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Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. In particular $G_0$ is the inertia group, and $G_1$ is the wild inertia group. When the extension is totally ramified, the uniformizer criterion for lower ramification groups permits the equivalent test on one uniformizer.

Put $\alpha =\sqrt[p]{p}$. The polynomial $X^p-p$ is Eisenstein, so ${\mathbb{Q}}_p(\alpha )/{\mathbb{Q}}_p$ is totally ramified of degree $p$. The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_p)/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$. Their coprime degrees make their intersection trivial, sohas degree $p(p-1)$ and is totally ramified. It is the splitting field of $X^p-p$, hence Galois.

Normalize $v_K$ by $v_K(K^{\times })=\mathbb{Z}$. Thensois a uniformizer. Write an automorphism aswhere $a\in {\mathbb{F}}_p^{\times }$ and $b\in {\mathbb{F}}_p$. Sincethe uniformizer criterion for lower ramification groups gives valuation one for $\sigma (\varpi )-\varpi$ when $a\ne 1$, and valuation $p+1$ when $a=1$, $b\ne 0$. Thereforeand $G_i=1$ for $i\ge p+1$. These are the ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Let $G=\mathrm{Gal}(L/K)$. The lower ramification groups are $G_{-1}=G$ and, for $i\ge 0$,

If $L/K$ is totally ramified and ${\pi }_L$ is a uniformizer, then ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. Factoring $F(\sigma {\pi }_L)-F({\pi }_L)$ by $\sigma {\pi }_L-{\pi }_L$ for $F\in {\mathcal{O}}_K[X]$ proves the uniformizer criterion for lower ramification groups

For $\sigma \in G_0$, defineThe inertia group $G_0$ acts trivially on $k_L$, so $\theta (\sigma \tau )=\theta (\sigma )\theta (\tau )$. Its kernel consists exactly of those $\sigma$ for which $\sigma ({\pi }_L)/{\pi }_L\equiv 1$ modulo the maximal ideal, namely $G_1$. The first isomorphism theorem therefore gives an injection