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In the context of generic embeddings, we fix a set $Z$, an ideal $I$ on $Z$, and $G$ a generic ultrafilter on $\mathcal{P}(Z) / I$. In $V[G]$, we can define the generic ultrapower $Ult(V, G)$ and an elementary embedding from $V$ into $Ult(V, G)$. If $I$ is precipitous then $Ult(V, G) \cong M$ for some transitive class $M$ of $V[G]$, so we also have $j : V \rightarrow M$ an elementary embedding. Now if $I$ has the disjointing property then in particular, $M^\omega \subset M$, which is a nice closure property to have.

I am interested in the scenario where one iterates this ultrapower construction in a larger forcing extension. Specifically, we consider $Z = \omega_1$ and $I = \mathrm{NS}_{\omega_1}$, the non-stationary ideal on $\omega_1$. If $I$ is precipitous then in $V^{Col(\omega, 2^{\omega_1})}$, $(V; \in, I)$ becomes generically iterable. Fix $G$ a $Col(\omega, 2^{\omega_1})$-generic filter over $V$ and consider a generic iteration $\mathfrak{I}$ of $(V; \in, I)$ of length $\omega_1^{V[G]}$ in $V[G]$. This yields an elementary embedding $j: V \rightarrow M$, where $M$ is the final iterate of $\mathfrak{I}$. Here $crit(j) = \omega_1^V$ and $j(\omega_1^V) = \omega_1^{V[G]}$. In this case, what additional properties does $I$ need to have, in order to guarantee $M^\omega \subset M$ in $V[G]$?

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It is impossible that $M^{\omega}\subseteq M$ in $V[G]$. The map $j$ is continuous at $\omega_2^V$, i.e. $\omega_2^M=\sup_{\alpha<\omega_2^V} j(\alpha)$ (this can be seen by induction along the length of the iteration). As $\omega_2^V$ is countable in $V[G]$, this means that $\omega_2^M$ has countable cofinality in $V[G]$, but of course $\omega_2^M$ is regular in $M$.

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