Decomposition inertia exact sequence

Given: The data and hypotheses of the statement.

Lifting residue frobenius by galois conjugates: Let L/K be finite Galois and $P\mid p$ nonzero primes. Put $q=\mathrm{\mid }\kappa (p)\mathrm{\mid }$. Some $\sigma \in D(P\mathrm{/}p)$ induces the arithmetic power map $x\mapsto x^q$ on $\kappa (P)$.

A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x\mapsto x^q$: Let ${\mathbb{F}}_q$ be a finite field of order $q$ and let $E$ be a finite field having ${\mathbb{F}}_q$ as a subfield, with $[E:{\mathbb{F}}_q]=n$ (def-extension-degree-and-finite-extension). Then $E\mathrm{/}{\mathbb{F}}_q$ is a finite Galois extension (def-finite-galois-extension-and-galois-group) and $\mathrm{Gal}(E\mathrm{/}{\mathbb{F}}_q)=⟨{\sigma }_q⟩$ is cyclic of order $n$, generated by the relative Frobenius ${\sigma }_q\text{\hspace{0.17em}⁣}:x\mapsto x^q$ (def-relative-frobenius-of-a-finite-field-extension).

Inertia group of a prime: For finite Galois L/K and a chosen nonzero prime $P\mid p$, set $\kappa (P)={\mathcal{O}}_L\mathrm{/}P$ and $\kappa (p)={\mathcal{O}}_K\mathrm{/}p$. Each $\sigma \in D(P\mathrm{/}p)$ induces a $\kappa (p)$-automorphism $\bar{\sigma }$ of $\kappa (P)$: the rule $\bar{a}\mapsto \overline{\sigma a}$ is independent of the representative because $\sigma P=P$. The inertia group is $I(P\mathrm{/}p)=\ker \text{\hspace{0.17em}⁣}(D(P\mathrm{/}p)⟶\mathrm{Gal}(\kappa (P)\mathrm{/}\kappa (p)))\mathrm{.}$ Equivalently, $\sigma \in I(P\mathrm{/}p)$ exactly when $\sigma \in D(P\mathrm{/}p)$ and $\sigma (a)-a\in P$ for every $a\in {\mathcal{O}}_L$. It is a normal subgroup of D.