TITLE: Cardinals below the critical point of a generic embedding
QUESTION [11 upvotes]: This may be an easy question.  Are the cardinals below the critical point of a precipitous ideal embedding always absolute between the generic ultrapower and the generic extension?
To focus on the simplest case, suppose $I$ is a precipitous normal ideal on $\omega_2$.  Let $G$ be generic for $\mathcal{P}(\omega_2)/I$.  Let $j : V \to M$ be the generic ultrapower.  Then $\omega_1^V$ is not moved by $j$, but could it be the case that $\omega_1^V$ is countable in $V[G]$?  Of course the answer is no if $I$ has enough presaturation.
Edit: As a subproblem towards a consistent negative answer, can we find (from some large cardinal assumption) a measurable $\kappa$ with measure $U$, and a $\kappa$-c.c. forcing $\mathbb{P}$ such that $\mathbb{P}$ preserves $\omega_1$, but $j_U(\mathbb{P})$ collapses $\omega_1$?  (Not in $Ult(V,U)$ of course, but in $V$.)

REPLY [2 votes]: Mohammad's answer here can be applied to give many negative examples.  Here's one of optimal consistency strength.  By Goldring's improvement on a result of Foreman-Magidor-Shelah, if we use the Levy collapse to turn a Woodin cardinal $\delta$ into $\aleph_{\omega+2}$, then $NS_{\aleph_{\omega+1}}$ is precipitous in the extension.  Forcing with $NS_{\aleph_{\omega+1}} \restriction cof(\omega_n)$ for $n \geq 1$ collapses $\aleph_{\omega+1}$ and $\delta$, and so must change $\delta$'s cofinality to some $\mu < \aleph_\omega$.  By Mohammad's argument, all cardinals in $(\mu,\delta]$ are collapsed.