# Precipitous ideal and inner model

Assume $$\kappa$$ is measurable, $$U$$ is its unique normal measure, $$V=L[U]$$. We levy collapse $$\kappa$$ to make it become $$\omega_1$$.

If we don't have the inner model condition, then we only know that $$\omega_1$$ has a precipitous ideal. In this case if we want the nonstationary ideal to be precipitous, we need a much more complex forcing notion. So my question is about the inner model case:

Do we have that, the nonstationary ideal on $$\omega_1$$ is precipitous?

No. Suppose otherwise. Let $$G$$ be the Levy collapse generic, and $$D$$ be the generic for forcing with the nonstationary ideal after that. Since $$\mathrm{Ult}(L[U,G],D)$$ is wellfounded, letting $$i_D:L[U,G]\to\mathrm{Ult}(L[U,G],D)$$ be the ultrapower map, we have that $$i_D(L[U])=L[i_D(U)]$$ is some iterate of $$L[U]$$. Let $$j:L[U]\to L[i_D(U)]$$ be the iteration map. Let $$\Gamma$$ be a proper class of ordinals fixed by both $$j$$ and $$i_D$$. We have $$j(U)=i_D(U)$$ and $$j(\alpha)=i_D(\alpha)$$ for all $$\alpha\in\Gamma$$. But $$L[U]=\mathrm{Hull}^{L[U]}_{\Sigma_1}(\Gamma)$$ (using that $$\Gamma$$ is proper class and definable in a generic extension of $$L[U]$$). It follows that $$j=i_D\upharpoonright L[U]$$. But working back in $$L[U]$$, because this Levy collapse (correction) has the $$\kappa$$-cc, the set $$X$$ of all $$\alpha<\kappa$$ of cofinality $$\omega$$ remains stationary in $$V[G]=L[U,G]$$. So we could have taken $$X\in D$$, giving that $$\kappa\in i_D(X)$$, whereas $$\kappa\notin j(X)$$, but $$X\in L[U]$$, a contradiction.
• One minor thing. We are using $\mathrm{Col}(\omega,<\kappa)$, which is not countably closed, but preserves stationary subsets of $\kappa$ by the $\kappa$-c.c. Dec 5, 2023 at 9:06