Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 136 / 2

For a finite extension $L/K$ of local fields, let $e=e(L/K)$ be its ramification index and $f=f(L/K)=[k_L:k_K]$ its residue-field degree. An unramified extension has $e=1$. A totally ramified extension has $e=[L:K]$, equivalently $f=1$. A tamely ramified extension has separable residue extension and ramification index coprime to the residue characteristic.

Let $\bar{\theta }$ generate the finite extension $k_L/k_K$, and let $\bar{g}$ be its minimal polynomial. Lift $\bar{g}$ to a monic $g\in {\mathcal{O}}_K[X]$ and choose any lift $t\in {\mathcal{O}}_L$ of $\bar{\theta }$. Since finite fields are perfect fields, ${\bar{g}}^{\prime }(\bar{\theta })\ne 0$. The simple-root form of Hensel lemma, applied inside $L$, gives $\theta \in {\mathcal{O}}_L$ withSet $K_0=K(\theta )$. Its residue field contains $k_K(\bar{\theta })=k_L$, soThe equation $g(\theta )=0$ gives the reverse inequality. Thus $[K_0:K]=f$, its residue-field degree is $f$, and $e(K_0/K)=1$; hence $K_0/K$ is unramified. Since $k_{K_0}=k_L$, the extension $L/K_0$ has residue-field degree one and is totally ramified. This constructs the maximal unramified subextension of a local field extension.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. In lower numbering,These are the ramification groups; $G_0$ is the inertia group and $G_1$ is the wild inertia group.

Because ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, every $x\in {\mathcal{O}}_L$ is $P(\alpha )$ for some $P\in {\mathcal{O}}_K[X]$. The polynomial identity $P(Y)-P(X)=(Y-X)Q(X,Y)$ has integral coefficients. Consequently the inequality for $\alpha$ implies it for every $x$, and the converse follows by taking $x=\alpha$. Therefore

Since $L/K$ is a Finite Galois extension, the minimal polynomial factors asDifferentiating and evaluating at $\alpha$ givesand henceFor a fixed nonidentity $\sigma$, its valuation is exactly the number of integers $s\ge 0$ for which $\sigma \in G_s$. Interchanging the two finite sums proves the ramification-group sum for a monogenic integer ring:

The extension is unramified exactly when $G_0=1$, which by this nonnegative sum is equivalent to $v_L(f^{\prime }(\alpha ))=0$, or $f^{\prime }(\alpha )\in {\mathcal{O}}_L^{\times }$. Moreover $|G_0|=e(L/K)$, so the $s=0$ term is $e(L/K)-1$. Equalityholds exactly when $G_1=1$, which is exactly tame ramification.