*I am not 100% certain this question is appropriate for MO; I may just be missing something obvious. Also, I vaguely recall a similar question being asked here a while ago, but I can't find it; if it turns out this is a duplicate, I'll delete this question. Anyways, apologies in advance if this is too easy or is a duplicate. Note that e.g. [Can measures be added by forcing?](https://mathoverflow.net/questions/110871/can-measures-be-added-by-forcing) prevents the obvious nuke from working.*


*Also, the "descriptive-set-theory" tag is purely a guess on my part, based on the surprising ubiquity of descriptive set theory in similar-sounding questions.*




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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 generic extension of $N$ (either by set or class forcing in $M$)?
> 
> 
> 


As mentioned above, I am almost certain the answer is "no", even if $M$ has loads of large cardinals, but I don't see how to prove this.


**EDIT: Danielle Ulrich answered the question for set forcing; I've asked the class forcing version [as a separate question](https://mathoverflow.net/questions/259628/can-an-ultrapower-be-undone-by-class-forcing).**




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A small observation:


Say (inside a model $W$) a cardinal $\mu$ is


* *potentially measurable* if $\mu$ is measurable in some forcing extension; and
* *reversibly measurable* if $\mu$ is measurable, and $W$ is a forcing extension of the transitive collapse of the ultrapower of $W$ by a measure on $\mu$ (that is, if $\mu$ is as above).


Then suppose we had such an $M, N, U, \kappa$, with $j$ the elementary embedding. Then $N$ satisfies "There is a potential measurable below $j(\kappa)$," so - pulling back along $j$ - $M$ satisfies "There is a potential measurable below $\kappa$." This shows that - in $M$ - the least potentially measurable is strictly less than the least reversibly measurable (otherwise we get a descending chain of measurable cardinals).


Now, it feels plausible to me that there's a clever trick that can be done here to outright build a descending sequence of reversibly measurables from a single reversibly measurable; but I don't see it.



Can an ultrapower be undone by forcing?
For set-forcing, the answer is no, see the following article


*Joel David Hamkins, Greg Kirmayer, and Norman Lewis Perlmutter*, [**Generalizations of the Kunen inconsistency**](http://dx.doi.org/10.1016/j.apal.2012.06.001), *Ann. Pure Appl. Logic* **163** (2012), no. 12, 1872--1890. (see also [arxiv.org/abs/1106.1951](http://arxiv.org/abs/1106.1951) and Hamkins's [blog post](http://jdh.hamkins.org/generalizationsofkuneninconsistency/))


The "Kunen Inconsistency" is the theorem that there is no nontrivial elementary embedding $j: V \to V$. The above article shows (among several other things) that if $V[G]$ is any set-forcing extension of $V$ then there is no nontrivial elementary embedding $j: V[G] \to V$. So even if you replaced ultrapowers by extenders, for example, the answer remains no.


