TITLE: If $L_\alpha \vDash ZFC$, then do we have $L_{\alpha+1} \vDash \alpha\text{ is inaccessible}$?
QUESTION [6 upvotes]: Here we choose the definition of "is a cardinal" as there is no surjective map from a smaller ordinal to it.
It's easy to prove that, if $L_{\alpha+1} \vDash\ \alpha\text{ is inaccessible}$, then $L_\alpha \vDash ZFC$. Also it's easy to prove that if $L_\alpha \vDash ZFC$, then α is a cardinal in $L_{\alpha+1}$. However, I don't know how to prove confinally many cardinals won't collapse at the next step of L.
If the answer is false, then is it true at the least such α?

REPLY [12 votes]: Yes. The elements of $L_{\alpha+1}$ are exactly those subsets of $L_\alpha$ which are definable from parameters over $L_\alpha$. But $L_\alpha\models\mathrm{ZFC}$, so from here we can just use the usual proof that second order ZFC implies inaccessibility. That is:
(i) every bounded subset of $L_\alpha$ which is in $L_{\alpha+1}$, is already in $L_\alpha$, and so $L_{\alpha+1}$ agrees with $L_\alpha$ regarding power sets and cardinalities of elements of $L_\alpha$; that is, for each $x\in L_\alpha$, we have $\mathcal{P}(x)^{L_{\alpha+1}}=\mathcal{P}(x)^{L_\alpha}$, and $\mathrm{card}(x)^{L_{\alpha+1}}=\mathrm{card}(x)^{L_\alpha}$;  in particular, $L_{\alpha+1}\models$"$\alpha$ is a strong limit cardinal", and
(ii) for every $\beta<\alpha$ and every $f:\beta\to\alpha$ which is in $L_{\alpha+1}$, $f$ is bounded in $\alpha$; that is, $L_{\alpha+1}\models$"$\alpha$ is regular".