Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 21 / 3

For a finite Galois extension of local fields, let $G=\mathrm{Gal}(L/K)$, $G_{-1}=G$, and, for integers $i\ge 0$, define the lower ramification groups byIn particular, $G_0$ is the inertia group. The identity satisfies every bound because $v_L(0)=\infty$.

If the extension is totally ramified, its residue fields agree. Choose representatives in ${\mathcal{O}}_K$ of their common residue field. Every $a\in {\mathcal{O}}_L$ has a convergent uniformiser expansion $a={\sum }_{j\ge 0}{c}_j{\pi }_L^j$ with these representatives $c_j$, all fixed by $G$. For $j\ge 1$,Each summand in the second factor has valuation $j-1$. Thus a bound $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$ implies the same bound on $\sigma (a)-a$, by convergence and the valuation inequality. Necessity follows by testing $a={\pi }_L$. This proves the uniformizer criterion for lower ramification groups:

Put $\zeta ={\zeta }_{p^n}$, with $n\ge 1$. The shifted cyclotomic polynomial ${\Phi }_{p^n}(1+X)$ is Eisenstein at $p$: its constant term is $p$, and modulo $p$ it is $X^{p^{n-1}(p-1)}$. Therefore $L/{\mathbb{Q}}_p$ is totally ramified of degree $p^{n-1}(p-1)$ and $\pi =\zeta -1$ is a uniformiser.

The Galois group is $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$, with ${\sigma }_a(\zeta )={\zeta }^a$. For $a\ne 1$, let $r=v_p(a-1)$, $0\le r<n$. Then ${\zeta }^{a-1}$ is a primitive $p^{n-r}$th root. Its difference from one is a uniformiser in the corresponding smaller cyclotomic field, and the relative ramification index is $p^r$. HenceThe uniformizer criterion for lower ramification groups now determines every group. Write $U_j=\{a\in (\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }:a\equiv 1(\mathrm{mod}{p}^j)\}$, with $U_n=\{1\}$. Then the lower ramification filtration of a prime-power cyclotomic extension isFor $n=1$ the middle range is empty and the extension is tame. The formula also includes $p=2$; then $U_1$ is already the whole group, so the first possible drop is later than in the odd-prime case. For $p=2,n=1$, the entire extension is trivial.

Write $\alpha =\sqrt[3]{3}$ and $\zeta ={\zeta }_3$. The cubic is Eisenstein and adjoining $\zeta$ adds at most a quadratic extension, so $[L:{\mathbb{Q}}_3]\le 6$. SetUsing ${\zeta }^2+\zeta +1=0$ and ${\alpha }^3=3$, compute ${\pi }^3=1+2\zeta$ and ${\pi }^6=-3$. Thus $\pi$ satisfies the Eisenstein polynomial $X^6+3$. This proves $L={\mathbb{Q}}_3(\pi )$, degree six, with $\pi$ a uniformiser and the extension totally ramified. It is the splitting field of $X^3-3$, so its Galois group is $S_3$. This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

The six automorphisms have $\sigma (\alpha )={\zeta }^a\alpha$, $a\in \{0,1,2\}$, and $\sigma (\zeta )={\zeta }^b$, $b\in \{1,2\}$. Since $v_L(3)=6$, we have $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$.

If $b=1$ and $a\ne 0$, then $\sigma (\pi )={\zeta }^{-a}\pi$, whence $v_L(\sigma (\pi )-\pi )=1+3=4$. These are the two nonidentity elements of the cyclic subgroup $C_3$.

If $b=2$, use ${\zeta }^2-1=(\zeta -1)(\zeta +1)=-(\zeta -1){\zeta }^2$. Then $\sigma (\pi )=-{\zeta }^{2-a}\pi$. The coefficient $-{\zeta }^{2-a}-1$ reduces to $-2\ne 0$ in the residue field ${\mathbb{F}}_3$, so the difference has valuation one. These three automorphisms are the transpositions.

The uniformizer criterion for lower ramification groups consequently givesAs a consistency check, the different exponent is $(6-1)+3(3-1)=11$. The derivative of the monogenic Eisenstein polynomial gives the same value, $v_L(6{\pi }^5)=6+5=11$. In particular, stopping the wild filtration at $G_1$ would give the wrong different.