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](https://doi.org/10.1016/0003-4843(70)90013-6). Ann. Math. Logic 1 (1970), 179–227.)



Measurable cardinals from saturated ideals
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.)


