Let $\kappa$ be an inaccessible cardinal and let $M \subseteq V\_{\kappa}$ be an inner model of $V\_{\kappa}$, i.e., a transitive model of $\mathsf{ZF}$ containing all the ordinals up to $\kappa$. 


My question is whether such a model is always a rank-truncation of an inner model of $\mathbf V$ (defined using $M$ as parameter). 
Equivalently this question can be phrased as follows: 


If $M \subseteq V\_{\kappa}$ is an inner model of $V\_\kappa$ is then $\mathbf L(M) \cap V\_\kappa = M$, where $\mathbf L(M)$ is the minimal inner model of $\mathbf V$ containing $M$ ? 


If the answer to this question is negative, I am curious whether there can be a counterexample $M$ which is definable in $V\_\kappa$ without parameters and/or which is a model of $\mathsf{ZFC}$. If the answer to this question is positive, I am curious whether it is still positive if we weaken the assumption that $\kappa$ is inaccessible to $\kappa$ being worldly. 



Is every transitive ZF-model of inaccessible height a truncation of an inner model?
**Theorem**: Let $\kappa$ be strongly inaccessible in $V$, such that $V \models ZFC$. If $M\models ZF$, then $L(M) \cap V\_\kappa = M$.


**Proof:** Let us prove by induction on $\alpha < \kappa$ that $L(M) \cap V\_\alpha = M \cap V\_\alpha$.


Let $x \in L(M) \cap V\_{\alpha + 1}$, so $x \subseteq M \cap V\_\alpha$. Let $\gamma$ be an ordinal such that $x\in L\_{\gamma}(M)$. 


Let us find $X \prec L\_\gamma(M)$, such that:


1. $x\in X$, $V\_\alpha \subseteq X$,
2. $X \cap \kappa \in \kappa$
3. $X \cap M$ is transitive
and equal to $M \cap V\_\beta$ for some $\beta$


This is possible by the strong inaccessiblity of $\kappa$: 


We define, by induction of $n < \omega$, $X\_n$. Let $X\_0 = M\cap V\_{\alpha} \cup \{x\}$. Extend $X\_0$ to $X\_0'$ such that $X\_0'\prec L\_\gamma(M)$ and $|X\_0'| = |X\_0| < \kappa$. Now, take $X\_1 \supseteq X\_0'$ such that assumptions 2 and 3 hold ($M \cap X\_1$ is transitive and $X \cap \kappa$ is ordinal). $|X\_1| < \kappa$, by the inaccessiblity of $\kappa$. Let $X\_1 \subseteq X\_1' \prec L\_\gamma(M)$, and so on. Let $X$ be $\bigcup X\_n$.


Let $\bar{X}$ be the transitive collapse of $X$. $\bar{X} = L\_{\bar{\gamma}}(\bar{M})$ where $\gamma < \kappa$ and $x\in \bar{X}$. Since $M \cap X$ is transitive, $\bar{M} = M \cap X$. Since $M\cap X = M \cap V\_\zeta$ for some $\zeta$, $\bar{M} = M \cap V\_\zeta \in M$ (as $\zeta \leq \bar{\gamma} \in \kappa$). In particular, $M$ can compute the model $L\_{\bar{\gamma}}(\bar{M})$ and compute $x$. We conclude that $x \in M$.


**Remark:** Without the inaccessibility assumption, one can get a model $M$ of ZFC such that $L(M) \cap V\_{\omega + 1} \neq M \cap V\_{\omega + 1}$. 


Let us start with $V = L$ and let $\kappa$ be a worldly cardinal of countable cofinality. Let $\langle \alpha\_n \mid n < \omega\rangle$ be the first cofinal sequence of singular cardinals with limit $\kappa$, in the canonical well order of $L$. Let us consider the following class forcing in $L\_\kappa$: for every singular cardinal $\mu\in L\_{\kappa}$, let $\lambda = \mu^{++}$, and force with the lottery sum of $Add(\lambda, \lambda^{++})$ and the trivial forcing. Take the $<\kappa$ support product of those forcing notions. This forcing does not collapse cardinals. Let $c$ be a Cohen real. Let $G$ be an $L\_\kappa$-generic filter such that $L\_\kappa[G] \models 2^{\alpha\_n^{++}} = \alpha\_n^{+4}$ iff $c(n) = 1$. This is possible by splitting the forcing into class forcing that only decides for which cardinal $\lambda$ we force $2^\lambda = \lambda^{++}$ (this class forcing does not add sets) and the second step in which we change the value of the corresponding $2^\lambda$-s. In $L[c]$ one can find a generic for the first step that codes $c$ as above. 


Let $M = L\_\kappa[G]$. $c \notin M$ (as $M$ and $L$ share the same reals), while $c \in L(M)$. 


