Suppose I have a transitive model $M$ of ZFC, and - in $M$ - $U$ is a measure on $\kappa$. Then the transitive collapse of the ultrapower of $M$ along $U$ is an inner model, $N\subset M$.


My question is:



> 
> Can we ever have $M$ be a class-generic extension of $N$?
> 
> 
> 


*Context*: The set-forcing version of this question [was answered by Douglas Ulrich](https://mathoverflow.net/questions/227049/can-an-ultrapower-be-undone-by-forcing/227052#227052), using a result of Hamkins-Kirmayer-Perlmutter; and Joel David Hamkins pointed out that the same result holds for many "nice" class forcings. However, the general class-forcing version remains open, and general class forcing is potentially *so* nasty that I thought this warranted its own question. 



Can an ultrapower be undone by class forcing?
The answer is negative, if one considers parameter-free definable ZFC-preserving class forcing satisfying the forcing theorem, which means that the forcing
relation is definable and true statements in the extension are
forced.


**Theorem.** Suppose that $\newcommand\P{\mathbb{P}}\P$ is a
ZFC-preserving definable class forcing notion in $M$ that satisfies the forcing
theorem for forcing over $M$. Then in the corresponding forcing
extension $V$ of $M$ by $\P$, there is no elementary embedding
$j:V\to M$ that is definable in $V$ from parameters.


**Proof.** Suppose not. Fix such a forcing notion $\P$ in $M$,
defined by $\varphi(\cdot)$, and suppose that $j(x)=y$ is forced to
be defined in $V$ by $\psi(x,y,\dot u)$.


Let $\kappa$ be least such that there is a condition forcing that
in the forcing extension of $M$, there is a parameter $u$ for which
the formula $\psi(x,y,u)$ defines an elementary embedding $j:V\to
M$. This is expressible in the first-order language of set theory. Consequently, we have given a parameter-free definition of $\kappa$ in $M$. This contradicts the fact that $j:V\to M$ is an
elementary embedding with critical point $\kappa$, since in this
case $\kappa$ will not be in the range of $j$, but every definable
element of $M$ will be in the range of $j$. Contradiction. **QED**


A similar argument works in the case that $\P$ is definable in $M$
from parameters below $\kappa$.


In the general case, let me now undertake a more general argument,
in the case that $\P$ is definable by a formula $\psi(\cdot,z)$
using a parameter $z$, but I shall use an extra assumption. Specifically, I want it to be first-order expressible in a parameter $z'$, whether the class forcing notion defined by $\psi(\cdot,z')$ satisfies the forcing theorem and what is the forcing relation.


**Theorem.** Suppose that $\P$ is a ZFC-preserving class forcing notion definable
from parameters in $M$ and satisfying the forcing theorem, such
that the definition $\psi(\cdot,z)$ of $\P$ has the property that
the property of whether $\psi(\cdot,z)$ defines such a class
forcing notion with the forcing theorem is itself expressible in
set theory, and the forcing relation is also definable from that parameter uniformly. Then in the forcing extension $V$ of $M$ by $\P$, there
is no elementary embedding $j:V\to M$ that is definable in $V$ from
parameters.


**Proof.** Let $\kappa$ be least among all parameters $z$ such that
it is the critical point of such an embedding. Now we don't need
the parameter $z$ to define $\kappa$ in $M$, and get the
contradiction as before. **QED**


I'm not sure in general how to express that a definable class has
the forcing theorem, uniformly in the parameter. Even in the case
of pre-tame forcing, this seems to involve quantifying over
arbitrary classes and has complexity $\Pi^1\_1$ over $M$.


