Let $V$ satisfy there exists a measurable cardinal. Let $\kappa$ be a measurable cardinal and $U$ be the normal measure on $\kappa$ witnessing this. Let $\mathbb{P}$ be a forcing of size less than $\kappa$.

Let $M$ be a countable transitive structure such that there exists an elementary embedding $\pi : M \rightarrow V$ (or a large $V_\Theta$). Suppose $U' \in M$ is such that $\pi(U') = U$. Suppose $\mathbb{Q} \in M$ is such that $\pi(\mathbb{Q}) = \mathbb{P}$.

It is known that $M$ can be iterated using $U'$ throughout the ordinals of $V$. In particular, $M$ can be iterated $\omega_1^V$ many times.

Since $M$ is countable, there is a $g \subseteq \mathbb{Q}$ which is $\mathbb{Q}$-generic over $M$ and $g \in V$.

My question is: Is $M[g]$ also $\omega_1^V$-iterable?

My ideas were: Since $|\mathbb{P}| < \kappa$, $\kappa$ would still be measurable in any forcing extension of $V$ by a generic of $\mathbb{P}$. So if there exists $G \subseteq \mathbb{P}$ which is $V$-generic and $\pi'' g \subseteq G$, then if one could lift $\pi$ to a map $M[g] \rightarrow V[G]$ which takes some measure $\tilde U' \in M$ to some measure $\tilde U \in V[G]$, then this should show $M[g]$ is iterable.

However, I am not sure such a $G$ can exists. Since $g \in V$, I do not think it can be used make $G$ in the ways that usual liftings are done.

Thanks for any information about this problem.

  • $\begingroup$ I guess one could use a "copy construction", since we have an elementary map $\pi: M \to M[g]$. Define it in such a way that everything commutes, i.e we need maps $\pi^{\mathcal{T}}_{\alpha}: M\to M_{\alpha}$, $i_{\alpha}:M_{\alpha} \to M[g]_{\alpha}$ etc...So if $\mathcal{T}$ is an iteration of $M$, $\pi\mathcal{T}$ is an iteration of $M[g]$. $\endgroup$ Sep 17 '15 at 17:12
  • $\begingroup$ @CarloVonSchnitzel Can you explain how do you get this elementary map $\pi : M \rightarrow M[g]$? $\endgroup$
    – William
    Sep 17 '15 at 18:25
  • $\begingroup$ I spoke too fast, we don't have an elementary map $\pi:M \to M[g]$. All you need to do is copy the iterations up to the generic extension as in Victoria's answer. $\endgroup$ Sep 17 '15 at 19:57

Yes, in general, the model $M[g]$ will have the same amount of iterability as $M$. To see this, we need the following lemma giving the necessary and sufficient condition for when an ultrapower of a generic extension is the lift of an ultrapower of the ground model.

Lemma: Suppose $M$ is a transitive model of ${\rm ZFC}^-$, $\mathbb P\in M$ is a poset and $G\subseteq \mathbb P$ is $M$-generic. Suppose further that $U\in M$ is an ultrafilter on a cardinal $\delta$ and $U^*\in M[G]$ is an ultrafilter on $\delta$ extending $U$, both with well-founded ultrapowers. Then the ultrapower by $U^*$ is a lift of the ultrapower by $U$ if and only if every $f:\delta\to M$ in $M[G]$ is $U^*$-equivalent to some $g:\delta\to M$ in $M$.

Let $\delta$ be the preimage of $\kappa$ under $\pi$ and let $M=M_0$. Let $j_{\xi\mu}:M_\xi\to M_\mu$ be the iterated ultrapowers of $M_0$ by $U'$. We can lift the entire iteration to the directed system consisting of $j_{\xi\mu}:M_\xi[g]\to M_\mu[g]$. So it suffices to argue that this is precisely the iteration of $M_0[g]$ by $U^*$, where $U^*$ is the ultrafilter extending $U'$ generated by using $\delta$ as a seed from the lift $j_{01}:M_0[g]\to M_1[g]$ (for $A\subseteq\delta$ in $M_0[g]$, we have $A\in U^*$ whenever $\delta\in j_{01}(A)$) . The embedding $j_{01}:M_0[g]\to M_1[g]$ is precisely the ultrapower by $U^*$. So, the conclusion for $j_{01}$ follows by the definition of $U^*$. Thus, $M_0[g]$ satisfies that "every new function from $\delta$ into $M$ is $U^*$-equivalent to an old function" by the lemma. But, then by elementarity $M_1[g]$ satisfies this for $j_{01}(U^*)$ and $M_1$, which means that the ultrapower of $M_1[g]$ by $j_{01}(U^*)$ is the lift of $j_{12}:M_1\to M_2$, namely $j_{12}:M_1[g]\to M_2[g]$ (lifts of small forcing extensions are unique). By elementarity, we can continue to propagate this statement along the entire iteration.

Indeed, the result is much more general. Even if $\mathbb Q$ is not small relative to $\delta$, but we are able to lift the first ultrapower $j_{01}:M_0\to M_1$ to $M_0[g]$, then every iterate by $U^*$ (obtained as before) will satisfy the "no new functions" statement and therefore it will be a lift of a ground model embedding and therefore well-founded. Thus, to lift the entire iteration, it suffices to lift the just the first step! Moreover, the lift of the iteration is the iteration of the lift. A more involved argument (and some more restrictions on the forcing) is needed in the case where the ultrafilter $U'$ is an $M$-ultrafilter that is external to $M$, because the statement whose elementarity we are using is not necessarily expressible in $M$ in that situation. This argument is given in my paper (with Arthur Apter and Joel Hamkins) Inner models with large cardinal features usually obtained by forcing.

  • $\begingroup$ Some very neat results in that paper! $\endgroup$ Sep 18 '15 at 4:18

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