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

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