# Measurable cardinals from saturated ideals

Assume that $$\omega<\kappa_1<\dotsb< \kappa_n$$ are infinite cardinals such that for each $$1\le i\le n$$ there is a $$\kappa_i$$-complete, $$\kappa_i^+$$-saturated ideal $$\mathcal I_i\subset \mathcal P(\kappa_i)$$. Can you obtain a ZFC model which contains $$n$$-many measurable cardinals? The natural candidate is $$L[\mathcal I_1,\dotsc, \mathcal I_n]$$.

It is well known that the answer is yes for $$n=1$$ (see Kunen: Some applications of iterated ultrapowers in set theory. Ann. Math. Logic 1 (1970), 179–227.)

Yes. And if the $$\mathscr{I}_k$$s are normal, then the suggested candidate works (I haven't really thought about whether the candidate still works without normality). Here is the argument, which is a typical one:

First, we may assume that each $$\mathscr{I}_k$$ is normal, by Jech Lemma 22.28.

Case 1. There is an proper class transitive inner model of ZFC with $$n+1$$ measurable cardinals (not just $$n$$).

Let $$M$$ be the minimal proper class transitive inner model satisfying ZFC with $$n$$ (not $$n+1$$) measurable cardinals $$\mu_1<\ldots<\mu_n$$, as witnessed by (unique) normal measures $$U_1,\ldots,U_n$$. Then (this uses the case hypothesis) $$\mu_1,\ldots,\mu_n$$ are countable. So we can iterate $$M$$ out at its measurables (using the $$U_k$$s and their images), eventually sending each $$\mu_k$$ to $$\kappa_k$$. Let $$\mathcal{T}$$ be this iteration (consisting of the sequence of iterates $$M^{\mathcal{T}}_\alpha$$, iteration maps $$i_{\alpha\beta}^{\mathcal{T}}$$, etc), and $$M'=M^{\mathcal{T}}_{\kappa_n}$$ be the final iterate and $$i:M\to M'$$ the iteration map. Then I claim that $$i(U_k)\subseteq\mathscr{F}_k$$, the filter dual to $$\mathscr{I}_k$$. For let $$j:V\to M$$ be a generic embedding given by $$\mathscr{I}_k$$. So $$j(\mathcal{T})$$ is an iteration with last model $$j(M')$$, and $$j(\mathcal{T})\upharpoonright(\kappa_k+1)=\mathcal{T}\upharpoonright(\kappa_k+1)$$, but the $$\kappa_k$$th measure used in $$j(\mathcal{T})$$ is $$i(U_k)$$, and $$j(\mathcal{T})$$ eventually sends $$\kappa_k$$ further out to $$j(\kappa_k)$$. Now standard calculations show that the iteration map $$i^{\mathcal{j(\mathcal{T})}}_{\kappa_kj(\kappa_k)}$$ agrees with $$j$$ over $$\mathcal{P}(\kappa_k)\cap M^{\mathcal{T}}_{\kappa_k}$$, which implies that $$j(U_k)\subseteq G$$, the generic filter, and since this is independent of $$G$$, therefore $$j(U_k)\subseteq\mathscr{F}_k$$, as desired.

Since $$i(U_k)$$ is an ultrafilter in $$M'$$ and $$i(U_k)\subseteq\mathscr{F}_k$$, it follows that $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[i(U_1),\ldots,i(U_n)]\subseteq M',$$ and that $$i(U_k)=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$; let $$\bar{\mathscr{F}}_k$$ denote this measure. Therefore $$L[\mathscr{F}_1,\ldots,\mathscr{F}_k]\models$$"$$\bar{\mathscr{F}_k}$$ is a $$\kappa_k$$-complete normal measure on $$\kappa_k$$". Of course $$L[\mathscr{I}_1,\ldots,\mathscr{I}_n]=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$, so we are done. (Using the minimality of $$M$$, can also show that $$M'=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$.)

Case 2: Otherwise.

Then the core model $$K$$ exists. Let $$j:V\to M$$ be a generic embedding given by $$\mathscr{I}_k$$. Then (using core model theory for this level) $$j(K)$$ is an iterate of $$K$$ and $$j\upharpoonright K$$ is the iteration map. But $$\mathrm{crit}(j)=\kappa_k$$. Therefore $$\kappa_k$$ is measurable in $$K$$ (as witnessed by its extender sequence), and because $$j\upharpoonright K$$ is the iteration map, letting $$D_k$$ be the unique normal measure on $$\kappa_k$$ in $$K$$ (uniqueness because otherwise we get a measure of Mitchell order 1), we have $$D_k\subseteq\mathscr{F}_k$$. It follows that $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[D_1,\ldots,D_n]\subseteq K$$, and that letting $$\bar{\mathscr{F}}_k=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$, then $$\bar{\mathscr{F}}_k$$ is a normal, $$\kappa_k$$-complete measure on $$\kappa_k$$ in $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$, as desired. (With the case hypotheses as they are, it might be that $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]\subsetneq K$$. I could have made the case hypothesis to case 1 be that the sharp for an inner model with $$n$$ measurables exists, and case 2 its negation. Then we would get that $$K$$ and $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$$ have the same universe.)

• Thank you very much. I should admit that I can not follow all the technival details, but i am convinced that your solution is correct. Mar 14, 2022 at 20:07