Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 123 / 4

Normalize the discrete valuation by $v_L({\pi }_L)=1$, and put $v_L(0)=+\infty$. For $\sigma \ne 1$, define its ramification numberand set $i_G(1)=+\infty$. The minimum exists: all finite values are nonnegative integers, and a nonidentity automorphism moves some integral element. The lower ramification numbering isThus $G_s=G_{\lceil s\rceil }$, $G_0$ is the inertia group, and $G_1$ is the wild inertia group. This states the convention at integer endpoints as well as between them. Automorphisms preserve $v_L$, so these are normal subgroups of $G$.

Define the Herbrand function and its inverse byThe integrand is positive and piecewise constant. Since $G_t=1$ for sufficiently large $t$, the function is continuous, strictly increasing and unbounded, and its inverse is defined on all nonnegative reals. The upper ramification numbering isIn particular, the denominator in the Herbrand integral involves $G_0$, even when $L/K$ is not totally ramified. These conventions are also given in Milne's Class Field Theory, Chapter I, §4.

Now use the monogenic hypothesis ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$. Since $L/K$ is Galois, its minimal polynomial has the distinct conjugates $\sigma (\alpha )$ as roots, soTaking valuations proves the first equality.

For every polynomial $P\in {\mathcal{O}}_K[T]$, the difference $P(\sigma (\alpha ))-P(\alpha )$ is divisible in ${\mathcal{O}}_L$ by $\sigma (\alpha )-\alpha$. Every element of ${\mathcal{O}}_L$ is such a polynomial value, and taking the element $\alpha$ attains the resulting lower bound. HenceFor an integer $i_G(\sigma )=m\ge 0$, the element $\sigma$ lies in exactly the groups $G_0,\dots ,G_{m-1}$; if $m=0$, it lies in none. Summing this count over the nonidentity elements and interchanging the two finite sums givesThis proves the ramification-group sum for a monogenic integer ring. Equivalently it is the different exponent from ramification groups, since the different ideal is generated by $f^{\prime }(\alpha )$ under the same monogenic hypothesis.

The polynomial is Eisenstein at $3$, so a root $\alpha$ generates a totally ramified cubic field $F={\mathbb{Q}}_3(\alpha )$ and is a uniformizer of $F$. Its cubic discriminant isIts $3$-adic valuation is $3$, so it is not a square. The cubic discriminant criterion therefore givesThe unique quadratic subfield is generated by $\sqrt{\Delta }$. Since the permitted square root $\sqrt{-5}$ is in ${\mathbb{Q}}_3$, this subfield isIt is totally ramified of degree two, by the Eisenstein polynomial $T^2-3$. The ramification index of $M/{\mathbb{Q}}_3$ is divisible by both the index $3$ of $F$ and the index $2$ of $E$. Since the whole degree is $6$, it follows thatIn particular $G_0=G$.

We use the following standard facts about totally ramified extensions: every uniformizer generates the valuation ring over the base valuation ring, and for a totally ramified Galois extensionThus we can use the uniformizer criterion for lower ramification groups. Put $s=\sqrt{3}$. Since $v_M(\alpha )=2$ and $v_M(s)=3$,is a uniformizer of $M$.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3$ be the three roots. The Galois group acts transitively on their unordered pairs and preserves the valuation, so all differences between distinct roots have the same valuation $d$. The discriminant identity giveshence $d=3$. Thus, for any $\sigma$, either $\sigma (\alpha )=\alpha$ or $\sigma (\alpha )-\alpha$ has valuation $3$.

A nontrivial three-cycle fixes $s$, since it fixes the quadratic subfield, and sends $\alpha$ to a different root. Also $v_M(\sigma (\alpha )+\alpha )=2$: write the sum as $2\alpha +(\sigma (\alpha )-\alpha )$, whose two terms have unequal valuations $2,3$. ThereforeA transposition changes the sign of $s$. ConsequentlyThe numerator has valuation $4$: it is $2{\alpha }^2$ plus terms of strictly higher valuation if the root changes, and exactly $2{\alpha }^2$ if the root is fixed. Thus every transposition has $v_M(\sigma (\pi )-\pi )=1$. We conclude that the ramification groups of the splitting field of T3 minus 3T plus 3 over Q3 are

As an independent check, the cubic field has ${\mathcal{O}}_F={\mathbb{Z}}_3[\alpha ]$ and different exponent $v_F(g^{\prime }(\alpha ))=v_F(3({\alpha }^2-1))=3$. The extension $M/F$ is tamely totally ramified of degree two, so its different exponent is $1$. The different in a tower gives $d(M/{\mathbb{Q}}_3)=1+2\cdot 3=7$, agreeing with the ramification sum $(6-1)+(3-1)=7$.