By the downward Lowenheim-Skölem theorem we can find two countable ordinals $\alpha < \beta$ such that $L\_\alpha \prec L\_{\omega\_1}$ and $L\_\beta \prec L\_{\omega\_1}$. That is, $L\_\alpha$ and $L\_\beta$ are elementary submodels of $L\_{\omega\_1}$. Consequently, $L\_\alpha \prec L\_\beta$.


Hence my question, the other way around:



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> Given countable ordinals $\alpha$ and $\beta$ such that $L\_\alpha \prec L\_\beta$, do we always have $L\_\alpha \prec L\_{\omega\_1}$ ? If not, how do construct/prove the existence of such a $\alpha$ and $\beta$?
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> 
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Elementary countable submodels in Gödel's universe
Very clearly not. Take some countable elementary submodel $M\_0$ of $L\_{\omega\_2}$, and take $M\_1$ to be another one, but with $M\_1$ a end extension of $M\_0$. We can find such models by first finding two uncountable $\gamma<\delta$ such that $L\_\gamma\prec L\_\delta\prec L\_{\omega\_2}$, and then taking $M\_1$ be a countable elementary submodel of $L\_\delta$ with $L\_\gamma$ added as a predicate to the language, and then letting $M\_0$ be $M\_1\cap L\_\gamma$.


Now, since both are well-founded models of enough set theory and $V=L$, their transitive collapses are some $L\_\alpha$ and $L\_\beta$ which are countable, and the collapses agree on $M\_0$.


Consequently we have that $L\_\alpha\prec L\_\beta$. But it is also clear that $L\_\alpha$ (and $L\_\beta$) think that they have an uncountable ordinal, whereas $L\_{\omega\_1}$ thinks that all the ordinals are countable.


