Let $A$ be a Dedeking ring with field of fraction $K$, $L$ be a Galois extension of $K$, $B$ the integral clousure of $A$ in $L$, $\mathfrak p$ a prime ideal of $A$ and $\mathfrak P$ a prime ideal of $B$ lying above $\mathfrak p$, $\bar B=B/\mathfrak P$ and $\bar A=A/\mathfrak p$ the residue fields. Let $G_{\mathfrak P}$ the decomposition group at $\mathfrak P$, $\phi:\sigma\to\bar\sigma$ the group homomorphism of $G_{\mathfrak P}$ onto the Galois group of $\bar B/\bar A$ (recall that $\bar B/\bar A$ is normal). The kernel of this homomorphism is the inertia group $T_{\mathfrak P}$.
The following argument shows that the order of $T_{\mathfrak P}$ is $e[\bar B:\bar A]_i$ where $e$ is the ramification index of $\mathfrak P|\mathfrak p$ and $[\bar B:\bar A]_i$ is the inseparability degree.
Since $\phi$ is onto we have $(G_{\mathfrak P}:T_{\mathfrak P})=[\bar B:\bar A]_s$ where $[\bar B:\bar A]_s$ denote the separability degree.
Thus $(T_{\mathfrak P}:1)=(G_{\mathfrak P}:1)/[\bar B:\bar A]_s$.
On the other hand the order of $G_{\mathfrak P}$ equals $ef$ where $f$ is the residue class field degree $[\bar B:\bar A]=f$ (see Lang Algebraic number theory Corollary 3 of Prop. 21 page 26). Since $[\bar B:\bar A]=[\bar B:\bar A]_s[\bar B:\bar A]_i$ the formula holds.
On the other hand on his Algebraic number theory Corollary 3 of Prop. 21 page 26, Lang say that the inertia group has order $e$. If my argumentation is correct this implies that $\bar B/\bar A$ is separable, but this is not true, in general.
So where is the mistake?
