The basic idea is that one uses the isomorphism there:
\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})
to calculate the quotient algebra P(Z)/J as \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})/e''H.
As discussed near the end of the last lecture, under certain assumptions, the statement of the duality theorem becomes somewhat simpler. The first examples will fall into this case.
Special Case: If I is \kappa-complete and \mathbb{P} is \kappa-c.c., then the hypothesis of the duality theorem holds, K=\{0\} and J=\bar{I}, the ideal generated by I in V^\mathbb{P}.
Exercise: Show that in the above case, if p\in \mathbb{P} and A\in (P(Z)/I)\cap V, then
\iota(p,\check{A})=(A,j(\dot{p})).
A further simplification will be that we will usually start with a measurable cardinal \kappa and take I to be the dual to the measure on \kappa, so P(\kappa)/I is the trivial Boolean algebra.
A measurable cardinal \kappa has a 2-saturated, \kappa-complete ideal, namely the dual to the measure on \kappa, and under GCH every cardinal \kappa carries a \kappa^{++}-saturated, \kappa-complete ideal, namely the ideal of bounded subsets. This motivates the following natural questions, which are the main focus of this lecture:
Question: Suppose \mu\le \kappa^+ is a regular cardinal. Is it consistent that there is a cardinal \kappa which is not measurable, but still \kappa carries a \mu-saturated, \kappa-complete ideal? (Here we want the amount of saturation to be exactly \mu).
Digression: does the answer change if we require \kappa to be a successor cardinal?
For the case where \kappa is a successor cardinal, \kappa^+-saturation is the strongest we can hope to achieve.
Exercise: Prove using the method of generic ultrapowers that if \kappa is a successor cardinal then there is no \kappa-complete, \kappa-saturated ideal on \kappa.
Kunen showed that if \kappa is a successor cardinal, then getting a \kappa^+-saturated ideal on \kappa requires large cardinals much stronger than a measurable, although we won't do this argument here (you can find it in this previous Specinar post. We now return to the original question.
Answer to question 1, if \mu<\kappa (\mu regular): We will use the basic technique of computing the quotient algebra P(\kappa)/J in V[H] using the duality theorem. Start with \kappa measurable in the ground model. Let \theta\ge \kappa, and consider \mathrm{Add}(\mu,\theta) which adds \theta Cohen subsets of \mu. \mathrm{Add}(\mu,\theta) is \kappa-c.c., and under the GCH it is even \mu^+-c.c. Let I=U^*, where U is a \kappa-complete normal ultrafilter on \kappa (here the star means taking the dual ideal). Let j:V\rightarrow M be the ultrapower embedding.
The duality theorem gives the isomorphism:
\mathrm{Add}(\mu,\theta)\ast P(\kappa)/\bar{I}\cong P(\kappa)/I\ast \mathrm{Add}(\mu,j(\theta))\cong\mathrm{Add}(\mu,j(\theta)),
since P(\kappa)/I is trivial.
If H is generic for \mathrm{Add}(\mu,\theta) over V, then
e''H=\{(1,j(p)):p\in H\}.
So
P(\kappa)/\bar{I}\cong \mathrm{Add}(\mu,j(\theta))/e''H\cong \mathrm{Add}(\mu,j(\theta)).
In V[H], P(\kappa)/\bar{I}\cong \mathcal{B}(\mathrm{Add}(\mu,j(\theta)), so \bar{I} is (\mu^{<\mu})^+ saturated. Furthermore, it is easy to check that \bar{I} is \kappa-complete, and 2^\mu\ge \kappa in V[H], so \kappa is not measurable. This answers question 1 for the case where \mu<\kappa. \Box
We might ask what large cardinal properties of \kappa are implied by this ideal hypothesis.
Proposition: If \kappa carries a \kappa-complete \mu-saturated ideal for some \mu<\kappa, then:
- \kappa is weakly Mahlo
- \kappa has the tree property.
Exercise: Prove (1) of the proposition using generic ultrapowers.
Proof of Proposition (2): Suppose T is a \kappa-tree. If G\subseteq P(\kappa)/I is generic, then in V[G], T has a branch b given by taking any member of level \kappa of j(T), where j is the generic ultrapower embedding. Now for each \alpha<\kappa, S_\alpha=\{x\in T_\alpha: \exists p(p\Vdash \check{x}\in \dot{b})\}<\mu by the saturation. Now \bigcap_{\alpha<\kappa} S_\alpha is a \kappa-tree all of whose levels have size <\mu<\kappa. It is well-known (or a good exercise) that such trees have cofinal branches. \Box
Definition: An ideal I is nowhere prime if there is no I-positive set A so that I\upharpoonright A is prime (i.e., dual to an ultrafilter).
Exercise: Show that if there is a nowhere prime, \kappa-complete, \mu^+-saturated ideal, where \mu<\kappa, then 2^\mu\ge \kappa.
We continue with Question 1 with other arrangements of \mu and \kappa.
Answer to question 1, if \mu=\kappa^+: Start with \kappa measurable with 2^\kappa=\kappa^+, U a normal ultrafilter and j:V\rightarrow M the ultrapower embedding. Let \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle be the Easton support iteration where for regular \alpha, \Vdash_{\mathbb{P}_\alpha} \dot{Q}=\mathrm{Add}(\alpha,1). It is straightforward to verify that \mathbb{P}_\kappa has the \kappa-c.c., so we are in the special case again.
Note that j(\mathbb{P}_\kappa)=\mathbb{P}_\kappa\ast \mathbb{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M (We use the notation \mathbb{P}_{\xi,j(\kappa)} for j(\mathbb{P}_\kappa)_{\xi,j(\kappa)}). The tail part is computed differently in M than in V, e.g., the support is on M-regular cardinals.
If G_\kappa\subseteq \mathbb{P}_\kappa is generic over V, then the special case of the duality theorem says that in V[G_\kappa], P(\kappa)/\bar{I}\cong (\mathbb{P}_{\kappa,j(\kappa)})^M. However, P(\kappa)/\bar{I}\cong(\mathbb{P}_{\kappa,j(\kappa)})^M does not have the \kappa^+-c.c. since there are M-regular cardinals between \kappa^+ and j(\kappa). So \bar{I} is not \kappa^+-saturated.
Now we could also satisfy the hypothesis of the duality theorem of adding a j(\mathbb{P}_\kappa)=P(\kappa)/J\cong \mathrm{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M generic filter \hat{H} over M in a different way. By a standard technique, a j(\mathbb{P})-generic over M exists in V[G_{\kappa+1}] (where G_{\kappa+1} is \mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)-generic). This is because we clearly get a generic for the initial part \mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1), just G_{\kappa+1} itself. For the tail, (\mathbb{P}_{\kappa+1,j(\kappa)})^{M[G_{\kappa+1}]} is is j(\kappa)-c.c. of size j(\kappa) in M[G_{\kappa+1}], so M[G_{\kappa+1}] thinks the poset has at most j(\kappa) maximal antichains. In V[G_{\kappa+1}], |j(\kappa)|=\kappa^+, and the poset is \kappa^+-closed (in M[G_{\kappa+1}], but also in V[G_{\kappa+1}] by the agreement between these models). So we can construct a generic by hand in V[G_{\kappa+1}] by enumerating all of the maximal antichains in M[G_{\kappa+1}]. This completes the construction of \hat{H} in the extension by \mathrm{Add}(\kappa,1).
In this construction, we have that j(\mathbb{P})/K\cong \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1), since the Boolean algebra homomorphism j(\mathbb{P}_\kappa)\rightarrow \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1) given by p\mapsto \|p\in \hat{H}\|. has kernel exactly K as defined in the last lecture, and the map is surjective since the codomain completely embeds into the domain. So in the duality theorem calculation, we obtain an ideal J so that P(\kappa)/J\cong \mathrm{Add}(\kappa,1). So J is a \kappa^+-saturated ideal on \kappa. \Box
Note that \kappa is inaccessible.
Exercise: Prove that \kappa is weakly compact in V[G_\kappa]. (Hint: use the tree property characterization.)
Exercise: Prove that \kappa is not measurable in V[G_\kappa], but it is measurable in V[G_{\kappa+1}].
Remark: By forcing with (\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]} instead of just \mathrm{Add}(\kappa,1) to add the j(\mathbb{P})-generic, we can get a nowhere prime \kappa complete \kappa^+-saturated ideal on \kappa in V[G_{\kappa+1}].
Answer to question 1, if \mu=\kappa: We will find an example so that \kappa is not weakly compact (compare to earlier results for saturation below \kappa), and in fact the quotient algebra is isomorphic to a \kappa-Suslin tree.
In the exercises, we will describe how to construct, for \alpha regular with \alpha^{<\alpha}=\alpha, a forcing \mathbb{Q}_\alpha which adds an \alpha-Suslin tree \dot{T}_\alpha so that \mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1). This is due to Kunen.
In the construction for \mu=\kappa^+ we got a model (there called V[G_\kappa]) where there was an inaccessible \kappa and an ideal J on \kappa so that
P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},
where \dot{T} is the \kappa-Suslin tree added by \mathbb{Q}_\kappa.
Now start with this to be our ground model V. Let H\subseteq \mathbb{Q}_\kappa be generic. We want to show that in V[H], there is an ideal J_1 on \kappa so that P(\kappa)/J_1\cong T.
If G\subseteq \mathrm{Add}(\kappa,1) is generic over V, then take in V[G] an embedding
j:V\rightarrow M
which was constructed before. We want to extend the embedding to V[H].
Now G\in M since M is closed under \kappa-sequences in V[G]. We can extend j to V[G] by constructing a generic \hat{G} for \mathrm{Add}(j(\kappa))^M over M with \hat{G}\upharpoonright \kappa=G (using the standard method; cf the second exercise following previous construction).
In V, by duality theorem there are J_1 and K so that
\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.
In this case, K is a maximal ideal since the j(\mathbb{Q}_\kappa)-generic over M is already just added by P(\kappa)/J. So in V[H], P(\kappa)/J_1\cong T.
Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing \mathbb{Q} are normal trees of successor ordinal height <\kappa which are homogeneous: for all t\in T not on the top level, T_t\cong T, where T_t is the tree \{s\in T: t\le_T s\} with the order inherited from T.
Exercise:
\Box
We will do one last application to construct a precipitous ideal on a cardinal \kappa which is not measurable so that its quotient algebra is \kappa^+ closed.
Start with \kappa measurable and 2^\kappa>\kappa^+. We will use the Easton support iteration \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle, where \dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)} for inaccessible \alpha<\kappa (and is trivial forcing otherwise).
Then \mathbb{P}_\kappa is \kappa-c.c., and forces that for all inaccessible \alpha<\kappa, 2^\alpha=\alpha^+ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).
If G_\kappa\subseteq \mathbb{P}_\kappa is generic, then j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}. Since M[G_\kappa] is closed under \le \kappa sequences in V[G_\kappa], this tail is \kappa^+-closed forcing over V[G_\kappa].
However, GCH holds at every inaccessible \alpha<\kappa and fails at \kappa in V[G_\kappa]. By a reflection argument, \kappa cannot be measurable in V[G_\kappa]. \Box
Exercise: Show that if H\subseteq P(\kappa)/\bar{I} is generic, then \kappa is measurable in V[G_\kappa\ast H].
Note that \kappa is inaccessible.
Exercise: Prove that \kappa is weakly compact in V[G_\kappa]. (Hint: use the tree property characterization.)
Exercise: Prove that \kappa is not measurable in V[G_\kappa], but it is measurable in V[G_{\kappa+1}].
Remark: By forcing with (\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]} instead of just \mathrm{Add}(\kappa,1) to add the j(\mathbb{P})-generic, we can get a nowhere prime \kappa complete \kappa^+-saturated ideal on \kappa in V[G_{\kappa+1}].
Answer to question 1, if \mu=\kappa: We will find an example so that \kappa is not weakly compact (compare to earlier results for saturation below \kappa), and in fact the quotient algebra is isomorphic to a \kappa-Suslin tree.
In the exercises, we will describe how to construct, for \alpha regular with \alpha^{<\alpha}=\alpha, a forcing \mathbb{Q}_\alpha which adds an \alpha-Suslin tree \dot{T}_\alpha so that \mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1). This is due to Kunen.
In the construction for \mu=\kappa^+ we got a model (there called V[G_\kappa]) where there was an inaccessible \kappa and an ideal J on \kappa so that
P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},
where \dot{T} is the \kappa-Suslin tree added by \mathbb{Q}_\kappa.
Now start with this to be our ground model V. Let H\subseteq \mathbb{Q}_\kappa be generic. We want to show that in V[H], there is an ideal J_1 on \kappa so that P(\kappa)/J_1\cong T.
If G\subseteq \mathrm{Add}(\kappa,1) is generic over V, then take in V[G] an embedding
j:V\rightarrow M
which was constructed before. We want to extend the embedding to V[H].
Now G\in M since M is closed under \kappa-sequences in V[G]. We can extend j to V[G] by constructing a generic \hat{G} for \mathrm{Add}(j(\kappa))^M over M with \hat{G}\upharpoonright \kappa=G (using the standard method; cf the second exercise following previous construction).
In V, by duality theorem there are J_1 and K so that
\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.
In this case, K is a maximal ideal since the j(\mathbb{Q}_\kappa)-generic over M is already just added by P(\kappa)/J. So in V[H], P(\kappa)/J_1\cong T.
Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing \mathbb{Q} are normal trees of successor ordinal height <\kappa which are homogeneous: for all t\in T not on the top level, T_t\cong T, where T_t is the tree \{s\in T: t\le_T s\} with the order inherited from T.
Exercise:
- Show that Kunen's forcing is \kappa-strategically closed. Hint: the strategy will go by choosing a particular branch through each of the small trees chosen in a play of the game so far.
- Show that \mathbb{Q}\ast \dot{T} has a \kappa-closed dense subset, and deduce that \mathbb{Q}\ast \dot{T}\cong \mathrm{Add}(\kappa).
- Show that \dot{T} is a Suslin tree.
\Box
We will do one last application to construct a precipitous ideal on a cardinal \kappa which is not measurable so that its quotient algebra is \kappa^+ closed.
Start with \kappa measurable and 2^\kappa>\kappa^+. We will use the Easton support iteration \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle, where \dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)} for inaccessible \alpha<\kappa (and is trivial forcing otherwise).
Then \mathbb{P}_\kappa is \kappa-c.c., and forces that for all inaccessible \alpha<\kappa, 2^\alpha=\alpha^+ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).
If G_\kappa\subseteq \mathbb{P}_\kappa is generic, then j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}. Since M[G_\kappa] is closed under \le \kappa sequences in V[G_\kappa], this tail is \kappa^+-closed forcing over V[G_\kappa].
However, GCH holds at every inaccessible \alpha<\kappa and fails at \kappa in V[G_\kappa]. By a reflection argument, \kappa cannot be measurable in V[G_\kappa]. \Box
Exercise: Show that if H\subseteq P(\kappa)/\bar{I} is generic, then \kappa is measurable in V[G_\kappa\ast H].
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