# amenable ultrafilters

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?

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

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 $$\kappa^+$$.

• Very nice! I can finally move on with my day... Dec 15, 2018 at 10:27
• 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? Dec 15, 2018 at 13:51
• @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.
– mbsq
Dec 15, 2018 at 13:55
• Maybe Woodin is enough using “surgery”?
– mbsq
Dec 15, 2018 at 14:25