# Can an ultrapower be undone by forcing?

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? 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.

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: Douglas Ulrich answered the question for set forcing; I've asked the class forcing version as a separate question.

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.

• The term "$\kappa$ is generically measurable" means that you can find a generic ultrapower embedding [into a transitive class] with $\kappa$ as a critical point. It just seemed relevant here. – Asaf Karagila Dec 26 '15 at 21:29
• @AsafKaragila It's not obvious to me, though, what the exact connection is; for instance, $\omega_1$ can be generically measurable, clearly never potentially measurable. (I suspect there is a connection in consistency strengths, but I don't quite see it.) – Noah Schweber Dec 26 '15 at 22:18
• I think there may be a mistake in the url of your link for the class forcing question. Probably you meant this one: mathoverflow.net/q/259628/1946. – Joel David Hamkins Jan 15 '17 at 11:14

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.