Does $L_\alpha\prec L_{\kappa^+}$?

Question

Let $\langle L_\alpha\mid\alpha\in On\rangle$ be the constructible hierarchy. Suppose $\alpha=(\kappa^+)^L<\kappa^+$. Is it the case that $L_\alpha\prec L_{\kappa^+}$?

Answer 1

As Andres points out, with $\kappa$ as a parameter we can distinguish between $L_{\alpha}$ and $L_{\kappa^+}$. Even more is true: we can consistently have $L_\alpha\not\equiv L_{\kappa^+}$.

One way for this to happen is for $\kappa^+$ to be a limit cardinal in $L$ (so that $L_\alpha$ satisfies "there is a largest cardinal" but $L_{\kappa^+}$ doesn't). This can be forced, for example, by (over $L$) collapsing everything below an inaccessible. (Phrased more directly: if there is a successor cardinal $\kappa^+$ which is inaccessible in $L$, then we have $L_{\kappa^{+L}}\not\equiv L_{\kappa^+}$.)