# Forcing as a Quotient

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!

• The equality of check-names is independent of the choice of $G$. So $\check M/G=M$. Feb 27 at 16:07
• @ASheard Yes. (Or they’re isomorphic, anyway.) Feb 27 at 16:41
• @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. Feb 27 at 17:12
• @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. Feb 27 at 17:16
• @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.) Feb 27 at 17:20