I have three three questions, the first two of which probably have the same answer and the third of which is more vague. 


For a set $A$ let $L\_\alpha(A)$ be the constructible universe up to $\alpha$, built from $A$ as a set (and not a predicate). Further let $X = (B, f)$ where $B$ is a transitive set and $f$ is a bijection from $\omega$ to $B$. 


Also assume that the background universe has whatever large cardinals you would like (or that would be helpful). In particular though there is at least one inaccessible cardinal in $L$. 


(1) Suppose $L\_\alpha\models ZFC$. Is it the case that $\omega\_1^L = \omega\_1^{L\_\alpha}$?


(2) Suppose $L\_\alpha(X)\models ZFC$. Is it the case that $\omega\_1^{L(X)} = \omega\_1^{L\_\alpha(X)}$?


(3) If the answer to (1), (2) is yes, is there any simpler way for $L\_\alpha$ to know that $\omega\_1^{L\_\alpha} = \omega\_1^L$ (other than $L\_\alpha\models ZFC$)?


Finally I will just make one observation to highlight why this question isn't trivial. If you replace $ZFC$ with $KP$ then there are many countable admissible sets $L\_\alpha\models KP$ with countable (in $L$) ordinals $\beta\in L\_\alpha$ such that $L\_\alpha \models \omega\_1 = \beta$. 


Thanks



Effect of large cardinals on the value of $\omega_1^L$ in $L$
The answer is no. If there is a transitive set model $M$ of set theory (and this is all you need), then if $\alpha$ is its height (that is, if $\alpha=\mathsf{ORD}^M$), then $L\_\alpha$ is a model of $\mathsf{ZFC}+V=L$. Note that the assumption is strictly stronger than the existence of an $\omega$-model of $\mathsf{ZFC}$, which in turn is strictly stronger than the mere consistency of $\mathsf{ZFC}$, but it is strictly weaker than the existence of inaccessible cardinals. 


Now work in $L$, and consider a countable elementary substructure $Y\preceq L\_\alpha$. Note that $Y$ is well-founded and satisfies $V=L$. Its collapse is then $L\_\beta$ for some $\beta$ that, by necessity, is countable in $L$.


(Note that this also addresses (2), by taking $X=\emptyset$.)




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By the way, this highlights some of the difficulties a challenger of "$V=L$" must face. Even if there are no inaccessibles, the model $M$ we began with may perfectly well be constructible, and satisfy that there are supercompact cardinals, or whatever. About this, you may also want to look at [this post](http://jdh.hamkins.org/multiverse-perspective-on-constructibility/) by Joel. 


