amenable ultrafilters

Question

Suppose $M$ is a transitive model of ZFC-powerset. If $\kappa \in M$ is a cardinal and $U$ is an ultrafilter on the boolean algebra $\mathcal P(\kappa)^M$, we say $U$ is amenable to $M$ if whenever $\{ X_\alpha : \alpha < \kappa \} \subseteq \mathcal P(\kappa)$ is in $M$, then $\{ \alpha : X_\alpha \in U \} \in M$. In these notes, Steel says that if $j : M \to N$ is elementary with critical point $\kappa$, and $U$ is the ultrafilter on $\mathcal P(\kappa)^M$ derived from $j$, then $U$ is amenable to $M$ if and only if $\mathcal P(\kappa)^M = \mathcal P(\kappa)^N$. I am able to show that $U$ is amenable to $M$ iff $\mathcal P(\kappa)^M = \mathcal P(\kappa)^{Ult(M,U)}$, but I don't see why necessarily $\mathcal P(\kappa)^{Ult(M,U)} = P(\kappa)^N$. Is the claim true, and how do you show it?

Answer 1

The claim is false.

Assume there is a measurable $\kappa$ and for every inaccessible $\delta < \kappa$, there is a precipitous ideal on $\delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.

Let $i : V \to M$ be the ultrapower embedding by a normal measure $U$ on $\kappa$. In $M$, there is a precipitous ideal $I$ on $\kappa^+$. Let $k : M \to N$ be a generic embedding coming from $I$. Note that $crit(k) = \kappa^+$. Let $j = k \circ i$.

$U$ is amenable to $V$, and for every $A \in \mathcal P(\kappa)^V$, $\kappa \in i(A)$ iff $\kappa \in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $\mathcal P(\kappa)^M \not= \mathcal P(\kappa)^N$, since forcing with $I$ collapses $\kappa^+$.