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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?

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  • $\begingroup$ Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $\kappa$ itself. $\endgroup$
    – Asaf Karagila
    Dec 15, 2018 at 6:09
  • $\begingroup$ @AsafKaragila I need to show the factor map critical point is above $2^\kappa$. Why is that? $\endgroup$
    – mbsq
    Dec 15, 2018 at 6:58

1 Answer 1

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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 $\kappa^+$.

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    $\begingroup$ Very nice! I can finally move on with my day... $\endgroup$ Dec 15, 2018 at 10:27
  • $\begingroup$ Do you really need a supercompact? Wouldn't a measurable with $o(\kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument? $\endgroup$
    – Asaf Karagila
    Dec 15, 2018 at 13:51
  • $\begingroup$ @AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill. $\endgroup$
    – mbsq
    Dec 15, 2018 at 13:55
  • $\begingroup$ Maybe Woodin is enough using “surgery”? $\endgroup$
    – mbsq
    Dec 15, 2018 at 14:25

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