Free finite index submodule of $\mathcal{O}_L$ Let $L/K$ be a finite Galois extension of $p$-adic fields, with Galois group $G$. I've read somewhere that it is well-known that $\mathcal{O}_L$ contains a free finite index $\mathcal{O}_K[G]$-module, where $\mathcal{O}_L$ and $\mathcal{O}_K$ are the valuation rings. I'm triyng to prove this result, but I'm stuck.
My attempt: by the normal basis theorem, $L=K[G]\cdot \alpha$, where $\alpha=\beta/\gamma\in L^\times$. It is not hard to prove that we can assume $\beta\in\mathcal{O}_L$ and $\gamma\in\mathcal{O}_K$.  I believe that the module we are looking for is $\mathcal{O}_K[G]\cdot \beta\subseteq \mathcal{O}_L$, but I don't know how to conclude...
 A: Let me try to answer my question.
FACT 1: $O_L$ is the integral closure of $O_K$ in $L$.
This is clear in the number field case, but it is also true for $p$-adic fields. See, for example, Neukirch, 'Algebraic number theory' (first step of the proof of Theorem 4.8 of Chapter II).
FACT 2: If $\alpha\in L$, then there exists an element $0\ne\gamma\in O_K$ such that $\gamma\alpha =\beta\in O_L$. In particular, $\alpha=\beta/\gamma$. This is Milne, 'Algebraic number theory', Proposition 2.6.
FACT 3: If $\alpha$ is a generator of a normal basis, then since $1/\gamma$ is fixed by every $\sigma\in G$, also $\beta$ is a generator of a normal basis.
FACT 4: $O_L$ is free of finite rank over $O_K$. This is Milne, 'Algebraic number theory', Proposition 2.29.
Now we show that $|O_L:O_K[G]\cdot \beta|<\infty$. Since $K\otimes_{O_K} O_L\cong L$ vía multiplication map, we have $$K\otimes_{O_K}O_L/(O_K[G]\cdot \beta)\cong L/(K[G]\cdot \beta)=0.$$ Nơw $O_L/(O_K[G]\cdot \beta)$ is a finitely generated over the PID $O_K$, so is has a free part isomorphic to some copies of $O_K$, and a finite torsion part. If it is infinite, then the tensor product with $K$ would contain some copies of $K$, and this would be a contradiction.
