Suppose $\alpha$ is admissible and $\beta<\alpha$. We know that $L_\alpha$ is an admissible set (by definition), but adding subsets of $\beta$ to $L_\alpha$ might break admissibility: while set forcing preserves admissibility, we might have e.g. $\alpha$ countable, $\beta=\omega$, and $X\subseteq\omega$ codes a well-ordering of length $>\alpha$ - then adding $X$ to $L_\alpha$ breaks admissibility.
My question is about when we can be certain this won't happen. Specifically:
Suppose $\alpha$ is admissible, $\beta<\alpha$. Under what conditions is there an admissible set $A$ containing $V_\beta$ with height $\alpha$?
(NOTE: I assume all admissible sets are transitive. I think that's standard, but I don't have Barwise' book on hand, so I just want to clarify.)
This is a really broad question, since there are of course three parameters kicking around:
What's $\alpha$?
What's $\beta$?
What's $V$? In particular, exactly how far is $V$ from $L$?
I'm especially interested in the following more specific twist on this question:
Suppose $\kappa$ is a cardinal in $V$ and $\alpha$ is the least admissible $>\kappa$. What sort of large cardinal properties can $\kappa$ satisfy and still have an admissible set containing $V_\kappa$ of height $\alpha$?
For instance, can $\kappa$ be measurable?
EDIT: A related question (which I meant to include originally): for any $\kappa$ whatsoever, of course there is an $\alpha$ which is the least height of an admissible set containing $V_\kappa$; and this $\alpha$ is admissible. Turning this around yields a potentially interesting function: say that the robustness of an admissible ordinal $\alpha$, $rob(\alpha)$, is the supremum of the $\beta<\alpha$ such that there is an admissible set of height $\alpha$ containing $V_\beta$. Then it's reasonable to ask what we can say about this function. That is:
How does $rob$ depend on the large cardinal structure of $V$?