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I'm reading Jech and following his Boolean algebra models approach to it. I'm wondering if I've got the right idea here.

Let $M \models \mathrm{ZFC}$ and $B \in \mathbf{CompBoolAlg}$. We construct $M^B$ as follows:

  • $M^B_0 = \emptyset$
  • $M^B_{\alpha+1} = \{ \text{partial functions} \ f : M^B_\alpha \to B \}$
  • $M^B_\Lambda = \bigcup_\lambda M^B_\lambda$ for limits $\Lambda$

and $M^B = \bigcup_\alpha M^B_\alpha$.

This gives an embedding of $B$ valued models $$\begin{align} \check{} : M &\to M^B \\ \emptyset &\mapsto \emptyset \\ x &\mapsto \{ \check y \mid y \in x \} \end{align}$$

Now let $G$ be a generic ultrafilter on $B$. Is it the case that $M[G] = \check M / G$? Any help appreciated!

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    $\begingroup$ The equality of check-names is independent of the choice of $G$. So $\check M/G=M$. $\endgroup$
    – Asaf Karagila
    Feb 27 at 16:07
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    $\begingroup$ @ASheard Yes. (Or they’re isomorphic, anyway.) $\endgroup$ Feb 27 at 16:41
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    $\begingroup$ @ASheard $M^B/G$ need not be well-founded in that case. But in terms of extracting relative consistency results, nothing bad happens… genericity (or even quotienting down to a two-valued model) is not necessary for that. $\endgroup$ Feb 27 at 17:12
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    $\begingroup$ @ASheard $M^B/G$ is always a model of ZFC, regardless of whether $G$ is generic or not. However, if $G$ is not generic, $M^B/G$ will not be a forcing extension of $M$ but of some larger model (essentially the model of all "names that look like check names"). See this paper by Hamkins and Seabold. $\endgroup$ Feb 27 at 17:16
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    $\begingroup$ @Asheard for an explicit, if a bit trivial example, when $B$ is a power set algebra, the quotient is isomorphic to the ultrapower. So if $G$ is an ultrafilter that is not countably complete, it won’t be well-founded. (This is covered in the paper I see Miha just provided.) $\endgroup$ Feb 27 at 17:20

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