andrés's comment can be turned into an answer, but there is a slight subtlety. Suppose $M$ is a countable iterable model of ZFC + $V = L[U]$. Let $\kappa$ be its measurable cardinal and $\alpha$ be its ordinal height. Then there is a transitive model $N\neq M$ of height $\alpha$ satisfying ZFC + $V=L[U]$ + $\kappa$ is measurable. The subtlety is that no such model $N$ can lie in $L$: since $\alpha$ and $\kappa$ are uncountable in $L$ (they're Silver indiscernibles), $N$ contains all the constructible reals, and since $N$ thinks there is a measurable cardinal, $N$ contains nonconstructible reals as well. The model $N$ will instead belong to a generic extension of $L$. Let $g\subseteq \text{Col}(\omega,\alpha)$ be $L$-generic. Then by $\Sigma^1_1$-absoluteness, $L[g]$ contains a model $N$ of height $\alpha$ satisfying satisfying ZFC + $V=L[U]$ + $\kappa$ is measurable. Note that $N$ is not equal to $M$ since $M$ does not belong to any generic extension of $L$: indeed $0^\#\in M$.

