Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 / 3

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

Let $\alpha$ be a root ofThe polynomial is Eisenstein at $3$, so $E={\mathbb{Q}}_3(\alpha )$ is totally ramified of degree three and ${\mathcal{O}}_E={\mathbb{Z}}_3[\alpha ]$. Its discriminant isIts odd valuation makes $\Delta$ nonsquare in ${\mathbb{Q}}_3$, so the Galois group of an irreducible cubic shows that the splitting field $L$ has Galois group $S_3$. The quadratic extension obtained by adjoining $\sqrt{\Delta }$ is ramified, so $L/{\mathbb{Q}}_3$ is totally ramified. Thereforebecause the wild inertia group is the unique Sylow $3$-subgroup of $S_3$.

It remains to find the wild break. Sinceand ${\alpha }^2+1$ is a unit, the different ideal of $E/{\mathbb{Q}}_3$ has exponent $v_E(f^{\prime }(\alpha ))=3$. The extension $L/E$ is a tamely ramified quadratic extension and has different exponent one. The different in a tower therefore gives different exponentOn the other hand, the different exponent from ramification groups iswhere $b$ is the last index for which $G_b=A_3$. Thus $b=1$, and