Is $V=L$ a single first-order sentence? Is $V=L$, the axiom of constructibility, a single first-order sentence?
Since $V=L$ really stands for $(\forall x)(\exists \alpha)[\alpha \in \mathrm{On} \wedge x \in L_\alpha]$, so my question might as well be "is $x \in L_\alpha$ a single first-order formula?".
I was confused about this since prima facie the definition of $L$ involves arbitrary first-order formulas and a satisfaction predicate for them.  Of course one can appeal to the recursive definition by using Goedel's operations, but I am still unsure whether all the details work out.

There are 15 very simple functions that generate $L$. See Schindler Set Theory textbook for more details. 
Examples of these function are the pairing function, set difference function, cartesian product, etc. 
Define a function $S$ which takes a set $x$ to it closure under these 15 functions. 
Define 
$S_0 = \emptyset$
$S_{\alpha + 1} = S (S_\alpha)$
For limits $\lambda$, let
$S_\lambda = \bigcup_{\alpha < \lambda} S_\alpha$
Then it can be shown that $L = \bigcup_{\alpha \in \text{ON}} S_\alpha$.
Since $S$ is the closure of a finite list of function, it may be more clear that $\alpha \rightarrow S_\alpha$ is definable using the usual transfinite recursion. 
This is similar to Godel and Jensen approach to constructibility. 
