index $i_0$ such that $\omega_i=\omega_0$ we have $r_{i_0}=R_{i_0}=R_0$. Taking the minimum over all $i$ now yields $$\begin{aligned}
\min_{\omega \in \mathcal{O}_L} v_K[\mathcal{O}_L : \mathcal{O}_K[G]   \omega]_{\mathcal{O}_K} & = \min_i r_i = \min_i \min\{r_i, r_{i_0}\} \\
& = \min_i \min\{R_i, R_0\} = \min_i R_i,
\end{aligned}$$ hence $m(L/K) = N\left( \pi_K^{\min_i R_i} \right)$. Since the finitely many quantities $R_i$ are effectively computable, so is $m(L/K)$, as claimed.

The associated order
--------------------

We briefly review the notion of *associated order* in the general setting of Dedekind domains that we will need later. We also show that the quantity $m(L/K)$ may be expressed in a natural way in terms of the associated order. The following result is well known.

\[associatedfree\] Let $R$ be a Dedekind domain with field of fractions $K$, and let $A$ be a finite-dimensional $K$-algebra. Let $N$ be an $R$-lattice such that $K  N:=K \otimes_{R} N \cong A$ as $A$-modules. Finally, suppose that $\Lambda$ is an $R$-order in $A$ such that $N$ is a free $\Lambda$-module. Then $\Lambda$ is equal to the associated order of $N$ in $A$: $$\Lambda=\mathfrak{A}(A, N) := \{ \lambda \in  A \mid \lambda N \subseteq N \}.$$

By assumption there exists $\alpha\in N$ such that $N=\Lambda \alpha$ is a 