Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 3

For $a\in k$ and each $n$, choose a lift $x_n\in {\mathcal{O}}_K$ of the unique $p^n$th root $a^{p^{-n}}$, which exists because $k$ is a perfect field. DefineChanging $x_n$ by an element of the maximal ideal changes its $p^n$th power by an element whose valuation tends to infinity, so the limit exists and is independent of all choices. In characteristic $p$, the Frobenius endomorphism satisfies $(x+y{)}^{p^n}=x^{p^n}+y^{p^n}$, making $[\cdot ]$ a ring homomorphism lifting the identity on $k$. If $s:k\to {\mathcal{O}}_K$ is any other such section, thenfor every $n$, and the same limiting construction forces $s(a)=[a]$. This proves uniqueness of the Teichmuller lift.

Choose a uniformizer $t$. Repeatedly subtracting the lift of the residue and dividing by $t$ gives every $x\in {\mathcal{O}}_K$ a unique convergent Teichmuller expansionBecause the lift is a ring map, this identifies ${\mathcal{O}}_K$ with $k[[t]]$ and its fraction field with the Laurent series field $k((t))$. This is the equal-characteristic complete discretely valued field classification.

If $K$ is locally compact, its compact valuation ring has only finitely many disjoint residue-class balls. Thus $k$ is finite, as also follows from the local compactness criterion for a complete non-Archimedean field.

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.

LetThe extension $K_2/K$ is the unramified quadratic extension. Since ${\mathbb{F}}_{p^2}^{\times }$ contains all $(p^2-1)$st roots of unity, $L/K_2$ is a cyclic, tamely and totally ramified extension of degree $p^2-1$, with automorphisms $u\mapsto \zeta u$.

The Frobenius automorphism of ${\mathbb{F}}_{p^2}/{\mathbb{F}}_p$ extends by fixing $u$ and conjugates $\zeta$ to ${\zeta }^p$. Hence $L/K$ is Galois, its inertia group isand its residue-field Galois group isThe tame ramification also gives $G_1=1$.