1st post (updated with some more facts about IA structures at the end)

(edited 12/10/2014 to improve clarity)

We'll explain the proof of Theorem 27 of FMS:

**Theorem 27:**Suppose $\kappa$ is a supercompact cardinal and $\mu<\kappa$ is regular. Then in $V^{\rm{Col}(\mu,<\kappa)}$, the nonstationary ideal on $\mu$ is precipitous.

We'll follow the proof in FMS, trying to use Foreman's chapter for some new insights. I think the material is generally covered much better in the Handbook chapter than in the article.

Now we have Lemma 1 for proving ideals are precipitous, we can carry out this gameplan.

Let $G$ be a generic for $\rm{Col}(\mu,<\kappa)$, and let's think in the generic extension $V[G]$. We will show:

*Main Claim:*Let $\mathfrak{A}$ be an expansion of $\langle H(\lambda),\in,\Delta\rangle$ (where $\lambda$ is some regular cardinal above, say, $2^{2^{2^\mu}}$). Then for almost all (in the sense of the nonstationary ideal) $N\prec\mathfrak{A}$ with $N\in \rm{IA}$, $|N|<\mu$, and $N\cap \mu$ an ordinal:

For all maximal antichains $\mathcal{A} \in N$ of $P(\mu)/NS$, there is an $A\in \mathcal{A}$ such that

- $N\cap \mu\in A$ and
- letting $N^A$ equal the Skolem hull of $N\cup \{A\}$, $N^A\cap \mu=N\cap \mu$ in $\mathfrak{A}$.

Let's explain the usefulness of these conditions. To apply our Lemma 1, let $\mathcal{A}=\langle A \rangle$ be a tree of maximal antichains for $P(\mu)/\rm{NS}$ with underlying tree $T\subseteq \langle\mu^+\rangle^{<\omega}$. We will construct a branch $f:\omega\rightarrow \mu^+$ so that $\bigcap_n A_{f\upharpoonright n}\neq \emptyset$.

We'll construct the branch $f$ in $\omega$ stages.

Take $N_0$ containing $\mathcal{A}$ as an element. Let $s_0=\langle\alpha_0\rangle$ be so that $A_{s_0}$ witnesses (1) and (2) with antichain equal to the first level of $\mathcal{A}$. From now on, use the notation $s_i:=\langle \alpha_0,\ldots, \alpha_{i}\rangle$.

At the $i+1$st stage, let $N_{i+1}:=(N_i)^{\alpha_i}$ (the Skolem hull of $N_i\cup\{\alpha_i\}$ in $\mathcal{L}$). Pick $\alpha_{i+1}$ so that (1) and (2) are satisfied for $N=N_{i+1}$ and $s=\langle \alpha_0,\ldots, \alpha_{i+1}\rangle$ in the maximal antichain consisting of $\langle A_s:s \textrm{ extends }s_i\rangle$ together with some set not containing the single point $N\cap \mu$ (in order to do this construction, we will need to show that the property that $N_i\in\rm{IA}$ is maintained, which is Lemma 2 from last time).

We have $N_i\cap \mu\in A_{s_i}$ for all $i$, but $N_i\cap\mu$ was always just equal to $N\cap \mu$, so $N\cap \mu$ would be in the intersection of all of the $A_{s_i}$. The key point of absorbing the index of the member of the antichain was so that the construction could continue in a coherent way, ensuring we are actually building a branch of $T$. (Foreman's chapter further develops these ideas, leading to the notion of "catching antichains".)

Let us now prove the main claim. Suppose there is a stationary $S\subseteq \rm{IA}$ of structures for which the conclusion of the main claim fails. The idea of the proof is that the supercompact gives us enough reflection of $S$ to catch the antichain. Let $\mathbb{P}=\rm{Col}(\mu,<\kappa)$ and let $j:V\rightarrow M$ be an $|H(\lambda)|$-supercompact embedding. Since $\mathbb{P}$ is $\kappa$-c.c., we can force to find $H\subseteq j(\mathbb{P})$ which contains $G$ as a subset, so we can extend $j$ to $\hat{j}:V[G]\rightarrow M[H]$ in $V[H]$.

*Subclaim:*$S$ reflects to a set of size $\mu$, i.e., there is $Y\subseteq \lambda$ so that $\mu\subseteq Y$ and $|Y|=\mu$ so that $S\cap P(Y)$ is stationary in $P(Y)$.

*Proof of Subclaim*: By Lemma 3 from last time, $\hat{j}(S)\cap P(\hat{j}``H(\lambda))$ is stationary in $V[H]$. Now because of the collapse, $|\hat{j}``H(\lambda)|=\mu$ and $\mu\subseteq \hat{j}``H(\lambda)$ since $\kappa$ is the critical point. By closure of $M$ (and, it can be seen, of $M[H]$, $\hat{j}``H(\lambda)\in M[H]$. So ($\hat{j}$ of) the statement of the subclaim holds in $M[H]$, so by elementarity the subclaim holds in $V[G]$.

Returning to the proof of the main claim, let $Y$ be given by the subclaim and let $f:\mu\rightarrow Y$ be a bijection. Let $T:=\{\delta<\mu: f``\delta\in S\textrm{ and }\delta=f``\delta\cap\mu\}$. Then it is easy to see that $T$ is stationary, using the fact that $S$ reflects to $Y$. By maximality, there is $A\in\mathcal{A}$ such that $T\cap A$ is stationary. So we can find an IA structure $N'\prec \mathfrak{A}$ containing $A, f$ and with $N'\cap \mu \in T\cap A$. Take $N$ to be $f``(N' \cap\mu)$, so $N\in S$ by definition of $T$. By the second property in the definition of $T$, we also have $N\cap \mu=N'\cap \mu$. But $N'$ contains $A$, so the Skolem Hull of $N\cup\{A\}$ is contained in $N'$ and therefore has the same supremum below $\mu$, and would show that $N$ cannot be in $S$.

This finishes the proof.