Let $K \subset L$ be number fields and $\mathcal{O_K}, \mathcal{O_L}$ the corresponding rings of algebraic integers. Further let dimension$(L/K)$ = n as a vector space.
If $K$ is a PID, then $\mathcal{O_L}$ is a free module of dimension n over $\mathcal{O_K}$ (I believe this is true, please correct me if I am wrong). Is this also true in the general case and if not, can we describe the structure of $\mathcal{O_L}$ over $\mathcal{O_K}$ as a module?
If $\mathcal{O_K}$ is a PID, then $\mathcal{O_L}$ is a free module over $\mathcal{O_K}$ (just use the fundamental theorem of f.g. modules over PID together with the fact that $\mathcal{O_L}$ is torsion free). In general, it is not true that $\mathcal{O_L}$ is free over $\mathcal{O_K}$. There is a similar theorem for f.g. modules over Dedekind domains, but there the torsion free part is a direct sum of fractional ideals (which are all free if the domain is actually a PID). See for example here.
For rings of integers, there are examples where the extension is not free. Moreover, in "On Integral Bases" by Henry B. Mann, it is shown that if $\mathcal{O}_K$ isn't a PID, then $K$ has a quadratic extension $L$ such that $\mathcal{O}_L$ is not free over $\mathcal{O}_K$, or in other words, $\mathcal{O}_K$ is a PID iff all the extensions $\mathcal{O}_L$ of integer rings are free over it.