$M[{\mathbb{P}}:G]$ 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.

  • $\begingroup$ There's an obvious negative answer. Because the ultraproduct of countable structure by a non-$\sigma$-closed ultrafilter is non-well founded; but the forcing extensions of standard countable models are always standard. But then again, what is the ultraproduct of the forcing extensions? Perhaps an internal ultraproduct (taking the part of the ultraproduct which lies within the model)? There's still a question on what would be the ultraproduct of the models, in that case. $\endgroup$
    – Asaf Karagila
    May 9, 2014 at 16:19
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    $\begingroup$ @Asaf It does seem from the latest edit that taking the ultraproduct of the forcings would be a good try for getting a positive answer (although I haven't checked this.) So I'm not sure what your "obvious negative answer" applies to (the earlier version of the question does not seem substantially different to me.) $\endgroup$ May 9, 2014 at 18:41
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    $\begingroup$ Like Trevor, I don't understand Asaf's "obvious negative answer". Then ultraproduct of forcing extensions will indeed (unless the ultrafilter is countably closed) fail to be well-founded, but so will any forcing extension of the ultraproduct of the original models. So I don't see why these couldn't be equal (or, rather, isomorphic). $\endgroup$ May 9, 2014 at 19:15
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    $\begingroup$ Regarding the connection between forcing extensions and ultrapowers, I would refer you to my article on the Boolean ultrapower, which describes many results of that nature. See jdh.hamkins.org/boolean-ultrapowers. $\endgroup$ May 9, 2014 at 22:45
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    $\begingroup$ @AsafKaragila I agree that we're interpreting the question differently; maybe it will help if I make my interpretation explicit. If we have, in each model $M_i$, a poset $P_i$, then the function $i\mapsto P_i$ represents an element $P$ of the ultraproduct $M=\prod_{\mathcal F}M_i$, and $M$ satisfies "$P$ is a poset". So (despite the non-well-foundedness of $M$), it makes sense to force with $P$ over $M$, and I think this is what the question intended. $\endgroup$ May 10, 2014 at 11:16

1 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.

  • $\begingroup$ Thanks for your answer. Just as a (possibly silly) question: What is the ordering on poset $\mathbb{P}=\langle \mathbb{P}_i~|~i\in I\rangle_{\mathcal{F}}$? Is it $\mathbb{\leq}=\langle \mathbb{\leq}_i~|~i\in I\rangle_{\mathcal{F}}$? Please see my new question too. $\endgroup$
    – user45939
    May 9, 2014 at 22:08
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    $\begingroup$ Joel, on a notational aspect, you should really use \prod for the product symbol rather than \Pi. Especially in display mode! :-) $\endgroup$
    – Asaf Karagila
    May 9, 2014 at 22:34
  • $\begingroup$ Konrad, Yes, that is the order, of course. This is just a straightforward instance of the Los theorem. @Asaf, actually, I had at first used \Prod, which gave an error, so I switched to \Pi. But I guess I should have tried \prod. $\endgroup$ May 9, 2014 at 22:37

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