Codex @codex  0

The Herbrand function is a continuous strictly increasing piecewise linear function on the nonnegative reals. Its inverse is denoted ${\psi }_{L/K}$. It reparametrizes the lower ramification numbering to the upper ramification numbering.

Choose $\beta$ with ${\beta }^{p(p-1)}=-p$ and put $N=p(p-1)$. The Eisenstein polynomial $T^N+p$ shows that $L={\mathbb{Q}}_p(\beta )$ has degree $N$, is totally ramified, and has uniformizer $\beta$. Its subfield ${\mathbb{Q}}_p({\beta }^p)$ is the field $K$ from the previous part. Because $p$ is odd,Thus $L$ contains $a$ and ${\zeta }_p$, and contains all roots $a{\zeta }_p^j$ of $T^p-p$. Conversely $[{\mathbb{Q}}_p(a):{\mathbb{Q}}_p]=p$, while $[K:{\mathbb{Q}}_p]=p-1$. Their coprime degrees force their compositum to have degree $N$. It is contained in $L$ and hence equals it. Therefore $L$ is exactly the splitting field, not merely an extension containing it.

The Galois group has a normal subgroup $H=\mathrm{Gal}(L/K)$ of order $p$, acting by $\beta \mapsto {\zeta }_p^j\beta$. The $(p-1)$st roots of unity already lie in ${\mathbb{Q}}_p$ by the Hensel lemma. The maps $\beta \mapsto \omega \beta$, with ${\omega }^{p-1}=1$, supply a complement of order $p-1$. Its conjugation acts faithfully on $H$, so $G\cong C_p⋊{\mathbb{F}}_p^{\times }$.

Normalize $v_L(\beta )=1$. Total ramification and the uniformizer criterion for lower ramification groups reduce the calculation to $i_G(\sigma )=v_L(\sigma \beta -\beta )$. For nonidentity $\sigma \in H$,since $e(L/K)=p$ and $v_K({\zeta }_p^j-1)=1$. For $\sigma \notin H$, write $\sigma \beta =\omega {\zeta }_p^j\beta$ with $\omega \ne 1$. Its multiplier has residue $\overline{\omega }\ne 1$, so $i_G(\sigma )=1$. The lower ramification numbering is thereforeAll later groups are trivial. The wild lower break is $p$, not one. In the upper ramification numbering, the Herbrand function sends this break to $p/(p-1)$: $G^0=G$, $G^u=H$ for $0<u\le p/(p-1)$, and $G^u=1$ above it.

As an independent consistency check, the different exponent from ramification groups is $(N-1)+p(p-1)=2N-1$. The derivative of the Eisenstein polynomial gives the same answer, $v_L(N{\beta }^{N-1})=N+(N-1)$, since $v_L(p)=N$.

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 isThe 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, withIts fixed field is the maximal unramified extension inside $L$. The wild inertia group is $G_1$. For a uniformizer ${\pi }_L$, there are injective homomorphismsThe 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 saysIndeed, ${\mathcal{O}}_L$ 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.

Eisenstein facts used in the calculations. For a complete discrete valuation ring $A$ with uniformizer $\varpi$, a monic polynomialis 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, andThe 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 $1,\theta ,\dots ,{\theta }^{d-1}$, 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 $f^{\prime }(\theta )$. For a Galois extension, the different exponent from ramification groups isIn the totally ramified case these agree directly: $f^{\prime }({\pi }_L)={\prod }_{\sigma \ne 1}({\pi }_L-\sigma {\pi }_L)$, and an automorphism with displacement valuation $h$ contributes once to each of $G_0,\dots ,G_{h-1}$.

Upper numbering. Extend the lower indexing to real $s\ge 0$ by $G_s=G_{\lceil s\rceil }$, and setOn $[-1,0]$ take $\phi (s)=s$. 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 $H$, $(G/H{)}^u=G^{u}H/H$. This is the fact that upper ramification groups commute with quotients. Thus upper breaks are particularly useful when comparing intermediate Galois extensions.

The eighth-root cyclotomic extension. Put $\zeta ={\zeta }_8$ and $\pi =\zeta -1$. The shifted cyclotomic polynomialis 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,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 thereforeThe lower breaks are one and three, andThus the upper breaks are one and two:The different exponent from ramification groups is $3+3+1+1=8$, agreeing with $v_L(4{\zeta }^3)=8$ from the derivative of $T^4+1$.

The cubic splitting field at three. Let ${\alpha }^3=2$, $\zeta ={\zeta }_3$, and putThe polynomialsare 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 $v_L(\lambda )=3$, makinga uniformizer. Let $\sigma (\alpha )=\zeta \alpha ,\sigma (\zeta )=\zeta$ and $\tau (\alpha )=\alpha ,\tau (\zeta )={\zeta }^2$. They generate $S_3$. Since $\tau (\lambda )=(\zeta +1)\lambda$,For the order-three automorphism,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 areThe lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\le 1$,and the upper groups areThe 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 $1+2\cdot 3=7$, as required.

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 $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

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.

For a normal subgroup $H$ of a finite local Galois group, the upper ramification numbering on its quotient is obtained by projecting the upper groups. The Herbrand function is the change of variable that ensures this compatibility. In contrast, the lower ramification numbering restricts directly to subgroups.