Is the critical point of an embedding of a model of set theory inaccessible in it?

Question

Can we find an elementary embedding $j:M\to N$ with $M,N$ transitive $ZFC$-models, $\kappa$ being the critical point, so that $\kappa$ is not inaccessible in $M$ ? ($\kappa$ is regular in $M$.) I would tend towards "probably yes", but at the moment I don't know where to start looking.

Answer 1

No, not necessarily.

Suppose $\kappa$ is measurable in $W$, let $M$ be $W[G]$ the model obtained after forcing with $\operatorname{Col}(\omega,<\kappa)$. If $j\colon W\to W'$ is an ultrapower embedding with critical point $\kappa$, then we can force over $M$ to obtain a model in which $j$ can be extended from $W$ to $M$.

Another important example is the stationary tower forcing, which can produce from a Woodin cardinal in $V$, an embedding [definable in $V[G]$] from $V\to M$, with critical point $\omega_1$.

Somewhere to look for this could be Matt Foreman's chapter in the Handbook "Ideals and Generic Elementary Embeddings". Although admittedly I just learned these facts from my advisor (I don't have notes to distribute, though, sorry).

Answer 2

Note that in Asaf's answer, the elementary embedding $j: M\rightarrow N$ does not "live" (that is, is not definable in) $M$.

By contrast, if we have an elementary embedding $j: M\rightarrow N$ which is definable in $M$ (from parameters in $M$), then $crit(j)$ is inaccessible, in fact measurable, in $M$: letting $\kappa=crit(j)$, we form the ultrafilter $$\mathcal{U}=\{S\subseteq \kappa: \kappa\in j(S)\},$$ and note that this ultrafilter is $<\kappa$-closed (since $j(\bigcup_{\alpha<\mu}A_\alpha)=\bigcup_{\alpha<\mu} j(A_\alpha)$ for $\mu<\kappa$, since $\kappa=crit(j)$ so $j(\mu)=\mu$); since $\mathcal{U}$ exists in $M$ (since $j$ is definable), we have $\kappa$ is measurable in $M$.

Even if $j$ is not definable in $M$, however, we may form the ultrafilter $\mathcal{U}$ described above; it might not live in $M$, is all that changes. In inner model theory, we are often interested in embeddings which, though maybe not definable over their domain model, nonetheless yield reasonably tame ("amenable") ultrafilters.

Answer 3

To complement Asaf's answer, let me point out that you don't need any large cardinals at all to get embeddings like this; they exist as long as there is an uncountable transitive model of ZFC+"there are no inaccessible cardinals".

If $N$ is such an uncountable model we can get an embedding by taking a countable elementary substructure $X\prec N$ and collapsing it to a transitive $M$. The inverse collapse map gives us the embedding $j\colon M\to N$. Since the critical point $\kappa$ has the same properties in $M$ as $j(\kappa)$ has in $N$ and $N$ had no inaccessibles, $\kappa$ cannot be inaccessible in $M$.

It is less clear to me what happens if there are no uncountable transitive models of this theory lying around. First of all, there might not be any uncountable transitive models of ZFC at all. And secondly, it doesn't seem impossible that every uncountable transitive model believes that there is an inaccessible (and when you chop it off at that inaccessible the model becomes countable). On the other hand, we are still ok in this second case, at least consistency-wise, since we can pass to $L$ where this cannot happen.