$L$ and $L_{\omega_2}$ agree on all $\Sigma_1$ formulae? I remember seeing this mentioned somewhere in a textbook I was looking at a while ago, but I can't seem to find the proof. Is anyone aware of the reference I'm looking for? Or can someone provide a quick sketch of the proof?
 A: The answer is: condensation!
Here's a quick sketch of a stronger theorem - that $\Sigma_1$ truth is absolute between $L$ and $L_\kappa$ for any uncountable cardinal $\kappa$:


*

*First, show that the $L$-hierarchy preserves bounded statements. That is, if $\varphi$ is a $\Sigma_0$-sentence with parameters from $L_\alpha$, and $\alpha<\beta$, then $L_\alpha\models\varphi$ iff $L_\beta\models\varphi$. (This is basically because $L_\beta$ is an end extension of $L_\alpha$.)

*Next, suppose $\psi$ is a $\Sigma_1$ sentence with parameters from $L_\kappa$ ($\kappa$ an uncountable cardinal); that is, $\psi\equiv \exists x\theta(x)$ for some $\Sigma_0$ formula $\theta$ with parameters from $L_\kappa$. If $L_\kappa\models\psi$, then $L_\kappa\models\theta(a)$ for some $a\in L_\kappa$; but this is a bounded statement, so true in $L$ as well.

*So finally, assume everything is as above but $L\models\psi$ instead of $L_\kappa\models\psi$. Here things get interesting. Let $b\in L$ be a witness to $\psi$ - that is, $L\models\theta(b)$. Now $b$ might not exist in $L_\kappa$; we'll fix that. Specifically, all the parameters in $\psi$ occur at some level $\theta$ of the $L$-hierarchy for $\theta<\kappa$ (why?); so take an elementary substructure $M$ of $L$, containing $L_\theta$ as a subset and $b$ as an element, of cardinality $<\kappa$ (why does this exist?). What does condensation tell you about $M$? Do you see how to use the absoluteness of bounded statements to finish the proof?

Here's a good exercise: clearly $L_{\omega_2}$ and $L$ disagree about $\Sigma_2$-truth - e.g. consider the sentence "there exist two uncountable cardinals." So where does the proof above break down if I try to get $\Sigma_2$-agreement?
