Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 1

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

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Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. If $K\subseteq M\subseteq L$, the same inequality defines $G_s(L/M)$ inside $\mathrm{Gal}(L/M)$, so directly

Now suppose $M/K$ is Finite Galois extension. The inertia group is the kernel of the action on the residue field. Restriction sends $G_0(L/K)$ into $G_0(M/K)$. Conversely, the maximal unramified subextension of $M/K$ is the intersection of $M$ with the maximal unramified subextension of $L/K$. The Galois correspondence therefore shows that the restriction image is all of $G_0(M/K)$.

For the explicit extension, take ${\alpha }^3=3$ and a primitive cube root of unity $\zeta$. The polynomial $X^3-3$ is Eisenstein over ${\mathbb{Q}}_3$, while ${\mathbb{Q}}_3(\zeta )/{\mathbb{Q}}_3$ is a ramified quadratic extension. Its splitting fieldis therefore a totally ramified extension of degree six with Galois group $S_3$. With $v_L(3)=6$, one has $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$, sois a uniformizer. Let $\tau (\alpha )=\zeta \alpha$, $\tau (\zeta )=\zeta$, and let $\sigma (\alpha )=\alpha$, $\sigma (\zeta )={\zeta }^{-1}$. ThenThe two nonidentity elements of $⟨\tau ⟩=A_3$ have ramification number four, whereas each transposition has ramification number one. HenceIn particular, $A_3$ is the wild inertia group.

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