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The following theorem of Jech, Magidor, Mitchell and Prikry is well-known.

Theorem. (1) If $\kappa$ is a regular cardinal that carries a precipitous ideal, then $\kappa$ is a measurable cardinal in an inner model of ZFC.

(2) If $\kappa$ is a measurable cardinal, then there is some $\mathbb{P}$ such that $\mathbb{P}$ forces that $\kappa$ carries a precipitous ideal.

Problem. Suppose there is a precipitous ideal on $\omega_{1}$. Then by the theorem, we know there is some $U$ such that in $L[U]$, $U$ witnesses that $\omega_{1}$ is measurable.

Fixing such a $U$, is there a poset $\mathbb{P}$ and $\mathbb{P}$-generic $G\in V$ (this is the rub) over $L[U]$ such that $L[U][G]$ satisfies that $\omega_{1}$ has a precipitous ideal?

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    $\begingroup$ Yes, just do the usual Levy collapse. The tricky part is if you want the original ideal to be generic over L[U], which I guess could be impossible. $\endgroup$
    – Asaf Karagila
    Mar 23, 2022 at 6:43
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    $\begingroup$ @AsafKaragila Why is there a Levy collapse filter in $V$? $\endgroup$
    – Farmer S
    Mar 23, 2022 at 10:26
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    $\begingroup$ @Farmer: Oh, I should write a book titled "On the Dangers of Posting Comments Immediately After Waking Up". Regardless, since $\omega_1$ is the measurable of $L[U]$, there's a filter for the point wise collapse of all the ordinals below it. I guess the only missing ingredient is to combine them together? $\endgroup$
    – Asaf Karagila
    Mar 23, 2022 at 10:34
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    $\begingroup$ Asaf, I think for some ground model and $U$, it's possible for there to be no generic filter for the Levy collapse in this situation. For example, start with $V_0=L[U]$ with a measurable $\kappa$ and normal measure $U$. Forcing to collapse it to $\omega_1$ with Silver collapse, then in $V=V_0[G]$, $\kappa$ is $\omega_1$ and has a precipitous ideal. If we happen to pick the $U$ we started with, then $L[U]$ here is just the ground model. But since $V$ is a Silver collapse extension, there is no $V_0$-generic filter for Levy collapse (see here). $\endgroup$ Mar 23, 2022 at 21:08
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    $\begingroup$ @JasonZeshengChen In fact the filter $U$ is uniquely determined by the critical point, assuming we take it to be normal in $L[U]$; anyway if we allow it to be non-normal, we still get $U'$ such that $L[U']=L[U]$. $\endgroup$
    – Farmer S
    Mar 24, 2022 at 0:36

1 Answer 1

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If more large cardinals exist, then there is such a filter. That is, suppose that the least $\kappa_0$ such that there is a proper class transitive model of "$V=L[U_0]$", is countable, (correction) and so is $\kappa_0^{+L[U_0]}$. (This holds if there is a proper class transitive model with two measurables, for example.) Then there is a such a filter.

(EDIT: I had ignored the possibility that $\kappa_0^{+L[U_0]}=\omega_1$ in the previous version. But if there is a precipitous filter on $\omega_1$ and $\kappa_0<\omega_1$ then $\kappa_0^{+L[U_0]}<\omega_1$, because otherwise if $j:V\to M$ is a generic embedding with $\mathrm{crit}(j)=\omega_1$ then $j(\kappa_0^{+L[U_0]})>\kappa_0^{+L[U_0]}$, although $j(\kappa_0)=\kappa_0$, which gives us two distinct $L[U]$-type models with the same critical point in $V[G]$, which is impossible.)

The construction is along the lines suggested by @AsafKaragila in his comment above. There are also similar such constructions in the literature; I think Sy Friedman used such constructions, for example. It suffices to find an $(L[U],\mathrm{Coll}(\omega,{<\omega_1}))$-generic filter. For this, first observe that because $\kappa_0^{+L[U_0]}<\omega_1$, the linear iteration of $L[U_0]$ of length $\omega_1$ does not send the measurable past $\omega_1$ at any stage ${<\omega_1}$, so the eventual $\omega_1$st iterate has measurable $\omega_1$, so it is just $L[U]$. Let $C\subseteq\omega_1$ is the club of critical points resulting from iterating $L[U_0]$ out to $\omega_1$, let $L[U_\alpha]$ be the $\alpha$th iterate of $L[U_0]$, with measurable $\kappa_\alpha\in C$. Choose a sequence $\left<G_\kappa\right>_{\kappa\in C}$ such that $G_\alpha$ is $\mathrm{Coll}(\omega,{<\kappa_\alpha})$-generic over $L[U_\alpha]$, and such that $G_\alpha\subseteq G_\beta$ when $\alpha<\beta$. We can proceed from $G_\alpha$ to $G_{\alpha+1}$ because

(i) $\mathrm{Coll}(\omega,{<\kappa_\alpha})^{L[U_\alpha]}=\mathrm{Coll}(\omega,{<\kappa_\alpha})^{L[U_{\alpha+1}]}$

(ii) $\mathcal{P}(\kappa_\alpha)\cap L[U_\alpha]=\mathcal{P}(\kappa_\alpha)\cap L[U_{\alpha+1}]$ is countable,

(iii) $\mathrm{Coll}(\omega,{<\kappa_{\alpha+1}})$ factors $\mathrm{Coll}(\omega,{<\kappa_\alpha})\times\mathrm{Coll}(\omega,[\kappa_\alpha,\kappa_{\alpha+1}))$

(iv) for limit $\lambda\leq\omega_1$, $\mathrm{Coll}(\omega,{<\kappa_\lambda})$ is just the direct limit of the $\mathrm{Coll}(\omega,{<\kappa_\alpha})$ under the iteration maps, which is the same thing as their finite support product, and $L[U_\lambda]$ is the direct limit of the $L[U_\alpha]$'s, so all dense subsets of $\mathrm{Coll}(\omega,{<\kappa_\lambda})$ in $L[U_\lambda]$ are met by some $G_\alpha$ where $\alpha<\lambda$.

In particular, $G_{\omega_1}$ is the desired $(L[U],\mathrm{Coll}(\omega,{<\omega_1}))$-generic filter.

But things seem more subtle if $\kappa_0=\omega_1$.

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    $\begingroup$ I first saw the construction in Greg Hjorth's "The size of the ordinal $u_2$". $\endgroup$ Mar 24, 2022 at 16:44
  • $\begingroup$ Thanks. That's a very pleasing construction. $\endgroup$ Mar 24, 2022 at 23:23
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    $\begingroup$ So, really you're just assuming $0^\dagger$, right? $\endgroup$
    – Asaf Karagila
    Mar 25, 2022 at 8:30
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    $\begingroup$ Right, $0^\dagger$ suffices to get $\kappa_0<\kappa_0^{+L[U_0]}<\omega_1$. $\endgroup$
    – Farmer S
    Mar 25, 2022 at 9:56

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