theorem shows that in this setting $L$ is a free $K[G]$-module of rank 1. It is then natural to ask whether $\mathcal{O}_L$ is also a free $\mathcal{O}_K[G]$-module of rank 1. The answer to this question is given by the following result.

\[tamepadic\] Let $L/K$ be a Galois extension of $p$-adic fields with Galois group $G$. Then $\mathcal{O}_L$ is free of rank $1$ as an $\mathcal{O}_K[G]$-module if and only if the extension is tamely ramified.

More generally, one may ask whether $\mathcal{O}_L$ is free over other subrings of $K[G]$. It turns out that there is a natural candidate for a ring over which $\mathcal{O}_L$ could be free, the so-called **associated order**.

The associated order (of $\mathcal{O}_L$ in $K[G]$) is the ring $$\mathfrak{A}_{L/K} = \{ \lambda \in K[G] : \lambda \cdot x \in \mathcal{O}_L \; \forall x \in \mathcal{O}_L \}.$$

It is well-known that the associated order is indeed an $\mathcal{O}_K$-order in $K[G]$. Furthermore, if $\mathcal{O}_L$ is free of rank 1 over an $\mathcal{O}_K$-order $\Lambda$ in $K[G]$, then necessarily $\Lambda=\mathfrak{A}_{L/K}$ (see Proposition \[associatedfree\]). In our setting, we have that $\mathfrak{A}_{L/K}$ coincides with $ \mathcal{O}_K[G]$ precisely when $L/K$ is at most tamely ramified.

In this paper, we investigate the situation when $\mathcal{O}_L$ is 