Spencinar #3: Precipitous ideals (part 1)

This week I'll continue the theme of Spencinar #1 by introducing some more concepts around the idea of generic ultrapowers by ideals. The emphasis will be on using large cardinals to construct ideals on small cardinals. The material comes mostly from Foreman's comprehensive chapter in the Handbook.

Prerequisites: Basic theory of large cardinals, forcing, and ultrapowers, some ideas in the first Spencinar post. The Boolean algebra approach to forcing will be helpful.

Recall the basic situation. If $I$ is an ideal on a set $Z$, then we can force with the poset $P(Z)/I$ as defined in Spencer's talk. Its elements are the mod $I$ equivalence classes of the positive subsets of $P(Z)$, ordered by the subset relation (also taken mod $I$). We denote the class of $S\subseteq Z$ by $[S]_I$. This poset is often called the quotient algebra of $I$. The generic object $G$, obtained by forcing with this poset is (equidefinable with) a $V$-ultrafilter, and so in the extension $V[G]$ we can take the $V$-ultrapower of $V$ by $G$. We denote this ultrapower by $V^Z/G$.

The basic definition, due to Jech and Prikry, is:

Definition: An ideal $I$ on a set $Z$ is precipitous if it is forced by $P(Z)/I$  that:

$V^Z/G$ is well-founded (in V[G]).

There are several combinatorial characterizations of precipitousness for an ideal $I$.

Let's give some nontrivial examples of this. A very basic result is that saturated ideals are precipitous.

Proposition: Suppose $I$ is $\kappa$-complete and $\kappa^+$-saturated. Then $I$ is precipitous.

Proof: Suppose that $I$ is not precipitous, so we can find $[S]_I$ forcing that $V^Z/G$ is ill-founded. Then let $\langle\dot{F}_n:n<\omega\rangle$ be $P(Z)/I$-names for functions $Z\rightarrow V$ in $V$ so that $[S]_I$ forces that for every $n$, $[\dot{F}_{n+1}]_G \in [\dot{F}_n]_G$ in the ultrapower.

The key is that one can replace the  $\langle\dot{F}_n:n<\omega\rangle$ by actual functions in $V$ without changing $S$. Let $n<\omega$. Find a maximal $P(Z)/I$ antichain $\mathcal{A}$ below $[S]_I$ deciding the value of $\dot{F}_n$. By a "disjointifying" argument using saturation and completeness, we can find a system of pairwise disjoint representatives $\langle A_i:i\in \mathcal{A}\rangle$ for the elements of this antichain. Define $f_n(z)$ to be the value of $\dot{F}_n$ forced by $[A_i]_I$ if $z\in A_i$, and 0 otherwise. Then we can see that $[S]_I$ forces $[\dot{F}_n]_G = [f_n]_G$. 

But now (using countable completeness of $I$), in $V$, for almost every $z\in S$, $f_{n+1}(z)\in f_n(z)$ for every $n$, giving an ill-foundedness in $V$. $\Box$

So the generic ultrapowers we considered last time (which didn't exist) were well-founded.

An immediate question is: can there be a precipitous ideal on small cardinals, e.g., $\omega_1$? The generic ultrapower construction and absoluteness of constructibility show that existence of any precipitous ideal implies that it is consistent that there is an embedding $L\rightarrow L$, so existence of a precipitous ideal has consistency strength at least that of the existence of $0^\#$.

We construct a precipitous ideal in a forcing extension starting from a measurable cardinal. The ideal is a fundamental one--start from a large cardinal ideal $J$, force to collapse the large cardinal to a small cardinal, and show that the ideal $I$ generated by $J$ in the extension retains some of the nice properties of $J$.

Actually, we'd like our precipitous idea to be a natural ideal, e.g., the nonstationary ideal on $\omega_1$. This is possible. However, today we will only get precipitousness for an induced (read: unnatural) ideal, that is an ideal which is defined to be the collection of subsets forced not to be in a particular ultrafilter of the form $U(\hat{j},i)=\{X:i\in \hat{j}(X)\}$ derived by a certain generic embedding $\hat{j}$. (We'll see this in action below).

Theorem: Suppose $\kappa$ is a measurable cardinal and let $U$ be a $\kappa$-complete normal ultrafilter on $\kappa$. Let $\mathbb{P}=\mathrm{Col}(\omega, <\kappa)$ be the Levy poset collapsing $\kappa$ to $\omega_1$. Then the ideal $I$ in $V^\mathbb{P}$ generated by the dual of $U$ is precipitous in $V^\mathbb{P}$.

Proof. By the basic theory of elementary embeddings and large cardinals, the ultrapower embedding $j:V\rightarrow M$ given by $U$ can be extended to $j^+:V[G]\rightarrow M[G\ast H]$, where $G$ is generic for $\mathbb{P}$ and $G\ast H$ is $V$-generic for $j(\mathbb{P})$ (here we used that $\mathbb{P}$ completely embeds into $j(\mathbb{P})=Col(\omega, <j(\kappa))^M$, and we'll write $j(\mathbb{P})=\mathbb{P}\ast \mathbb{Q}$). The extension takes the object $\mathbb{P}$-named by $\dot{\tau}$ (evaluated by $G$) to the object $j(\mathbb{P})$-named by $j(\dot{\tau})$ (evaluated by $G\ast H$).

The correct way to think of this is that forcing with $\mathbb{Q}$ over $V$ adds the embedding $j^+$.

First we show that (in $V[G]$) 

Claim: $I$ is equal to the ideal $\{X \subset \kappa : 1\Vdash  \kappa \not\in  j^+(X) \}$. 

Clearly, $I$ is contained in this ideal. Conversely, work in $V$ and suppose $\dot{X}$ is a $\mathbb{V}$ name for a subset of $\kappa$, and $p\in \mathbb{P}$ forces that $1\Vdash_\mathbb{ Q} \kappa \not\in  j^+(X)$. Define $A=\{\alpha:p\Vdash\alpha\not\in \dot{X}\}$. Then $A\in U$ (and so $p$ forces that $\dot{X}\in \dot{I}$, finishing the proof), otherwise we can define $F:(\kappa \setminus A)\rightarrow \mathbb{P}/p$ so that $F(\alpha)$ forces $\alpha\in X$. But then in $M$, $F$ represents a condition below $j(p)=p$ which forces $\kappa$ into $j^+(X)$, contradicting the choice of $p$. This proves the claim.

Now we show the precipitousness. Work in $V[G]$. Define a Boolean algebra embedding $P(\kappa)\rightarrow \mathcal{B}(\mathbb{Q})$ (here the codomain is the Boolean completion of $\mathbb{Q}$) which sends a set $X$ to the truth value $\lVert \kappa\in j^+(X)\rVert$. The use of the Boolean completion is so that this map is defined--in general, many different members of $\mathbb{Q}$ could force $\kappa\in j^+(X)$, but in the Boolean completion there is a greatest such member. Since $I$ is the kernel of this embedding, we see that the embedding factors to $\iota:P(\kappa)/I\rightarrow \mathcal{B}(\mathbb{Q})$. We will show that $\iota$ gives a dense embedding from the quotient algebra of $I$ into $\mathbb{Q}$. 

Claim: The range of $\iota$ is dense in $\mathcal{B}(\mathbb{Q})$. 

Since $\mathbb{Q}$ is densely embedded in $\mathcal{B}(\mathbb{Q})$, it suffices to show that for any $q\in\mathbb{Q}$ there is $X\subseteq \kappa$ with $\iota(X)=q$. Let $q=j(F)(\kappa)$, where $F\in V$ is a function such that for all $\alpha$, $F(\alpha)\in \mathrm{Col}(\omega, <\kappa)$. Define $X=\{\alpha: F(\alpha)\in G\}$. Now for any $H$ which is $V[G]$ generic for $\mathbb{Q}$, we have

 $\iota(X)\in H$ iff $\kappa \in j^+(X)$ (definition of $\iota$)

 iff $j(F)(\kappa)\in j^+(G)$ (definition of $X$)

iff $q\in j^+(G)$ (definition of $F$)

iff $q\in H$

so by separativity, $\iota(X)=q$, proving the claim.

So forcing with the quotient algebra of $I$ is equivalent to forcing with $\mathbb{Q}$; generic filters are equidefinable using the $\iota$ function. So let $H$ be obtained by applying $\iota$ image to an arbitrary $V[G]$-generic filter for $P(Z)/I$.

Now in $V[G\ast H]$, let $U^+=\{X\subseteq \kappa: \kappa\in j^+(X)\}$ be the ultrafilter derived from $j^+$, so we know from general considerations that the ultrapower of $V[G]$ by $U^+$ embeds into $M[G\ast H]$ (the embedding takes the element represented by a function $f$ to $j^+(f)(\kappa)$). But $U^+$ is exactly the filter generated the $\iota$-preimage of $H$, by the definition of $\iota$, so it is the original generic for $P(Z)/I$.