Removing large cardinals from an uncountable transitive model The usual way of removing large cardinals from a given model of set theory is to cut off the model below the least large cardinal of interest. But this method may have dramatic effects on the external properties of the model, the simplest of which is cardinality.
Question: Suppose there is an uncountable transitive model of ZFC. Is there also an uncountable transitive model of ZFC + "There are no inaccessibles" ?
I don't have a strong intuition about which way this should go, but I would expect the answer to be positive. Having spent some time thinking about the problem, here are some observations.
Fix an uncountable transitive model $M$ and write $M_\alpha$ for $V_\alpha^M$. We may assume that for any $\eta$ inaccessible in $M$, the initial segment $M_\eta$ is countable, for if not we can replace $M$ by the first uncountable such initial segment. Since we have excluded the possibility of cutting off the model $M$ and passing to an inner model clearly will not help with getting rid of the inaccessibles, the only alternative that seems to be left is some sort of forcing construction.
The problem now splits into a number of cases:


*

*$M$ has boundedly many inaccessibles. If we let $\eta$ be the supremum of the inaccessibles of $M$ then $\eta$ (and $M_\eta$) is countable. To destroy the inaccessibles we could now try to force with a poset of size $\eta$ in $M$, for example $\mathrm{Add}(\omega,\eta)$. The issue, of course, is that, for this to have any bearing on the original question, the $M$-generic for this forcing needs to exist in $V$. This is fine if $(2^\eta)^M$ is countable, since then $M$ only has countably many dense subsets of the forcing (as seen in $V$), but if this is not the case we seem to be stuck. Alternatively, if we could at least arrange that $(2^\eta)^M<\mathfrak{c}^V$ then an appropriate fragment of MA in $V$ would still give an $M$-generic.

*$M$ has class many inaccessibles and $M\models\mathrm{Ord\ is\ not\ Mahlo}$. In this case we can fix a definable over $M$ class club $C$ in $\mathrm{Ord}^M$ not containing any of $M$'s inaccessibles. We now force over $M$ with a class-length Easton product $\mathbb{P}$ of the posets $\mathbb{Q}_\delta=\mathrm{Add}(\delta^+,\delta^*)^M$ where $\delta\in C$ and $\delta^*$ is the next point of $C$ above $\delta$. The key point now is that $V$ has $M$-generics for $\mathbb{P}$. This is because $M$ is $\omega_1$-like, meaning that all of its elements are countable, and remains such through all of the initial stages of the forcing $\mathbb{P}$. This means that, given any $\delta\in C$, the extension $M^{\mathbb{P}_\delta}$ constructed thus far only has countably many
dense subsets of $\mathbb{Q}_\delta$, which allows $V$ to build a $M^{\mathbb{P}_\delta}$-generic $G_\delta\subseteq\mathbb{Q}_\delta$. Finally, it follows that $G=\prod_\delta G_\delta\subseteq \mathbb{P}$ is $M$-generic. As it is well known that forcing with $\mathbb{P}$ preserves ZFC, the model $M[G]$ is an uncountable transitive model without inaccessibles.

*$M\models\mathrm{Ord\ is\ Mahlo}$. In this case we would like to undertake the same construction as in the previous bullet point after first forcing over $M$ to add a class club avoiding the inaccessibles. The problem again is with the generic club existing in $V$. The model $M$ is $\omega_1$-like again, so the club shooting forcing has $\omega_1$ many definable dense classes. The forcing is $\alpha$-strategically closed in $M$ for any $\alpha$, but I don't see that this would allow $V$ to build a generic.
 A: i added an answer and then realized it was a complete nonsense, i was getting at the following question.
Suppose kappa is an inaccessible cardinal. Is there a transitive model M of ZFC such that

*

*M has height kappa


*for some stationary in kappa set S, for every alpha in S, M computes alpha^+ correctly,


*for some stationary set D, for every alpha in D, alpha is inaccessible in M


*there are no inacessibles below kappa.
One can show that if the answer is yes then there is an inner model with a proper class of measurables.
Assume M is as in the statement. Assume there is no inner model with a proper class of measurables. Then clause 2 implies that K^M and K must coiterate (K is the core model). This means that K has stationary set of inaccessibles, so by Trevor's trick V must also have inaccessibles. So we must have covering fails in V which then implies that K doesn't exist.
A: I think the answer is (consistently) no.
Following Asaf's comment, let $\kappa$ be Mahlo in $L$ and let  $G \subset \text{Col}(\omega,\mathord{<}\kappa)$ be an $L$-generic filter.  Let $\eta$ be the least ordinal such that $\kappa < \eta$ and $L_\eta \models \mathsf{ZFC}$, and consider the model $N = L_\eta[G]$.
Then in $N$ there is an uncountable transitive model of $\mathsf{ZFC}$, namely $L_\kappa$.  It suffices to show that every such model has an inaccessible cardinal.
Let $M$ be an uncountable transitive model of $\mathsf{ZFC}$ in $N$.  Note that $\text{Ord}^M \ge \kappa$ by uncountability and in fact $\text{Ord}^M = \kappa$ because otherwise $L^M$ would violate the minimality of $\eta$.
The set of $L$-inaccessibles below $\kappa$ is stationary in $L$ by definition, and this stationarity is preserved by the Levy collapse forcing that adds $G$ to obtain $N$.  So the model $M$ satisfies "the class of $L$-inaccessibles is definably stationary."
Therefore (because $M \models \mathsf{ZFC}$) there is an $L$-inaccessible $\alpha < \kappa$ that is a strong limit cardinal in $M$.
Then $\alpha$ is inaccessible in $M$ by Jensen's covering theorem (note that $0^\sharp$ could not have been added by forcing over $L$.)
