case that we now recall.

\[imageberge\] The following equality holds: $$\label{intersection}
 \mathcal{O}_{L^{nr}}\otimes_{\mathcal{O}_K}(\mathfrak{A}_{L/L^{nr}}\cap K[G_0])=\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}.$$ In particular, if $G$ is abelian then (\[injection\]) is a surjection.

Note that $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$ is an $\mathcal{O}_{L^{nr}}$-order in $L^{nr}[G_0]$ contained in $\mathfrak{A}_{L/L^{nr}}$, hence $\mathcal{O}_L$ has the structure of a module over it. It is natural to ask whether there is a relation between freeness of $\mathcal{O}_L$ over $\mathfrak{A}_{L/K}$ and over $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$ (which would imply $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}=\mathfrak{A}_{L/L^{nr}}$ by Proposition \[associatedfree\]). In full generality we only know the following.

\[bergeprojective\] Keeping the above notation, $\mathcal{O}_L$ is projective over $\mathfrak{A}_{L/K}$ if and only if it is projective over $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$.

Fortunately there are situations in which knowing that $\mathcal{O}_L$ is projective is sufficient to conclude that it is free, for example when the relevant orders are commutative. More precisely, the next proposition shows that in several cases the minimal index $m(L/K)$ is controlled purely by the totally ramified extension $L/L^{nr}$.

\[unramifiedindex\] Assume that $G_0$ is abelian and that $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/K}$. Then $\mathcal{O}_{L}$ is free over $\mathfrak{A}_{L/L^{nr}}$ and $$m(L/K)=m(L/L^{nr})=[\mathfrak{A}_{L/K}:\mathcal{O}_K[G]]=[\mathfrak{A}_{L/L^{nr}}:\mathcal{O}_{L^{nr}}[G_0]].$$ Conversely, if $G$ is abelian and $\mathcal{O}_{L}$ is free over $\mathfrak{A}_{L/L^{nr}}$, then $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/K}$ and as before $$m(L/K)=m(L/L^{nr})=[\mathfrak{A}_{L/K}:\mathcal{O}_K[G]]=[\mathfrak{A}_{L/L^{nr}}:\mathcal{O}_{L^{nr}}[G_0]].$$

For the forward implication, assume 