For which $\alpha$ does $L_\alpha\models\omega_1$ is a cardinal? i'm reading the proof of Jech of Ketonen's theorem, that if there is a uniform nonregular ultrafilter over $\omega_1$, then Zero Sharp exists.
At some point (Lemma 38.13, (Pure and Applied Mathematics, Vol. 79) Thomas Jech - Set theory-Academic Press,  Elsevier (1978)) he proves, that $\omega_1$ is inaccessible in L.
My Question is, for which $\alpha$ does $L_\alpha\models\omega_1^V$ is a cardinal?
I would have assumed, that this is the case for all $\alpha>\omega_1$, but in the proof it seems, that $L_{\omega_2}\models$"$\omega_1^V$ is a cardinal", but it's not always the case for $L_\gamma$ for $\omega_1<\gamma<\omega_2$.
I thought, being a cardinal is downwards absolute.
Maybe i'm just misinterpreting something in the proof.
Let D be the nonregular ultrafilter which extends the club filter and $\omega_1<\gamma<\omega_2$ s.t. $L_\gamma\prec L_{\omega_2}$. We then can define elementary embeddings $j_{\gamma}^\alpha:L_{f_\gamma(\alpha)}\to L_\gamma$ with $j_{\gamma}^\alpha(\alpha)=\omega_1$ for club many $\alpha$.
We then assume, that $\omega_1$ is not inaccessible in $L$. This means there is an $\alpha_0<\omega_1$ s.t. for all $\alpha_0<\alpha<\omega_1$ there is an $f(\alpha)<\omega_1$ s.t. $L_{f(\alpha)}\models \alpha$ is not a cardinal.
Then we get by the nonregularity of the ultrafilter, that there is a $\gamma$ s.t. $f<f_\gamma$ mod D. So for almost all $\alpha$, $L_{f_\gamma(\alpha)}\models \alpha$ is not a cardinal.
This should be a contradiction because then there is an $\alpha$ s.t we have this elementary embedding and $L_{f_\gamma(\alpha)}\models \alpha$ is not a cardinal. So $L_\gamma\models\omega_1$ is not a cardinal. Contradiction.
By this, it seems, that it's not important to have, that $L_\gamma$ is an elementary submodel of $L_{\omega_2}$.
 A: (Caveat lector: I do not have a copy of this specific edition of the book. I can only guess what is inside of it based on the information given in the question.)
What does it mean for $\omega_1$ to not be an inaccessible in $L$? Well, it means that there is some $\alpha_0<\omega_1$ such that $(\alpha_0^+)^L=\omega_1$.
That means that for all $\alpha\in(\alpha_0,\omega_1)$, $\alpha$ is not a cardinal in $L$, and therefore there is some $f(\alpha)$ witnessing that.
Now using the properties of $D$, we get that for almost all $\alpha$, $f_\gamma(\alpha)$ works as an example of that.
Because $D$ extends the club filter, we can also assume that $\gamma$ is such that $L_\gamma\prec L_{\omega_2}$. Then for a club of $\alpha<\omega_1$ we have elementary embeddings $j_\gamma^\alpha\colon L_{f_\gamma(\alpha)}\to L_\gamma$ such that $j_\gamma^\alpha(\alpha)=\omega_1$. Simply take a countable elementary submodel of $L_\gamma$ with $\omega_1$ an element of the model (for simplicity we may assume it contains $\alpha_0$ as well). 
But now we have a contradiction, since $L_{f_\gamma(\alpha)}\prec L_\gamma$, with $\alpha$ being mapped to $\omega_1$. But $L_\gamma\models\omega_1\text{ is a cardinal}$, while $L_\gamma\models j_\gamma^\alpha(\alpha)\text{ is not a cardinal}$.
A: You are indeed right - being a cardinal is downwards absolute. In particular, if $L_\alpha$ thinks that $\beta$ isn't a cardinal, that means that there are $f,\gamma\in L_\alpha$ such that $L_\alpha$ thinks $\gamma$ is an ordinal $<\beta$ and $f$ is a surjection from $\gamma$ to $\beta$. But all of that data is upwards absolute: we would have in $V$ that $\gamma$ is an ordinal $<\beta$ and $f$ is a surjection from $\gamma$ to $\beta$.
It might help if you can explain why the proof suggests that this fails.
