End elementary extension in infinitary logic of some $L_\alpha$ producing a $L_\beta$

Question

Let $L_\alpha$ be some admissible level of the constructible hierarchy and $M \supseteq L_\alpha$ an extension of $L_\alpha$. I am looking for conditions under which $M \simeq L_\beta$. It is not enough to ask that $M$ be an end elementary extension of $L_\alpha$ as $M$ may not be well-founded.

However I suspect that this may be provable when $M$ is an $\mathcal{A}$-elementary end extension of $L_\alpha$ for some fragment $\mathcal{A}$ of infinitary logic $L_{\omega_1 \omega}$. In the general case, well-orderdness is not definable in $L_{\omega_1 \omega}$ but here we also have the structure of $L_\alpha$ to help us.

The idea is to do as follows to show that $M$ is well-founded:

1. In $M$ we can define ordinals as hereditarly transitive sets and they are well-founded by foundation.
2. Then we can define by $\Sigma$-recusion (here we need $\alpha$ to be admissible and likely even $L_\alpha$ to be a model for infinitary $KP$) a predicate $L(x, \alpha)$ defined as: \begin{align*} L(x, \alpha+1) &\longleftrightarrow \bigvee_{n \in \omega} \bigvee_{\varphi} \exists p_1, \ldots, p_n \, \forall y \in x (L(y, \alpha) \wedge \varphi(y, p_1, \ldots, p_n)) \\ L(x, \alpha) &\longleftrightarrow \exists \beta \in \alpha \, L(x, \beta) \text{ for } \alpha \text{ limit} \end{align*} With it, we define $L_\alpha = \left\{ x \mid L(x, \alpha) \right\}$.
3. With this definition, the two following infinitary sentences are true in $L_\alpha$ (observe that the second sentence is not usually true when $L(x, \alpha)$ is defined in the finite setting using rudimentary functions instead of the definibality predicate) : \begin{align*} \begin{cases} \forall x \, \exists \alpha \, x \in L_\alpha \\ \forall \alpha \, (x \in L_\alpha \implies \forall y \in x \, \exists \beta \in\alpha \, y \in L_\beta) \end{cases} \end{align*}
4. Those two sentences can be reflectd in $M$ by $\mathcal{A}$-elementarity and with those, any infinite decreasing sequence $a_0 \ni^M a_1 \ni^M a_2 \ldots$ would yield an infinite decreasing sequence of ordinals $\alpha_0 \ni^M \alpha_1 \ni^M \alpha_2 \ldots$ which in turn would contradict the fact that ordinals are well-founded w.r.t. $\in^M$.

Now: do we need $L_\alpha$ to be a model for $KP$ for infinitary logic for the definition by recursion? And is there a more direct argument to show this?

Answer 1

(Turning some comments into an answer)

The definition of $L(x,\alpha+1)$ was wrong, instead it should have been $$L(x,\alpha+1)\leftrightarrow\bigvee_{n\in\omega}\bigvee_{\varphi}\exists p_1,\ldots,p_n \, \forall y(y\in x\leftrightarrow (L(y,\alpha)\land\varphi(y,p_1,\ldots,p_n)))$$, and as this is not $\Sigma_1$, $\Sigma$-recursion is not applicable with the corrected definition.

Additionally, from step #1 on, the argument is dependent on stating an axiom of foundation in $\mathcal L_{\omega_1\omega}$ such that $M$ satisfying the axiom means that even just the ordinals of $M$ are externally well-ordered. But Karp ("Finite-Quantifier Equivalence", in The Theory of Models, J. Addison, L. Henkin, and A. Tarski (eds.), 1965) and Lopez-Escobar ("On Defining Well-Orderings, Fundamenta Mathematicae vol. 59, 1966) showed that external well-orderedness is not definable in $\mathcal L_{\omega_1\omega}$ nor in $\mathcal L_{\kappa\omega}$ for any cardinal $\kappa$.