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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 $q$ is compatible with $k(p)$. 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?

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?

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 $q\in k(\mathbb{P}$ such that for every dense set $D\subseteq \mathbb{P}$ with $D\in M$, there is a condition $p\in D$ such that $q$ is compatible with $k(p)$. 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?

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 $q$ is compatible with $k(p)$. 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?

Suppose we have an elementary embedding $j: M\to N$, a forcing notion $\mathbb{P}\in M$, and a strong master condition $q\in j(\mathbb{P})$. 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?

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 $q\in k(\mathbb{P}$ such that for every dense set $D\subseteq \mathbb{P}$ with $D\in M$, there is a condition $p\in D$ such that $q$ is compatible with $k(p)$. 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?