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Suppose we have an elementary embedding $j: M\to N$, a forcing notion $\mathbb{P}\in M$, and a strong master condition $q\in j(\mathbb{P})$. A strong master condition for $j$ and $\mathbb{P}$ is a condition $q\in j(\mathbb{P})$ such that for every dense set $D\subseteq \mathbb{P}$ with $D\in M$, there is a condition $p\in D$ such that $q\leq j(p)$. In his chapter of the handbook on page 814, James Cummings claims that the set $G=\{p\in\mathbb{P} : q\leq j(p)\}$ is an $M$-generic filter. In some cases I find this totally believable, but putting some mild conditions on $j$, a problem arises that I have a question about.

My concern is that you should not be able to build a generic over $M$ in $M$. Now, I understand that in general, $q$ will not be a member of $M$, but sometimes I think this should be the case. For example, if $j$ is merely definable over $M$ (even though I know this is not a hypothesis given) and $M$ and $N$ are class transitive models, then I see issues. In particular, $k$ restricted to any $V_{\alpha}$ will be a member of $M$, i.e. every initial segment of $N$ is a member of $M$, so work in a big enough $V_{\alpha}$ to contain $k(\mathbb{P}).$ For example, if $\kappa$ is measurable, go up to $\aleph_{\kappa+\omega}$, and take the image of that under the ultrafilter. This will be enough.

So, now that I've given some particulars, I can state some specific questions.

  1. If $j$ is definable and $M,N$ are class models, will $q$ never be definable over $M$ or in $M$? When is the strong master condition in $M$ and when is it not?
  2. Maybe just in general some more context on this matter?
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  • $\begingroup$ Please add the definition of a strong master condition. $\endgroup$ Commented Mar 12, 2023 at 8:24
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    $\begingroup$ @MonroeEskew Added! $\endgroup$
    – Connor W
    Commented Mar 12, 2023 at 8:26

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You’re right; this kind of strong master condition cannot exist in the usual context where $j : M \to N$ is a class of $M$. The relevance is when we’re in an intermediate stage of lifting an embedding. Suppose $j : V \to N$ is definable from parameters in $V$ and $\mathbb P$ is $\mathrm{crit}(j)=\kappa$-c.c. and has the property that for some $\kappa$-directed-closed $\mathbb Q$ in $V^{\mathbb P}$, $\mathbb P * \dot{\mathbb Q}$ completely embeds into $j(\mathbb P)$ in $N$, where the complete embedding follows $j \restriction \mathbb P$. If we force with $j(\mathbb P)$, then we get a lifted embedding $j : V[G] \to N[G’]$, and $G’$ also generates a $\mathbb Q$-generic $H$. If $N$ is closed enough, then there is a strong master condition $q$ for $j$ and $\mathbb Q$ in $N[G’] \subseteq V[G’]$. Notice that $V[G’]$ is not a subclass of $V[G]$.

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  • $\begingroup$ To clarify, at the end, when you say that $q$ is a strong master condition for $j$ and $\mathbb{Q}$, is this $j$ the lifted embedding? $\endgroup$
    – Connor W
    Commented Mar 12, 2023 at 9:13
  • $\begingroup$ Yes, lifted through G. $\endgroup$ Commented Mar 12, 2023 at 9:55
  • $\begingroup$ Thanks, Monroe. Is there a more direct way to see that a strong master condition can't exist for a definable class embedding? I don't mean directly in the precise sense, I mean an argument which refers to the actual embedding and poset, rather than a "Oh, if there were a strong master condition, then you could build an $M$-generic internally." Or, at least, some intuitive reasoning. (By the way, I notice in your first sentence that you don't mention definability, where I actually use this in my argument. Is this not necessary for the nonexistence of a strong master condition?) $\endgroup$
    – Connor W
    Commented Mar 12, 2023 at 10:06
  • $\begingroup$ I don’t see any way to show it in general besides noting that generic filters for nontrivial forcings don’t exist (in the ground model). $\endgroup$ Commented Mar 12, 2023 at 10:16

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