Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 123 / 2

Normalize the discrete valuation on $L$ by $v_L(L^{\times })=\mathbb{Z}$, set $v_L(0)=+\infty$, and write ${\mathcal{O}}_L$ for its valuation ring. For a finite Galois extension of local fields, define $G_{-1}=G=\mathrm{Gal}(L/K)$ and, for integers $i\ge 0$,Equivalently, $G_i$ is the kernel of the action on ${\mathcal{O}}_L/{\mathfrak{m}}_L^{i+1}$. These are the lower ramification numbering of the ramification groups. In particular, $G_0$ is the inertia group, the kernel of the action on the residue field, and $G_1$ is the wild inertia group. They form a decreasing sequence of normal subgroups, eventually trivial.

For a real index $t\ge -1$, extend by $G_t=G_{\lceil t\rceil }$. Define the Herbrand function byIt is continuous, strictly increasing and piecewise linear; denote its inverse by ${\psi }_{L/K}$. The upper ramification numbering isThe definition includes the placement of the groups at break endpoints: the group at a break is the group before the drop. Lower numbering is compatible with subgroups; upper numbering is the one compatible with quotients. In a totally ramified extension, the uniformizer criterion for lower ramification groups permits testing the defining valuation inequality on one uniformizer instead of on every integral element.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.