Order of inertia group on Lang

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?

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Doesn't the "$\phi$ is onto" part use Proposition 14 of §5, which requires $\mathfrak p$ to be maximal? –  darij grinberg Jan 13 '12 at 20:35
Yes, but in a Dedekind ring each prime ideal is maximal, see paragraph 6 chapter I. –  Fabio Lucchini Jan 13 '12 at 21:22
Each non-zero prime ideal is maximal. At any rate, @f.lucchini, you're correct, in general the order of inertia is e times the inseparable degree of the residue extension. I can't see where Lang makes any assumption that would imply that the residue extension is separable (e.g. that the residue fields of the DD in question are perfect), so, I guess it's an oversight on his part. –  Keenan Kidwell Jan 13 '12 at 21:26
Do you know an example of $A$ and $B$ such that $\bar B/\bar A$ is not separable? –  Fabio Lucchini Jan 13 '12 at 22:09
In order to conclude that B is a Dedekind ring you need the residual extension $\bar B/\bar A$ to be separable, I think. –  Tommaso Centeleghe Jan 13 '12 at 22:15