that the $\mathfrak{A}_{L/K}$-module $\mathcal{O}_L$ is free, hence also projective. By Theorem \[bergeprojective\] we have that $\mathcal{O}_L$ is projective over $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$. The latter is a commutative order since $G_0$ is abelian, so it is “clean”, that is, every projective $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$-lattice that spans the algebra $L^{nr}[G_0]$ is free (see [@MR0175950] or [@MR0283014 IX Corollary 1.5]). It follows that $\mathcal{O}_L$ is free over $\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$, so the latter is the associated order of $\mathcal{O}_L$ in $L^{nr}[G_0]$, that is, $$\bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}=\mathfrak{A}_{L/L^{nr}}$$ and $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/L^{nr}}$. By Proposition \[imageberge\] we have that (\[injection\]) is a bijection. Therefore, by Proposition \[indexassociated\] and Proposition \[productindex\], we have $$m(L/K)=[\mathfrak{A}_{L/K}:\mathcal{O}_K[G]]=[\mathfrak{A}_{L/L^{nr}}:\mathcal{O}_{L^{nr}}[G_0]]=m(L/L^{nr}).$$ As for the converse, assume that $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/L^{nr}}$, which coincides with the intersection $ \bigcap_{g\in G}g\mathfrak{A}_{L/L^{nr}}g^{-1}$ since $G$ is abelian. It follows from Theorem \[bergeprojective\] that $\mathcal{O}_L$ is projective over $\mathfrak{A}_{L/K}$, which is commutative, hence clean. We obtain that $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/K}$, and the remaining equalities are proved as above.

The case when the associated order is maximal {#sec:assab}
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When $\mathfrak{A}_{L/K}$ is a maximal order in $K[G]$ the problem of studying $m(L/K)$ simplifies considerably. In this section we study this situation and prove Theorem \[thm:IntroAbsolutelyAbelian\]. We start with the 