Ultraproduct of Forcing Extensions & Forcing Extension of Ultraproduct

Question

Notation:

$M[{\mathbb{P}}:G]$ denotes the forcing extension of $M$ by $\mathbb{P}$-generic filter $G$.

$\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle$ denotes the ultraproduct of models using the ultrafilter $\mathcal{F}$ on the index set $I$.

---

Consider the index set $I$, the ultrafilter $\mathcal{F}$ on it, the family $\{M_i~|~i\in I\}$ of countable transitive models of $ZFC$, the collection $\{\mathbb{P}_i~|~i\in I\}$ of forcing notions such that $\forall i\in I~~~\mathbb{P}_{i}\in M_i$ and the collection $\{\mathbb{G}_i~|~i\in I\}$ of sets such that $\forall i\in I~~~G_i$ is a $\mathbb{P}_{i}$-generic filter over $M_i$.

For all $i\in I$, $M_i$ and $M_i[\mathbb{P_i}:G_i]$ are models of $ZFC$.

Also $\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle$ and $\prod_{\mathcal{F}}\langle M_i[\mathbb{P}_i:G_i]~|~i\in I\rangle$ are models of $ZFC$.

Question 1: Is there any natural (categorical) relation between the ultraproduct of forcing extensions of some ground models $\prod_{\mathcal{F}}\langle M_i[\mathbb{P}_i:G_i]~|~i\in I\rangle$ and the forcing extension of a ground model produced by ultraproduct of component ground models $\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle$?

For example, is there a forcing notion $\mathbb{P}$ and a $\mathbb{P}$-generic filter $G$ over $\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle$ such that:

$$(\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle)[\mathbb{P}:G]\cong \prod_{\mathcal{F}}\langle M_i[\mathbb{P}_i:G_i]~|~i\in I\rangle$$

Question 2: Let $M$ be a model of $ZFC$ and $I, \mathcal{F}\in M$ and ($\mathcal{F}$ is an ultrafilter on $I$)$^M$ and $\{\mathbb{P}_i~|~i\in I\}$ is a family of forcing notions such that $\forall i\in I~~~\mathbb{P}_i\in M$ and $\{G_i~|~i\in I\}$ a family of sets such that $\forall i\in I~~~G_i$ is a $\mathbb{P}_i$ - generic filter over $M$.

Note that $\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle$ is a forcing notion in $M$ also $\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle\subseteq \prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle$.

(a) Under what conditions the subset $\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle$ is a $\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle$-generic filter over $M$?

(b) If $\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle$ is a $\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle$-generic filter over $M$, consider the generic extension $M[\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle:\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle]$, is there a natural relation between this generic extension of ground model and generic extensions $\{M[\mathbb{P}_i:G_i]~|~i\in I\}$?

For example, is $\prod_{\mathcal{F}}\langle M[\mathbb{P}_i:G_i]~|~i\in I\rangle$ isomorphic to some ultrapower of $M[\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle:\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle]$? i.e. is there an index set $J$ and an ultrafilter $\mathcal{U}$ on $J$, such that:

$$\prod_{\mathcal{U}}\langle M[\prod_{\mathcal{F}}\langle \mathbb{P}_i~|~i\in I\rangle:\prod_{\mathcal{F}}\langle G_i~|~i\in I\rangle]~|~j\in J\rangle\cong \prod_{\mathcal{F}}\langle M[\mathbb{P}_i:G_i]~|~i\in I\rangle$$

In the other words, is ultraproduct of a family of generic extensions of a given ground model isomorphic to an ultrapower of generic extension of ground model using ultraproduct forcing notion?

Remark: Forcing is a Boolean-valued ultraproduct itself. It seems reasonable to have some sort of natural interactions between ultraproducts.

Answer 1

The answer is that the two models are related in the most natural way: The ultraproduct and forcing extension constructions commute, in the sense that the ultraproduct of a sequence of forcing extensions is a forcing extension of the ultraproduct of the ground models.

Specifically, the ultraproduct of the forcing extensions $\Pi_i M_i[G_i]/U$ is precisely the forcing extension of the ultraproducts of the original ground models $\Pi_i M_i/U$, using the poset $\mathbb{P}$ which is represented by the sequence of original posets $\mathbb{P}=[\langle \mathbb{P}_i\mid i\in I\rangle]_U$ and having the generic filter $G$ represented by the sequence of original filters $G=[\langle G_i\mid i\in I\rangle]_U$.

$$\Pi_i M_i[G_i]/U \qquad = \qquad \left(\strut\Pi_i M_i/U\right)\left[ [\langle G_i\mid i\in I\rangle]_U\right].$$

Well, perhaps we should say $\cong$ by the natural map, rather than $=$. The map associates $[\langle x_i\mid i\in I\rangle]_U$, where $x_i=\text{val}(\dot x_i,G_i)$ has name $\dot x_i$, with the object obtained by interpreting the $\mathbb{P}$-name $[\langle \dot x_i\mid i\in I\rangle]_U$ via $G$.

This is an immediate consequence of the Los theorem, applied to the full ultraproduct $\Pi_i M_i[G_i]/U$. That is, it is true for each $i$ that $G_i$ is $M_i$-generic for $\mathbb{P}_i$, and so by Los the ultraproduct thinks that it is a forcing extension of $\Pi_i M_i$ by $[\langle G_i\mid i\in I\rangle]_U$ using $[\langle \mathbb{P}_i\mid i\in I\rangle]_U$. For example, if $D_i$ is the collection of all open dense subsets of $\mathbb{P}_i$ in $M_i$, then since each $G_i$ meets every set in $D_i$, it follows by Los that $[\langle G_i\mid i\in I\rangle]_U$ meets every set in $[\langle D_i\mid i\in I\rangle]_U$, which is the collection of open dense subsets of $[\langle\mathbb{P}_i\mid i\in I\rangle]_U$ in $\Pi_i M_i/U$.

The conclusion becomes even more obvious when you think in terms of the ground-model definability theorem, which asserts that each $M_i$ is uniformly definable in $M_i[G_i]$, and so we may simply apply that definition inside $\Pi_iM_i[G_i]/U$. The ground model of an ultraproduct is the ultraproduct of the ground models, simply because they are each uniformly definable.