= Solution
\b[The filtration and its meaning.] Let $L/K$ be a finite <Galois extension> of non-Archimedean <local fields>, with group $G$, normalized <valuation> $v_L$, and <residue fields> $k_L/k_K$ of characteristic $\ell$. The <lower ramification numbering> is
$$
G_{-1}=G,\qquad
G_j=\{\sigma\in G:v_L(\sigma a-a)\geq j+1
\text{ for every }a\in\mathcal O_L\}\quad(j\geq0).
$$
The <higher ramification groups> are normal, decrease with $j$, and eventually become trivial: for each nonidentity automorphism some integral element is moved by a nonzero amount. The first group $G_0$ is the <inertia group>, with
$$
G/G_0\cong\operatorname{Gal}(k_L/k_K).
$$
Its fixed field is the maximal <unramified extension> inside $L$. The <wild inertia group> is $G_1$. For a <uniformizer> $\pi_L$, there are injective homomorphisms
$$
G_0/G_1\hookrightarrow k_L^\times,\quad
\sigma\longmapsto\overline{\sigma(\pi_L)/\pi_L},
\qquad
G_j/G_{j+1}\hookrightarrow(k_L,+),\quad
\sigma\longmapsto\overline{\frac{\sigma(\pi_L)-\pi_L}{\pi_L^{j+1}}}
\quad(j\geq1).
$$
The kernels are the indicated next groups; the additive homomorphism assertion follows by expanding a product of automorphisms modulo $\pi_L^{j+2}$. These maps show that $G_0/G_1$ is cyclic of order prime to $\ell$, while $G_1$ is an $\ell$-group. Thus <tame ramification> is exactly the case $G_1=1$.
To calculate the groups, the <uniformizer criterion for lower ramification groups> says
$$
\boxed{G_j=\{\sigma\in G_0:v_L(\sigma\pi_L-\pi_L)\geq j+1\}.}
$$
Indeed, $\mathcal O_L$ is generated by $\pi_L$ over the integers of the maximal unramified subextension, which $G_0$ fixes. Differences of powers of $\pi_L$ are divisible by $\sigma\pi_L-\pi_L$, so the one test implies all the defining inequalities.
\b[Eisenstein facts used in the calculations.] For a complete <discrete valuation ring> $A$ with <uniformizer> $\varpi$, a monic polynomial
$$
f(T)=T^d+a_{d-1}T^{d-1}+\cdots+a_0
$$
is an <Eisenstein polynomial> if $\varpi\mid a_j$ for every $j<d$ and $\varpi^2\nmid a_0$. The <Eisenstein criterion> makes it irreducible. If $\theta$ is a root, its <valuation> relative to the base normalization is $1/d$, the extension has degree and <ramification index> $d$, its <residue field> is unchanged, and
$$
\boxed{\mathcal O_{K(\theta)}=A[\theta],\qquad
\theta\text{ is a uniformizer}.}
$$
The root <valuation> follows by comparing the terms of its equation. The <ramification index> must then be at least $d$, hence exactly $d$. In the basis $1,\theta,\ldots,\theta^{d-1}$, the <valuations> of $b_j\theta^j$ are distinct modulo $d$, so an integral linear combination has every $b_j\in A$. This proves the integer-ring assertion. Conversely, a <uniformizer> in a <totally ramified extension> generates the field, and its minimal polynomial is <Eisenstein>; this follows from its value $1/d$, equal <valuations> of its conjugates, and the <valuation> one of its norm.
For a monogenic separable integer ring $A[\theta]$, the <different ideal> is generated by $f'(\theta)$. For a <Galois extension>, the <different exponent from ramification groups> is
$$
\boxed{v_L(\mathfrak D_{L/K})=\sum_{j\geq0}(|G_j|-1).}
$$
In the totally ramified case these agree directly: $f'(\pi_L)=\prod_{\sigma\ne1}(\pi_L-\sigma\pi_L)$, and an automorphism with displacement <valuation> $h$ contributes once to each of $G_0,\ldots,G_{h-1}$.
\b[Upper numbering.] Extend the lower indexing to real $s\geq0$ by $G_s=G_{\lceil s\rceil}$, and set
$$
\varphi_{L/K}(s)=\int_0^s\frac{dt}{[G_0:G_t]},\qquad
\psi_{L/K}=\varphi_{L/K}^{-1},\qquad
G^u=G_{\psi_{L/K}(u)}.
$$
On $[-1,0]$ take $\varphi(s)=s$. This <Herbrand function> slows the indexing as the groups shrink. Lower numbering is compatible with subgroups, while <upper ramification numbering> is compatible with quotients: for normal $H$, $(G/H)^u=G^uH/H$. This is the fact that <upper ramification groups commute with quotients>. Thus upper breaks are particularly useful when comparing intermediate <Galois extensions>.
\b[The eighth-root cyclotomic extension.] Put $\zeta=\zeta_8$ and $\pi=\zeta-1$. The shifted cyclotomic polynomial
$$
(1+T)^4+1=T^4+4T^3+6T^2+4T+2
$$
is <Eisenstein> at two. Hence $L=\mathbb Q_2(\zeta)$ is totally ramified of degree four, $v_L(2)=4$, and $\pi$ is a <uniformizer>. Its <Galois group> consists of $\sigma_a(\zeta)=\zeta^a$ for $a=1,3,5,7$, and is $C_2\times C_2$.
Since $\zeta$ is a unit,
$$
v_L(\sigma_a\pi-\pi)=v_L(\zeta^{a-1}-1).
$$
For $a=3,7$, this is the <valuation> of $i-1$ or $-i-1$, namely two, because their squares are units times two. For $a=5$, it is $v_L(-2)=4$. The <ramification groups of the eighth-root cyclotomic extension of the 2-adic field> are therefore
$$
\boxed{
G_{-1}=G_0=G_1=C_2\times C_2,\qquad
G_2=G_3=\{1,\sigma_5\}\cong C_2,\qquad
G_j=1\ (j\geq4).}
$$
The lower breaks are one and three, and
$$
\varphi(s)=
\begin{cases}
s,&0\leq s\leq1,\\
1+(s-1)/2,&1\leq s\leq3,\\
2+(s-3)/4,&s\geq3.
\end{cases}
$$
Thus the upper breaks are one and two:
$$
\boxed{
G^u=
\begin{cases}
C_2\times C_2,&-1\leq u\leq1,\\
\{1,\sigma_5\},&1<u\leq2,\\
1,&u>2.
\end{cases}}
$$
The <different exponent from ramification groups> is $3+3+1+1=8$, agreeing with $v_L(4\zeta^3)=8$ from the derivative of $T^4+1$.
\b[The cubic splitting field at three.] Let $\alpha^3=2$, $\zeta=\zeta_3$, and put
$$
\beta=\alpha+1,\qquad\lambda=\zeta-1.
$$
The polynomials
$$
\beta^3-3\beta^2+3\beta-3=0,\qquad
\lambda^2+3\lambda+3=0
$$
are <Eisenstein> at three. They give totally ramified subextensions of degrees three and two. Their intersection is the base field, so $L=\mathbb Q_3(\alpha,\zeta)$ has degree six. Its <ramification index> is divisible by both three and two, hence equals six; therefore $L/\mathbb Q_3$ is totally ramified. It is the splitting field of $T^3-2$, with <Galois group> $S_3$.
Normalize $v_L$ so that $v_L(3)=6$. Then $v_L(\beta)=2$ and $v_L(\lambda)=3$, making
$$
\boxed{\pi=\frac{\lambda}{\beta}}
$$
a <uniformizer>. Let $\sigma(\alpha)=\zeta\alpha,\ \sigma(\zeta)=\zeta$ and $\tau(\alpha)=\alpha,\ \tau(\zeta)=\zeta^2$. They generate $S_3$. Since $\tau(\lambda)=(\zeta+1)\lambda$,
$$
\tau(\pi)-\pi=\zeta\pi,\qquad v_L(\tau(\pi)-\pi)=1.
$$
For the order-three automorphism,
$$
\sigma(\pi)-\pi
=-\frac{\alpha\lambda^2}{(\alpha+1)(\zeta\alpha+1)},
\qquad v_L(\sigma(\pi)-\pi)=6-2-2=2.
$$
Here $\alpha$ is a unit and $v_L(\zeta\alpha+1)=v_L(\sigma\beta)=2$. The same calculation gives <valuation> two for $\sigma^2$. The three transpositions are conjugate and the defining filtration is normal, so they all have displacement <valuation> one. The <ramification groups of the splitting field of T3 minus 2 over Q3> are
$$
\boxed{G_{-1}=G_0=S_3,\qquad
G_1=\langle\sigma\rangle\cong C_3,\qquad
G_j=1\ (j\geq2).}
$$
The lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\leq1$,
$$
\varphi(s)=
\begin{cases}s/2,&0\leq s\leq1,\\
1/2+(s-1)/6,&s\geq1,
\end{cases}
$$
and the upper groups are
$$
\boxed{
G^u=
\begin{cases}
S_3,&-1\leq u\leq0,\\
C_3,&0<u\leq1/2,\\
1,&u>1/2.
\end{cases}}
$$
The <different exponent from ramification groups> is $5+2=7$. As an independent check, the cubic subfield has different exponent $v_{\mathbb Q_3(\alpha)}(3\alpha^2)=3$, and the <quadratic extension> above it is tame with different exponent one. Transitivity of the <different ideal> gives $1+2\cdot3=7$, as required.
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