In $V$, let me call a set theory structure A is a $\omega\_1$ model if the $\omega\_1$ of $A$ is the same as the $\omega\_1$ in $V$ (up to isomorphism). The question I would like to ask is the following: Given any transitive structure $(B,\in)$ containing $\omega\_1$ in $V$, is it always possible to find an $\omega\_1$ model $A$ such that $B\prec A$ and $B\cap H(\omega\_2)$ is in the well-founded part of $A$. 


Some remark: Lowenheim-Skolem or Compactness theorem can be employed to construct an extension. But in general this extension is not an $\omega\_1$ model. To get one $\omega\_1$ model, one can either use some large cardinal or force a generic ultrafilter to construct an ultrapower which is well founded below $\omega\_1$. But I am not sure how to get this using ZFC alone. On the other hand, it is known that the existence of $\omega$-model is correct between transitive ZFC models and thus can be constructed under ZFC.


I know very little about model theory of set theory. So it would be very helpful if anyone can provide some reference about this topic.



A ZFC construction to get a proper extension which is a $\omega_1$-model
In this generality, this is impossible, if you want $B\prec A$ to be a proper extension.


For a counterexample, let $B=V\_{\omega\_2}$, the rank initial segment of the universe of height $\omega\_2$. I claim that there can be no proper extension $\langle B,\in\rangle\prec\langle A,\hat\in\rangle$, where the well-founded part of $A$ includes $H\_{\omega\_2}$. Suppose that we have such an $A$, and let $\beta$ be any of the new ordinals in $A$, that are not in $B$. (There must be such a new ordinal, since $B$ thinks every set has an ordinal rank, and for the ranks up to $\omega\_2$, the original structure $B$ already had all the sets of that rank.) So $\beta$ is larger than every ordinal in $B$. But since $B$ thinks every ordinal has size $\omega\_1$, it follows that $A$ thinks it has a bijection of $\beta$ with $\omega\_1$. But it is correct about $\omega\_1$, and so we can form an actual bijection of the objects that $A$ thinks are below $\beta$ with the actual $\omega\_1$. This is impossible, since the former set includes all of $\omega\_2$. Contradiction.


In the comments, you inquire whether one can find a counterexample $B$ of size $\omega\_1$, and the answer is yes. Let $B=L\_{\omega\_1+1}$, which is a transitive set that contains $\omega\_1$ as an element (I mean to use true $\omega\_1$ here, which might be larger than $\omega\_1^L$), and indeed $\omega\_1$ is the largest ordinal of $B$. Suppose that $\langle B,\in\rangle\prec \langle A,\hat\in\rangle$ is a nontrivial extension, where $A$ is an $\omega\_1$-model and $B\cap H\_{\omega\_2}=B$ is in the well-founded part of $A$. It follows that $A$ is well-founded up to and including $\omega\_1$, and by elementarity, since $\omega\_1$ is the largest ordinal of $B$, it must also be the largest ordinal of $A$. So the ordinals of $A$ are precisely $\omega\_1+1$. Since also $A\models V=L$, it follows by elementarity that $A=L\_{\omega\_1+1}$, which is to say, that $A=B$, and so this isn't a nontrivial extension after all. Contradiction. Perhaps this example shows that you haven't quite asked the question you intended to ask.


