# Ramification groups of some finite Galois extensions of maximal nilpotency
class
over local fields of positive characteristic

Koto Imai  Graduate School of Mathematical Sciences
The University of Tokyo
3–8–1 Komaba Meguro-ku Tokyo 153–8914 Japan

##  0 Introduction

Let $K$ be a complete discrete valuation field of equal characteristics $p>0$.
Assume that the residue field $k$ of $K$ is perfect. For a finite Galois
extension of $K$, upper ramification groups are defined in [7], VI, §3. We are
interested in the jumps of the upper ramification groups.

Let $L/K$ be an Artin-Schreier extension defined by $x^{p}-x=a$ for $a\in K$
such that the valuation $v(a)$ of $a$ is negative and prime to $p$. Then it is
well known that the unique jump of the upper ramification groups is located at
$-v(a)$. The number $-v(a)$ is the conductor of the extension $L/K$.
Furthermore, the calculation of the conductor was generalized by Brylinski [4]
to the cases where the extension is abelian. We give an analog in some non-
abelian cases. Our result is related to that of Abrashkin [2], but our
approach is totally different and the proofs are more elementary.

Fix an integer $2\leq n\leq p$. Let $A$ be the nilpotent matrix of order $n$
defined by

$A=\left(\delta_{i,j-1}\right)_{ij}\in M(n)$ (0.1)

where $\delta$ denotes the Kronecker delta and $M(n)$ denotes the group of
matrices of order $n$. Let
$\displaystyle{R(t,x)=\sum_{i=0}^{p-1}\binom{x}{i}t^{i}\in\mathbb{F}_{p}[t,x]}$
be the $(p-1)$-st Maclaurin polynomial of $(1+t)^{x}$ with respect to $t$.
Define a morphism $\mathbf{A}:\mathbb{G}_{a}\to GL(n)$ of algebraic groups by
$\mathbf{A}(x)=R(A,x)\in GL(n)$. Let $G\subset GL(n+1)$ be the unipotent
algebraic group of dimension $n+1$ over $\mathbb{F}_{p}$ defined by

$G=\left\\{\begin{pmatrix}\mathbf{A}(x)&y\\\
0&1\end{pmatrix}\;\middle|\;x\in\mathbb{G}_{a},\;y\in{\mathbb{G}_{a}}^{n}\right\\}.$
(0.2)

The order of the group $G(\mathbb{F}_{p})$ is $p^{n+1}$ and $G$ is of
nilpotency class $n$. When $n=2$, $G$ is nothing but the algebraic group of
upper triangular unipotent matrices of order 3. The group $G(\mathbb{F}_{p})$
is one of the two non-abelian $p$-groups of the smallest order up to
isomorphism.

We study a totally ramified Galois extension $M_{n}/K$ whose Galois group
$G_{M_{n}/K}$ is isomorphic to $G(\mathbb{F}_{p})$. Let
$F:{\mathbb{G}_{a}}^{n}\to{\mathbb{G}_{a}}^{n}$ denote the component-wise
Frobenius map. We show in Lemma 1.9 that there exist

$a\in K,\;\alpha\in M_{n},\;b\in K^{n},\;\gamma=\begin{pmatrix}\gamma_{n}\\\
\vdots\\\ \gamma_{1}\end{pmatrix}\in{M_{n}}^{n},$ (0.3)

such that

$\alpha^{p}-\alpha=a\in
K,\;F(\gamma)-\gamma=\mathbf{A}(-\alpha^{p})b=:c=\begin{pmatrix}c_{n}\\\
\vdots\\\ c_{1}\end{pmatrix}\in K(\alpha)^{n}.$ (0.4)

and $M_{n}=K(\alpha,\gamma_{1},\ldots,\gamma_{n})$. Let $r_{n}$ denote the
largest rational number such that the $r_{n}$-th upper ramification group of
$M_{n}/K$ is non-trivial. We may take $a,b$ satisfying the conditions of Lemma
3.2 without loss of generality. The aim of this paper is to express $r_{n}$ in
terms of these $a,b$.

This extension is of maximum nilpotency class, i.e., the descending central
series of its Galois group is the longest among the groups of the same order.
Combined with the abelian cases, the calculation of the ramification groups
for this case should give some insight when calculating the ramification
groups of non-abelian Galois extensions in general. This is a motivation for
us to consider the extension $M_{n}/K$.

This paper consists of three parts: Section 1, Section 2, and Section 3. In
the first two sections, we give some preliminary results and settings. In the
last section, we prove our main theorem, Theorem 3.3.

In Section 1, we give some basic results on Galois cohomology required to
prove the main theorem. Using Galois cohomology, we give a combination of
$a\in K,b\in K^{n}$ defining $M_{n}/K$. At the last of this section, we will
give some calculations required in the last section.

In Section 2, we introduce filtrations
$F_{\bullet}K,F_{\bullet}\Omega_{K}^{1},F_{\bullet}H^{1}(K)$ on
$K,\Omega_{K}^{1},\;H^{1}(K):=H^{1}(K,\mathbb{F}_{p})$, and the graded modules
$\operatorname{Gr}K,\operatorname{Gr}\Omega_{K}^{1},\operatorname{Gr}H^{1}(K)$
associated to these filtrations as in [1]. We investigate the relationship
between the graded modules defined for $K$ and $L$, where $L/K$ is an Artin-
Schreier extension. This is required since $M_{n}/K(\alpha)$ is a composition
of $n$ Artin-Schreier extensions.

In Section 3, we give our main result, i.e., the calculation of $r_{n}$. We
reduce the problem to the calculation of the conductor $m^{\prime}_{n}$ of the
Artin-Schreier extension of $L=K(\alpha)$ defined by $x^{p}-x=c_{n}$ under
some conditions on $a,b$ as in Lemma 3.2. Take
$c^{\prime}_{n},c^{\prime\prime}_{n}\in M_{n}$ such that

$c_{n}-{c^{\prime\prime}_{n}}^{p}+c^{\prime\prime}_{n}=c^{\prime}_{n},\;-v_{L}(c^{\prime}_{n})=m^{\prime}_{n}.$
(0.5)

Let

$s_{n}=\max\left(-v_{L}\left(t\frac{dc_{n}}{dt}\right),-v_{L}\left(t\frac{dc^{\prime\prime}_{n}}{dt}\right)\right),$
(0.6)

where $t$ denotes a uniformizer of $L$. First, we express $s_{n}$ in terms of
$a,b$ in Corollary 3.7(b). The main ingredient of the proof of Theorem 3.3 is
to prove that the inequality in

$m^{\prime}_{n}=-v_{L}\left(t\frac{dc^{\prime}_{n}}{dt}\right)\leq s_{n}$
(0.7)

is actually an equality, by calculating the image of $c^{\prime}_{n}$ in
$\operatorname{Gr}_{s_{n}}H^{1}(L)$ using the results from Section 2.

The author wishes to thank Professor Takeshi Saito for suggesting this
problem, giving helpful advice and ideas, and pointing out mistakes during the
preparation of this paper.

##  1 Preliminaries

Let $B,C$ be groups, $\Gamma$ a profinite group acting continuously on $B$ and
$C$, and $f,g:B\to C$ group homomorphisms preserving the actions of $\Gamma$.
Assume that the map $h:B\to C$ defined by $y\mapsto f(y)g(y)^{-1}$ is a
surjection. Let $A$ denote the inverse image of $\\{1\\}$ by $h$. Then $A$ is
a subgroup of $B$ with a continuous action of $\Gamma$.

We consider the following sequence:

$\\{1\\}\to A\hookrightarrow B\overset{h}{\to}C\to\\{1\\}.\\\ $ (1.1)

This is “exact” in the sense that the images of the maps coincide with the
inverse image of $\\{1\\}$ by the next map. Furthermore, the map $B\times A\to
B\times_{C}B$ defined by $(y,x)\mapsto(y,yx)$ is a bijection. Nevertheless,
this is not an exact sequence, because $h$ is not in general a group
homomorphism. However, we can still consider the “long exact sequence of
cohomology” for this “exact” sequence as in [7], VII, Annex.

######  Proposition 1.1.

  1. (a)

For $y\in B$ such that $h(y)\in C^{\Gamma}$, the map $\zeta_{y}:\Gamma\to B$
defined by $\sigma\mapsto y^{-1}\sigma(y)$ is a 1-cocycle of $A$. Moreover, we
can define a map $\delta:C^{\Gamma}\to H^{1}(\Gamma,A)$ as follows:

$\delta(z)=\bar{\zeta_{y}},$ (1.2)

where $h(y)=z\in C^{\Gamma}$, and $\bar{\zeta_{y}}$ denotes the class of
$\zeta_{y}$ as a 1-cocycle of $A$.

  2. (b)

The image of the map $\delta:C^{\Gamma}\to H^{1}(\Gamma,A)$ coincides with the
inverse image of $\\{1\\}$ by $H^{1}(\Gamma,A)\to H^{1}(\Gamma,B)$.

  3. (c)

Take $z,z^{\prime}\in C^{\Gamma}$. Let $\Gamma_{z}$ and $\Gamma_{z^{\prime}}$
denote the intersections of the stabilizer subgroups of $\Gamma$ with respect
to the elements in $h^{-1}(\\{z\\})$ and $h^{-1}(\\{z^{\prime}\\})$
respectively. Then $\Gamma_{z}=\Gamma_{z^{\prime}}$ if and only if there
exists $y\in B^{\Gamma_{z}\Gamma_{z^{\prime}}}$ such that
$z^{\prime}=f(y)zg(y)^{-1}$.

###### Proof.

(a) For any $y\in B$ such that $h(y)\in C^{\Gamma}$, we have

$\\{yx\in B|x\in A\\}=h^{-1}(\\{h(y)\\})\supset\\{\sigma(y)|\sigma\in G\\}.$
(1.3)

Thus, for any $\sigma\in\Gamma$, we have $y^{-1}\sigma(y)\in A$. Since for any
$\sigma,\tau\in\Gamma$,

$y^{-1}\sigma(y)\sigma(y^{-1}\tau(y))=y^{-1}\sigma\tau(y),$ (1.4)

$\zeta_{y}$ is a 1-cocycle of $A$. For any $x\in A$, $\zeta_{yx}$ is
cohomologous to $\zeta_{y}$, since

$\zeta_{yx}(\sigma)=x^{-1}\zeta_{y}(\sigma)\sigma(x).$ (1.5)

Thus $\delta$ is well-defined.

(b) Take a 1-cocycle $s:\Gamma\to A$ of $A$.

Assume that $H^{1}(\Gamma,A)\to H^{1}(\Gamma,B)$ sends the class $\bar{s}$ of
$s$ to 1. Since $A\subset B$, this implies that there exists $y\in B$ such
that $s(\sigma)=y^{-1}\sigma(y)$. Since $s(\sigma)\in A$, we have

$h(y)=f(y)g(y)^{-1}=f(y)h(s(\sigma))g(y)^{-1}=f(y)f(y^{-1}\sigma(y))g(y^{-1}\sigma(y))^{-1}g(y)^{-1}$
$=f(\sigma(y))g(\sigma(y))^{-1}=\sigma(f(y)g(y)^{-1})=\sigma(h(y)).$ (1.6)

Hence $h(y)\in C^{\Gamma}$ and $\delta(h(y))=\bar{s}$.

Conversely, Take $z\in C^{\Gamma}$. Let $y$ denote an element of $B$
satisfying $h(y)=z$ and let $s=\zeta_{y}$. Then we have
$s(\sigma)=y^{-1}\sigma(y)$ for all $\sigma\in\Gamma$ and $\delta(z)=\bar{s}$.
Since this is a $B$-coboundary, we have that $H^{1}(\Gamma,A)\to
H^{1}(\Gamma,B)$ sends $\bar{s}$ to 1.

(c) Assume $\Gamma_{z}=\Gamma_{z^{\prime}}$. Then for all
$\sigma\in\Gamma_{z}\Gamma_{z^{\prime}}=\Gamma_{z}=\Gamma_{z^{\prime}}$,
$y_{0}\in h^{-1}(\\{z\\})$, and $y_{0}^{\prime}\in h^{-1}(\\{z^{\prime}\\})$,
we have $(y_{0}^{\prime}y_{0}^{-1})\sigma(y_{0}^{\prime}y_{0}^{-1})=1$ in $B$.
Therefore we have $y_{0}^{\prime}{y_{0}}^{-1}\in
B^{\Gamma_{z}\Gamma_{z^{\prime}}}$ for all $y_{0}\in h^{-1}(\\{z\\})$ and
$y_{0}^{\prime}\in h^{-1}(\\{z^{\prime}\\})$. Meanwhile, we have

$f(y_{0}^{\prime}{y_{0}}^{-1})zg(y_{0}^{\prime}{y_{0}}^{-1})^{-1}=f(y_{0}^{\prime})f(y_{0})^{-1}h(y_{0})g(y_{0})g(y_{0}^{\prime})^{-1}$
$=f(y_{0}^{\prime})g(y_{0}^{\prime})^{-1}=h(y_{0}^{\prime})=z^{\prime}.$ (1.7)

Therefore setting $y=y_{0}^{\prime}{y_{0}}^{-1}$ for some $y_{0}\in
h^{-1}(\\{z\\})$ and $y_{0}^{\prime}\in h^{-1}(\\{z^{\prime}\\})$ yields
$z^{\prime}=f(y)zg(y)^{-1}$.

Conversely, assume that there exists $y\in B^{\Gamma_{z}\Gamma_{z^{\prime}}}$
such that $z^{\prime}=f(y)zg(y)^{-1}$. Take $y_{0}\in h^{-1}(\\{z\\})$ and let
$y_{0}^{\prime}=yy_{0}$. Then $y_{0}^{\prime}\in h^{-1}(\\{z^{\prime}\\})$,
since $h(y_{0}^{\prime})=f(y)h(y_{0})g(y)=z^{\prime}$. Since $y$ is fixed by
$\Gamma_{z}\cup\Gamma_{z^{\prime}}\subset\Gamma_{z}\Gamma_{z^{\prime}}$, we
have ${y_{0}^{\prime}}^{-1}\sigma(y_{0}^{\prime})={y_{0}}^{-1}\sigma(y_{0})=1$
for all $\sigma\in\Gamma_{z}\cup\Gamma_{z^{\prime}}$. Thus,
$\Gamma_{z}\supset\Gamma_{z}\cup\Gamma_{z^{\prime}}$. Hence,
$\Gamma_{z}\supset\Gamma_{z^{\prime}}$. By symmetry, we also have
$\Gamma_{z^{\prime}}\supset\Gamma_{z}$. Thus, we have
$\Gamma_{z}=\Gamma_{z^{\prime}}$. ∎

######  Proposition 1.2.

Let $G$ be a group. Assume that $G$ admits a descending normal series of
subgroups
$G=G_{0}\mathrel{\ooalign{\raisebox{-2.15277pt}{\reflectbox{\rotatebox{90.0}{$\nshortmid$}}}\cr$\triangleright$\cr}\mkern-3.0mu}G_{1}\mathrel{\ooalign{\raisebox{-2.15277pt}{\reflectbox{\rotatebox{90.0}{$\nshortmid$}}}\cr$\triangleright$\cr}\mkern-3.0mu}\cdots\mathrel{\ooalign{\raisebox{-2.15277pt}{\reflectbox{\rotatebox{90.0}{$\nshortmid$}}}\cr$\triangleright$\cr}\mkern-3.0mu}G_{r}=\\{1\\}$.
For $0\leq i<r$, let $\pi_{i}$ denote the canonical projection $G_{i}\to
G_{i}/G_{i+1}$. Let $f,g:G\to G$ be group homomorphisms satisfying
$f^{-1}(G_{i})=g^{-1}(G_{i})=G_{i}$ for all $i$. Define a map $h:G\to G$ by
$y\mapsto f(y)g(y)^{-1}$, and for all $i$, let
$h^{\prime}_{i}:G_{i}/G_{i+1}\to G_{i}/G_{i+1}$ be the morphism induced by $h$
and $\pi_{i}$. Then $h$ is surjective if $h^{\prime}_{i}$ is surjective for
all $i$.

###### Proof.

We prove this proposition by induction on the length $r$ of the descending
normal series of $G$. If $r=1$, then we have $G_{1}=\\{1\\}$ and $\pi$ is the
identity map. Hence $h$ is surjective if $h^{\prime}_{0}$ is surjective.

Suppose $n>1$. Since $G_{1}$ admits a descending normal series of length
$r-1$, it suffices to show that $h$ is surjective if $h|_{G_{1}}$ and
$h^{\prime}_{0}$ are surjective.

Suppose $h|_{G_{1}}$ and $h^{\prime}_{0}$ are surjective. Consider the
following commutative diagram with exact rows.

$\textstyle{\\{1\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h|_{G_{1}}}$$\textstyle{G\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{h}$$\textstyle{G/G_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h^{\prime}_{0}}$$\textstyle{\\{1\\}}$$\textstyle{\\{1\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{G/G_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\\{1\\}}$
(1.8)

We will prove that $h$ is surjective by a technique similar to that used to
prove the five lemma. Note that we cannot simply apply the five lemma, because
$h$ is not in general a homomorphism.

Take $c^{\prime}\in G$. Since $h^{\prime}_{0}$ and $\pi$ are surjective, there
exists $c\in G$ satisfying $h^{\prime}_{0}(\pi(c))=\pi(c^{\prime})$. By the
commutativity of the diagram, $\pi(f(c)g(c)^{-1})=\pi(c^{\prime})$. Since
$\pi$ is a homomorphism, $\pi(f(c)^{-1}c^{\prime}g(c))=1$. Then by the
exactness of the lower row, we have $f(c)^{-1}c^{\prime}g(c)\in G_{1}$. Since
$h|_{G_{1}}$ is surjective, there exists $b\in G_{1}$ satisfying
$h(b)=f(c)^{-1}c^{\prime}g(c)$. We have

$h(cb)=f(c)h(b)g(c)^{-1}=f(c)f(c)^{-1}c^{\prime}g(c)g(c)^{-1}=c^{\prime}.$
(1.9)

Thus, $h$ is surjective. ∎

For a field $K$ of characteristic $p>0$, let $K_{s}$ denote a separable
closure of $K$ respectively, and $G_{K}=\operatorname{Gal}(K_{s}/K)$ the
absolute Galois group of $K$. Let $G$ be a unipotent group over $K$, i.e., an
algebraic subgroup of the group of unitriangular $n\times n$ matrices over $K$
for some $n$.

######  Definition 1.3.

We say that a unipotent group $G$ over $K$ is split if it admits a finite
descending normal series of subgroups whose quotients are isomorphic to the
additive group $\mathbb{G}_{a}$ (cf. [3], Definition 15.1).

######  Remark 1.4.

By [3], Theorem 15.5(ii), every connected unipotent group is split if $K$ is
perfect. However, this is not true if $K$ is not perfect. According to [6],
V.3.4, if $K$ is not perfect and $t$ is an element of $K-K^{p}$, then the
algebraic subgroup $\\{(x,y)|x^{p}-x-ty^{p}=0\\}$ of
$\mathbb{G}_{a}\times\mathbb{G}_{a}$ is not split.

######  Proposition 1.5.

  1. (a)

(Nguyễn) [8] A connected unipotent group $G$ over $K$ is split if and only if
$H^{1}(G_{L},G)=\\{1\\}$ for every extension $L/K$.

  2. (b)

Let $G$ be a split unipotent group over $\mathbb{F}_{p}$. Let $F:G\to G$
denote the morphism defined by the absolute Frobenius and let $P:G\to G$
denote the map defined by $y\mapsto F(y)y^{-1}$. Then $P$ is surjective.

###### Proof.

(a) See [8].

(b) Since $K_{s}\to K_{s}$ defined by $x\mapsto x^{p}-x$ is surjective, by
Proposition 1.2, $P$ is surjective. ∎

######  Definition 1.6.

Fix an integer $2\leq n\leq p$.

  1. (a)

Let $A$ be the nilpotent matrix of order $n$ defined by

$A=\left(\delta_{i,j-1}\right)_{ij}\in M(n)$ (1.10)

where $\delta$ denotes the Kronecker delta and $M(n)$ denotes the group of the
matrices of order $n$. Let $R(t,x)$ be the $(p-1)$-st Maclaurin polynomial of
$(1+t)^{x}$ with respect to $t$, i.e.,

$R(t,x)=\sum_{i=0}^{p-1}\binom{x}{i}t^{i}\in\mathbb{F}_{p}[t,x].$ (1.11)

Define a morphism $\mathbf{A}:\mathbb{G}_{a}\to GL(n)$ of algebraic groups by
$\mathbf{A}(x)=R(A,x)\in GL(n)$.

  2. (b)

Let $G\subset GL(n+1)$ be the unipotent algebraic subgroup of dimension $n+1$
over $\mathbb{F}_{p}$, defined by

$G=\left\\{\begin{pmatrix}\mathbf{A}(x)&y\\\
0&1\end{pmatrix}\;\middle|\;x\in\mathbb{G}_{a},\;y\in{\mathbb{G}_{a}}^{n}\right\\}.$
(1.12)

Let

$G\supsetneq Z_{1}G\supsetneq\cdots\supsetneq Z_{n}G=\\{1\\}$ (1.13)

be the lower central series of $G$.

######  Remark 1.7.

We have for all $1\leq j\leq n$,

$Z_{j}G=\left\\{\begin{pmatrix}\mathbf{A}(x)&y\\\ 0&1\end{pmatrix}\in
G\;\middle|\;y=\begin{pmatrix}y_{n}\\\ \vdots\\\
y_{1}\end{pmatrix},\;x=y_{1}=\cdots=y_{j}=0\right\\}.$ (1.14)

Moreover, $G$ is split since $\mathbb{F}_{p}$ is perfect.

######  Definition 1.8.

Fix an integer $2\leq n\leq p$. Let $K$ be a field. Let $M_{n}/K$ denote a
Galois extension whose Galois group is isomorphic to $G(\mathbb{F}_{p})$. Let
$K\subset M_{1}\subset\cdots\subset M_{n}$ be the Galois subextensions of
$M_{n}/K$ corresponding to the lower central series of $G(\mathbb{F}_{p})$.

Assume that $K$ is of characteristic $p>0$. We will apply Proposition 1.1 to
$B=C=G(K_{s}),\;\Gamma=G_{K},\;f=F,\;g=\operatorname{id}_{G(K_{s})}$, where
$G_{K}$ denotes the absolute Galois group over $K$, and $F:G(K_{s})\to
G(K_{s})$ denotes the Frobenius map. By Proposition 1.5(b), the map
$h:G(K_{s})\to G(K_{s})$ defined by $y\mapsto F(y)y^{-1}$ is a surjection.
Thus we can apply Proposition 1.1.

We have $A=h^{-1}(\\{1\\})=G(\mathbb{F}_{p})$. Since $G_{K}$ acts on
$G(\mathbb{F}_{p})$ trivially, we can identify
$H^{1}(G_{K},G(\mathbb{F}_{p}))$ with the set of conjugacy classes of
$\operatorname{Hom}(G_{K},G(\mathbb{F}_{p}))$ by $G(\mathbb{F}_{p})$.

######  Lemma 1.9.

  1. (a)

There exist $a\in K,\;b\in K^{n}$ such that the extension $M_{n}/K$ is defined
by

$\begin{pmatrix}\mathbf{A}(x^{p})&F(y)\\\
0&1\end{pmatrix}=\begin{pmatrix}\mathbf{A}(a)&b\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(x)&y\\\ 0&1\end{pmatrix},$ (1.15)

where $F:{\mathbb{G}_{a}}^{n}\to{\mathbb{G}_{a}}^{n}$ denotes the component-
wise Frobenius map.

  2. (b)

Take

$\alpha\in K_{s},\;\gamma=\begin{pmatrix}\gamma_{n}\\\ \vdots\\\
\gamma_{1}\end{pmatrix}\in{K_{s}}^{n}$ (1.16)

such that

$\alpha^{p}-\alpha=a,\;F(\gamma)-\gamma=\mathbf{A}(-\alpha^{p})b.$ (1.17)

Then for all $1\leq j\leq n$, we have
$M_{j}=K(\alpha,\gamma_{1},\ldots,\gamma_{j})$.

###### Proof.

(a) Let $\pi:G_{K}\to G_{M_{n}/K}$ denote the canonical projection. Let
$\phi:G_{M_{n}/K}\to G(\mathbb{F}_{p})$ be any isomorphism. Define
$\delta:G(K)\to H^{1}(G_{K},G(\mathbb{F}_{p}))$ as in Proposition 1.1. By
Proposition 1.1(b) and Proposition 1.5(a), $\delta$ is a surjection. Thus,
there exists

$T=\begin{pmatrix}\mathbf{A}(a)&b\\\ 0&1\end{pmatrix}\in G(K)$ (1.18)

such that $\delta(T)$ equals the conjugacy class of $\phi\circ\pi$. Then by
definition of $\delta$ in Proposition 1.1(a), $M_{n}/K$ is defined by (1.15).

(b) By applying Proposition 1.1(c) to $K(\alpha)$, we may replace $T$ with

$\begin{pmatrix}\mathbf{A}(-\alpha^{p})&0\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(a)&b\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(-\alpha)&0\\\
0&1\end{pmatrix}^{-1}=\begin{pmatrix}\mathbf{A}(0)&A(-\alpha^{p})b\\\
0&1\end{pmatrix}$ (1.19)

Thus $M_{n}=K(\alpha,\gamma_{1},\ldots,\gamma_{n})$. Note that
$G_{M_{n}/M_{j}}$ acts trivially on $\alpha,\gamma_{1},\ldots,\gamma_{j}$, but
not trivially on $\gamma_{j+1},\ldots,\gamma_{n}$. Therefore,
$M_{j}=K(\alpha,\gamma_{1},\ldots,\gamma_{j})$. ∎

######  Lemma 1.10.

Let $l_{1},\ldots,l_{n}$ and $\lambda_{1},\ldots,\lambda_{n}$ be sequences of
integers and $\lambda$ an integer. Assume that for all $2\leq i\leq n$, they
satisfy the following conditions:

  1. (i)

If $l_{i}<-(p-1)\lambda+p\lambda_{i}$, then we have
$\lambda_{i-1}=-\lambda+\lambda_{i}$.

  2. (ii)

We have $l_{i}\leq-(n-i)\lambda+\lambda_{n}$.

  3. (iii)

We have $\lambda<\lambda_{i}$.

Then we have

$l_{i}<-(p-1)\lambda+p\lambda_{i},$ (1.20)

and

$\lambda_{i}=(i-1)\lambda+\lambda_{1}$ (1.21)

for all $2\leq i\leq n$.

###### Proof.

We prove this lemma by induction on $n$. It is clear that the lemma holds for
$n=1$.

Suppose $n=j$. Assume that the lemma holds for $n=j-1$.

By (ii) for $i=j$, we have

$l_{j}\leq\lambda_{j}.$ (1.22)

By (iii) for $i=j$, we have

$\lambda_{j}<-(p-1)\lambda+p\lambda_{j}.$ (1.23)

Thus we get (1.20) for $i=j$. By (i) for $i=j$, we have

$-(j-1-i)\lambda+\lambda_{j-1}=-(j-i)\lambda+\lambda_{j}.$ (1.24)

The integer $\lambda$ and the sequences $l_{1},\ldots,l_{j-1}$ and
$\lambda_{1},\ldots,\lambda_{j-1}$ clearly satisfy conditions (i) and (iii) of
the lemma for $n=j-1$. We will show that they also satisfy condition (ii).

By (1.24) and (ii) for $2\leq i\leq j-1$, we have

$l_{i}\leq-(j-1-i)\lambda+\lambda_{j-1}$ (1.25)

for all $2\leq j\leq i-1$. Therefore, by induction hypothesis, we have (1.20)
and (1.21) for all $2\leq j\leq i-1$. By (1.21) for $i=j-1$ and (1.24), we
have (1.21) for $i=j$.

Hence, the lemma also holds for $n=j$. ∎

######  Lemma 1.11.

We have an equality

$\sum_{i=1}^{j}\frac{(-1)^{i-1}y}{(j-i)!(i-1)!(x+(i-1)y)}=\prod_{i=1}^{j}\frac{y}{x+(i-1)y}$
(1.26)

in
$\displaystyle{\mathbb{Z}\left[\frac{1}{(j-1)!},x,y,\frac{1}{\prod_{i=1}^{j}x+(i-1)y}\right]}$.

###### Proof.

We may assume $y=1$. Then it follows from the fact that the polynomial of
degree $j-1$,

$f(x)=\sum_{i=1}^{j}\left(\frac{(-1)^{i-1}}{(j-i)!(i-1)!}\prod_{\begin{subarray}{c}1\leq
i^{\prime}\leq j\\\ i^{\prime}\neq i\end{subarray}}(x+(i^{\prime}-1))\right)$
(1.27)

satisfies $f(0)=f(-1)=\cdots=f(-(j-1))=1$. ∎

##  2 Filtrations

Let $K$ be a complete discrete valuation field of characteristic $p>0$ with
perfect residue field $k$ and $K_{s}$ a separable closure of $K$. Define
filtrations $F_{\bullet}K$ of $K$ as $F_{n}K={\mathfrak{m}_{F}}^{-n}$.

Let $P:K_{s}\to K_{s}$ be the surjective map defined by $P(x)=x^{p}-x$. Then
by Proposition 1.1, Proposition 1.2 and Proposition 1.5(a), we can identify
the cokernel of $P|_{K}:K\to K$ with
$H^{1}(K):=H^{1}(G_{K},\mathbb{F}_{p})=\operatorname{Hom}(G_{K},\mathbb{F}_{p})$
by the following isomorphism:

$\begin{array}[]{ccc}\operatorname{Coker}P|_{K}&\to&\operatorname{Hom}(G_{K},\mathbb{F}_{p})\\\
x&\mapsto&(\sigma\mapsto\sigma(y)-y)\end{array}$ (2.1)

where $y\in K_{s}$ satisfies $P(y)=x$.

Consider the map $K\to H^{1}(K)$ defined by the projection $K\to
K/P(K)=\operatorname{Coker}P|_{K}$. Let $F_{n}H^{1}(K)$ be the image of
$F_{n}(K)$ by this map in $H^{1}(K):=H^{1}(K,\mathbb{F}_{p})$.

Let
$\operatorname{Gr}_{n}K=F_{n}K/F_{n-1}K,\operatorname{Gr}_{n}H^{1}(K)=F_{n}H^{1}(K)/F_{n-1}H^{1}(K)$
denote the graded quotients, and define the graded algebra
$\operatorname{Gr}K:=\oplus_{n\in\mathbb{Z}}\operatorname{Gr}_{n}K$. The
graded algebra $\operatorname{Gr}K$ is isomorphic to $k[t,t^{-1}]$.

The space $\Omega_{K}^{1}$ of Kähler differentials is a 1-dimensional
$K$-vector space, and its submodule $\Omega_{\mathcal{O}_{K}}^{1}$ is a free
$\mathcal{O}_{K}$-module of rank 1. Let $d:K\to\Omega_{K}^{1}$ denote the
canonical derivation. Let
$F_{n}\Omega_{K}^{1}=\mathfrak{m}_{K}^{-n-1}\Omega_{\mathcal{O}_{K}}^{1}$ and
define the graded quotient $\operatorname{Gr}_{n}\Omega_{K}^{1}$ and the
graded module $\operatorname{Gr}\Omega_{K}^{1}$ as above. Note that this
graded module is a $\operatorname{Gr}K$ module. For $\chi\in\Omega_{K}^{1}$,
let $v_{K}(\chi)$ denote the smallest integer $n$ such that $\chi\in
F_{-n}\Omega_{K}^{1}$.

######  Lemma 2.1.

  1. (a)

Let $t$ denote a uniformizer of $K$. The multiplication $K\to\Omega_{K}^{1}$
by $t^{-1}dt$ induces an isomorphism
$\mu:\operatorname{Gr}_{n}K\to\operatorname{Gr}_{n}\Omega_{K}^{1}$. This
isomorphism does not depend on the uniformizer $t$.

  2. (b)

The derivation $d$ induces a morphism
$\partial:\operatorname{Gr}_{n}K\to\operatorname{Gr}_{n}\Omega_{K}^{1}$. We
have $\partial=-n\cdot\mu$.

###### Proof.

(a) The multiplication by $t^{-1}dt$ induces an isomorphism $\mu$, since the
multiplication is clearly an isomorphism and
$F_{n}\Omega_{K}^{1}=t^{-1}dtF_{n}K$. For any uniformizer $t,t^{\prime}$ of
$K$, there exist $0\neq a\in k$ and $b\in\mathcal{O}_{K}$ satisfying
$t^{\prime}=at+bt^{2}$. Hence we have

$t^{\prime-1}dt^{\prime}=t^{-1}(a+bt)^{-1}(adt+d(bt^{2})).$ (2.2)

Since $(a+bt)^{-1}\in a^{-1}+\mathfrak{m}$ and $d(bt^{2})\in\mathfrak{m}dt$,
we have

$t^{\prime-1}dt^{\prime}\equiv t^{-1}dt\mod\mathcal{O}_{K}dt.$ (2.3)

Therefore, $\mu$ does not depend on the uniformizer.

(b) Let $t$ be a uniformizer of $K$. Since $d(t^{-n})=-nt^{-n-1}dt$, $d$
induces $\partial$, and we have $\partial=-n\cdot\mu$. ∎

######  Lemma 2.2.

Let $L/K$ be a finite separable totally ramified extension of complete
discrete valuation fields with residue field $k$, $e$ the ramification index
of $L/K$, and $\delta$ the valuation of the different of $L/K$. For any
integer $n$, let $n^{\prime}=e(n+1)-\delta-1$.

The canonical morphisms $K\to L$ and $\Omega_{K}^{1}\to\Omega_{L}^{1}$ induce
$F_{n}K\to F_{en}L$, $F_{n-1}K\to F_{en-1}L$, $F_{n}\Omega_{K}^{1}\to
F_{n^{\prime}}\Omega_{L}^{1}$, and $F_{n-1}\Omega_{K}^{1}\to
F_{n^{\prime}-1}\Omega_{L}^{1}$. There exists a unique non-zero element
$\theta\in\operatorname{Gr}_{e-\delta-1}L\simeq k$ such that for all $n$, the
diagram below is commutative, where $\mu$ is the morphism in Lemma 2.1(a)

$\textstyle{\operatorname{Gr}_{n}K\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\textstyle{\operatorname{Gr}_{en}L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta\cdot}$$\textstyle{\operatorname{Gr}_{n^{\prime}}L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\textstyle{\operatorname{Gr}_{n}\Omega_{K}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Gr}_{n^{\prime}}\Omega_{L}^{1}.}$
(2.4)

Moreover, for all $\chi\in\Omega_{K}^{1}$ such that $-v_{K}(\chi)=n$, we have
$-v_{L}(\chi)=n^{\prime}=e(n+1)-\delta-1$.

###### Proof.

Let $t_{K}$ and $t_{L}$ denote a uniformizer of $K$ and $L$ respectively. Let

$\lambda=\frac{{t_{K}}^{-1}dt_{K}}{{t_{L}}^{-1}dt_{L}}\in F_{e-\delta-1}L.$
(2.5)

The following diagram is commutative:

$\textstyle{K\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda\cdot}$$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Omega_{K}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Omega_{L}^{1},}$
(2.6)

where the lower horizontal arrow denotes the canonical morphism, and the left
and right vertical arrows denote the multiplication by ${t_{K}}^{-1}dt_{K}$
and ${t_{L}}^{-1}dt_{L}$ respectively. Thus setting
$\theta\in\operatorname{Gr}_{e-\delta-1}L$ as the image of $\lambda$ makes the
diagram (2.4) commutative. It follows from Lemma 2.1(a) that $\theta$ does not
depend on the choices of $t_{K}$ and $t_{L}$. ∎

######  Lemma 2.3.

Let $n$ be an integer. Consider
$\nu:\operatorname{Gr}_{n}K\to\operatorname{Gr}_{n}H^{1}(K)$ induced by the
canonical morphism $K\to H^{1}(K)$. If $n>0$ and $p\nmid n$, the morphism
$\nu$ is an isomorphism. If $n=0$, the morphism $\nu$ is a surjection and
$\operatorname{Gr}_{0}H^{1}(K)$ is isomorphic to $H^{1}(k)$. Otherwise, the
morphism $\nu$ is the zero-map.

###### Proof.

Suppose $n>0$ and $p\nmid n$. Since there is no $x\in K$ satisfying
$v_{K}(P(x))=-n$, we have $P(K)\cap(F_{n}K)\subset F_{n-1}K$. Thus, the
morphism $\nu$ is an isomorphism.

Suppose $n=0$. The surjectivity follows from the definition of $\nu$. Since
$F_{-1}K\subset P(K)$, we have

$\operatorname{Gr}_{0}H^{1}(K)\simeq(F_{0}K/F_{-1}K)/((F_{0}K\cap
P(K))/F_{-1}(K))$ $=k/(P(\mathcal{O}_{K})/\mathfrak{m}_{K})=k/P(k)\simeq
H^{1}(k).$ (2.7)

Suppose $n>0$ and $p\mid n$. Take $n^{\prime}$ such that $n=pn^{\prime}$. For
all $x\in F_{n}K$, there exists $y\in F_{n^{\prime}}K$ such that
$-v_{K}(x-P(y))<n$. Thus, $\nu$ is the zero-map.

Suppose $n<0$. Since $F_{-1}H^{1}(K)=0$, $\nu$ is the zero-map. ∎

For a finite Galois extension $L/K$ such that $K$ is a complete discrete
valuation field of characteristic $p>0$, let $G_{L/K}$ denote its Galois
group, and $L_{L/K},U_{L/K}$ the sets of indices at jumps of the lower and
upper ramification groups of $G_{L/K}$ respectively. For $i\geq-1$, Let
$G_{L/K,i},{G_{L/K}}^{i}$ denote the $i$-th lower and upper ramification group
of $G_{L/K}$ respectively. Define the Herbrand function $\psi_{L/K}$ as in
[7], IV, §3.

######  Lemma 2.4.

Let $K$ be a complete discrete valuation field and $M/K$ a finite Galois
extension. Let $L/K$ be a Galois subextension of $M/K$. Let $G=G_{M/K}$,
$H=G_{M/L}$, and $\psi=\psi_{L/K}$. Then we have $G^{i}\cap H=H^{\psi(i)}$ for
all $i\geq-1$.

###### Proof.

By definitions of the lower and upper ramification groups, we have
$G^{i}=G_{\psi_{M/K}(i)}$ and $H^{\psi(i)}=H_{\psi_{M/L}\circ\psi(i)}$. By
[7], IV, §3, Proposition 15, we have $\psi_{M/K}=\psi_{M/L}\circ\psi$. By [7],
IV, §1, Proposition 2, $G_{\psi_{M/K}(i)}\cap H$ coincides with
$H_{\psi_{M/K}(i)}$. ∎

######  Proposition 2.5.

Let $L/K$ be a ramified Artin-Schreier extension defined by $P(x)=a\;(a\in
K-P(K))$. Let $m_{a}>0$ be the smallest integer such that the image of $a$ is
in $F_{m_{a}}H^{1}(K)$.

  1. (a)

The valuation of the different of $L/K$ is $(m_{a}+1)(p-1)$.

  2. (b)

We have $U_{L/K}=\\{m_{a}\\}$.

  3. (c)

We have

$\psi_{L/K}(i)=\max(i,pi-(p-1)m_{a}).$ (2.8)

  4. (d)

For an integer $n$ such that $p\nmid n$, let $n^{\prime}=pn-(p-1)m_{a}$ and
$n^{\prime\prime}=\max(n,n^{\prime})=\psi_{L/K}(n)$. The canonical morphism
$H^{1}(K)\to H^{1}(L)$ induces $F_{n}H^{1}(K)\to F_{n^{\prime\prime}}H^{1}(L)$
and
$\operatorname{Gr}_{n}H^{1}(K)\to\operatorname{Gr}_{n^{\prime\prime}}H^{1}(L)$.
Define $\theta$ as in Lemma 2.2. Then the $\mathbb{F}_{p}$-linear map

$\begin{array}[]{cccc}u:&\operatorname{Gr}_{pn}L&\to&\operatorname{Gr}_{n^{\prime\prime}}L\\\
&x&\mapsto&\left\\{\begin{array}[]{ll}\sqrt[p]{x}&(n<m_{a})\\\
\sqrt[p]{x}+\theta x&(n=m_{a})\\\ \frac{n}{m_{a}}\theta
x&(n>m_{a})\end{array}\right.\end{array}$ (2.9)

makes the diagram below commutative, where $\nu$ is defined as in Lemma 2.3

$\textstyle{\operatorname{Gr}_{n}K\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{\operatorname{Gr}_{pn}L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\textstyle{\operatorname{Gr}_{n^{\prime\prime}}L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{\operatorname{Gr}_{n}H^{1}(K)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Gr}_{n^{\prime\prime}}H^{1}(L).}$
(2.10)

If $n\neq m_{a}$, then the map $u$ is an isomorphism. If $n=m_{a}$, then the
kernel of $u$ is generated by the image of $a$.

###### Proof.

(a) The claim follows from the beginning (p. 42) of Section b) of [5].

(b) By (a) and [7], IV, §1, Proposition 4, we have $U_{L/K}=\\{m_{a}\\}$.

(c) This follows from the definition of $\psi_{L/K}$.

(d) Define $\mu$ as in Lemma 2.1(a). Take $y\in K$ such that $v_{K}(y)=-n$.
Let $\eta$ be an element of $K_{s}$ such that $P(\eta)=y$. When $y\notin
P(L)$, applying Lemma 2.4 to $L(\eta)/K$ we get $y\in
F_{n^{\prime\prime}}H^{1}(L)$ by (b). When $y\in P(L)$, we have $y\equiv 0\in
H^{1}(L)$. Thus $y\in F_{n^{\prime\prime}}H^{1}(L)$ also in this case. Thus
the canonical morphism $H^{1}(K)\to H^{1}(L)$ induces $F_{n}H^{1}(L)\to
F_{n^{\prime\prime}}H^{1}(L)$ and
$\operatorname{Gr}_{n}H^{1}(K)\to\operatorname{Gr}_{n^{\prime\prime}}H^{1}(L)$.

Let $\bar{y}$ denote the image of $y$ in $\operatorname{Gr}_{n}K$. By (a) and
Lemma 2.2, we have $-v_{L}(dy)=pn-(p-1)m_{a}=n^{\prime}$. Since $k$ is
perfect, there exists $s\in L$ such that $v_{L}(y-s^{p})>-np$. We have
$v_{L}(s)=-n$. Let ${s^{\prime}}=y-s^{p}$. Since $dy\neq 0$ in
$\Omega_{L}^{1}$, we may assume $p\nmid v_{L}({s^{\prime}})$. Then we have
$v_{L}({s^{\prime}})=v_{L}(d{s^{\prime}})$. Since $d{s^{\prime}}=dy$, we have
$v_{L}({s^{\prime}})=-n^{\prime}$.

We have ${s^{\prime}}+s=y-P(s)\equiv y\in H^{1}(L)$. We will now write the
images of ${s^{\prime}}$ and $s$ in $\operatorname{Gr}_{n^{\prime}}L$ and
$\operatorname{Gr}_{n}L$ respectively in terms of $\bar{y}$. By Lemma 2.1(b)
and Lemma 2.2, the image of $d{s^{\prime}}=dy$ in
$\operatorname{Gr}_{n^{\prime}}L$ equals

$-n^{\prime}\mu({s^{\prime}})=\mu(-n\theta\bar{y}).$ (2.11)

Therefore, the image of ${s^{\prime}}$ equals
$\displaystyle{\frac{n}{n^{\prime}}\theta\bar{y}}$ in
$\operatorname{Gr}_{n^{\prime}}L$. By definition, the image of $s$ in
$\operatorname{Gr}_{n}L$ is $\sqrt[p]{\bar{y}}$. Thus the image of
${s^{\prime}}+s$ equals

$\left\\{\begin{array}[]{ll}\sqrt[p]{\bar{y}}&(n<m_{a})\\\
\sqrt[p]{\bar{y}}+\theta\bar{y}&(n=m_{a})\\\
\frac{n}{n^{\prime}}\theta\bar{y}&(n>m_{a})\end{array}\right.$ (2.12)

in $\operatorname{Gr}_{n^{\prime\prime}}L$. Since $n^{\prime}\equiv m_{a}\mod
p$, the diagram (2.10) is commutative.

If $n<m_{a}$, then the map $u$ is an isomorphism, since $k$ is perfect.

If $n>m_{a}$, then the map $u$ is an isomorphism, since $n$ is prime to $p$.

If $n=m_{a}$, then the kernel of $u$ is generated by
$(-\theta)^{-\frac{p}{p-1}}$. On the other hand, the image of $a$ in
$H^{1}(L)$ equals 0. Thus, there exists $i\in\mathbb{F}_{p}^{\times}$ such
that $a\equiv i(-\theta)^{-\frac{p}{p-1}}$ in $\operatorname{Gr}_{pm_{a}}L$,
and the kernel of $u$ is generated by the class of $a$. ∎

##  3 Calculation of the Ramification Groups

Let $2\leq n\leq p$. Recall the algebraic group $G\subset GL(n+1)$ over
$\mathbb{F}_{p}$ and its lower central series

$G\supsetneq Z_{1}G\supsetneq\cdots\supsetneq Z_{n}G=\\{1\\}$ (3.1)

of Definition 1.6(b).

Let $K$ be a complete discrete valuation field, and $K_{s}$ a separable
closure of $K$. Define $P:K_{s}\to K_{s}$ by $x\mapsto x^{p}-x$. Assume that
the residue field $k$ of $K$ is perfect of characteristic $p>0$. Take
$K\subset M_{1}\subset\cdots\subset M_{n}$ as in Definition 1.8. Assume that
$M_{n}/K$ is totally ramified. Recall that $L_{E/K},U_{E/K}$ denote the sets
of indices at jumps of the lower and upper ramification groups respectively of
the Galois group $G_{E/K}$ for a field extension $E/K$. Then we have

$U_{M_{1}/K}\subset\cdots\subset U_{M_{n}/K}$ (3.2)

by [7], IV, §3, Proposition 14.

######  Lemma 3.1.

There exists a sequence $r_{1}<\cdots<r_{n}$ of rational numbers such that for
all $q>r_{1}$, we have

$G_{M_{n}/K}^{q}=G_{M_{n}/M_{j}}\simeq Z_{j}G(\mathbb{F}_{p}),$ (3.3)

where $j$ is the largest integer satisfying $q>r_{j}$. Moreover, for $1\leq
j\leq n$, $r_{j}$ is the largest element of $U_{M_{j}/K}$.

###### Proof.

We prove this lemma by descending induction on $j$.

Suppose $j=n$. Let $r_{n}$ be the largest element of $U_{M_{n}/K}$. Then we
have (3.3) for $q>r_{n}$.

Suppose $1\leq j\leq n-1$. Assume that we have (3.3) for $q>r_{j+1}$. Let
$l,r_{j}$ be the largest elements of $L_{M_{j+1}/K},U_{M_{j}/K}$ respectively.
Then we have $G_{M_{n}/K}^{q}\subset G_{M_{n}/M_{j}}$ for $q>r_{j}$ by [7],
IV, §3, Proposition 14. It suffices to show
$G_{M_{j+1}/K}^{r_{j+1}}=G_{M_{j+1}/K,l}=Z(G_{M_{j+1}/K})$, since
$Z(G_{M_{j+1}/K})=G_{M_{n}/M_{j}}/G_{M_{n}/M_{j+1}}$. By [7], IV, §2,
Proposition 10, we have

$[G_{M_{j+1}/K},G_{M_{j+1}/K,l}]\subset G_{M_{j+1}/K,l+1}=\\{1\\},$ (3.4)

since $l$ is the largest element of $L_{M_{j+1}/K}$. Thus,
$G_{M_{j+1}/K,l}\subset Z(G_{M_{j+1}/K})$. Since $G_{M_{j+1}/K,l}$ is not
trivial and $Z(G_{M_{j+1}/K})$ is isomorphic to $\mathbb{F}_{p}$, we have
$G_{M_{j+1}/K,l}=Z(G_{M_{j+1}/K})$. ∎

Recall $A\in
M(n),\;R(t,x)\in\mathbb{F}_{p}[t,x],\;\mathbf{A}:\mathbb{G}_{a}\to GL(n)$ of
Definition 1.6(a).

######  Lemma 3.2.

Assume that $K$ is of characteristic $p>0$ and $M_{n}/K$ is totally ramified.
There exist $a\in K,\;b\in K^{n}$ satisfying the conditions of Lemma 1.9(a)
and conditions (i)–(iii) below. Let

$b=\begin{pmatrix}b_{n}\\\ \vdots\\\ b_{1}\end{pmatrix},$ (3.5)

and let $m_{a},m_{j}\;(1\leq j\leq n)$ denote $-v_{K}(a),-v_{K}(b_{j})$
respectively.

  1. (i)

$m_{a},m_{1}$ are positive and prime to $p$.

  2. (ii)

For all $2\leq j\leq n$, we have $p\nmid m_{j}$ if $m_{j}>0$.

  3. (iii)

If $n\leq p-1$ and $m_{a}=m_{1}$, then the images $\bar{a},\bar{b}_{1}$ of
$a,b_{1}$ in $\operatorname{Gr}_{m_{a}}K=\operatorname{Gr}_{m_{1}}K$
respectively are linearly independent over $\mathbb{F}_{p}$.

###### Proof.

Take $a\in K,\;b\in K^{n}$ as in Lemma 1.9(a). By Proposition 1.1(c), we may
replace

$\begin{pmatrix}\mathbf{A}(a)&b\\\ 0&1\end{pmatrix}$

by

$\begin{pmatrix}\mathbf{A}(s^{p})&F(t)\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(a)&b\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(s)&t\\\ 0&1\end{pmatrix}^{-1}$
$=\begin{pmatrix}\mathbf{A}(a+P(s))&\mathbf{A}(s^{p})b-\mathbf{A}(a+P(s))t+F(t)\\\
0&1\end{pmatrix}$ (3.6)

for $s\in K,\;t\in{M_{n}}^{n}$, if
$\mathbf{A}(s^{p})b-\mathbf{A}(a+P(s))t+F(t)\in K^{n}$.

When $n\leq p-1$, let

$S(t,x)=\frac{R(t,x)-1}{t}=\sum_{i=0}^{p-2}\binom{x}{i+1}t^{i}\in\mathbb{F}_{p}[t,x],$
(3.7)

and define a morphism $\mathbf{v}:\mathbb{G}_{a}\to{\mathbb{G}_{a}}^{n}$ of
algebraic varieties by

$\mathbf{v}(x)=S(A,x)\begin{pmatrix}0\\\ \vdots\\\ 0\\\
1\end{pmatrix}\in{\mathbb{G}_{a}}^{n}.$ (3.8)

Then for any $x,y\in\mathbb{G}_{a}$, the morphism $\mathbf{v}$ satisfies

$\mathbf{v}(x+y)=\mathbf{v}(x)+\mathbf{A}(x)\mathbf{v}(y).$ (3.9)

In particular, we have

$\mathbf{v}(x^{p})-\mathbf{A}(P(x))\mathbf{v}(x)=F(\mathbf{v}(x))-\mathbf{A}(P(x))\mathbf{v}(x)=\mathbf{v}(P(x)).$
(3.10)

We have the following fact.

  1. Fact (*)

For any

$t=\begin{pmatrix}t_{n}\\\ \vdots\\\ t_{1}\end{pmatrix}\in K^{n},$ (3.11)

the first component from the bottom of $(1-\mathbf{A}(a))t$ is $0$, and the
$j$-th component from the bottom of $(1-\mathbf{A}(a))t$ depends only on
$t_{1},\ldots,t_{j-1}$ for all $2\leq j\leq n$.

We may define operations (I), (II) and (III) as follows.

  1. (I)

Choose appropriate $s\in K$, set $t=0$, and replace $(a,b)$ by
$(a+P(s),\mathbf{A}({s}^{p})b)$, so that we have $p\nmid m_{a}$ if $m_{a}>0$.
We know by Lemma 2.3 that such $s$ exists.

  2. (II)

Choose appropriate $t_{j}\in K\;(1\leq j\leq n)$ successively, set $s=0$, and
replace $(a,b)$ by $(a,b+(1-\mathbf{A}(a))t+(F(t)-t))$, so that we have
$p\nmid m_{j}$ if $m_{j}>0$ for all $1\leq j\leq n$. We know by Lemma 2.3 and
Fact (*) that such $t_{j}\;(1\leq j\leq n)$ exist.

  3. (III)

Assume $n\leq p-1$, $m_{a}=m_{1}$, and
$i=\bar{a}^{-1}\bar{b}_{1}\in\mathbb{F}_{p}$. Take $\alpha\in M_{n}$ such that
$\alpha^{p}-\alpha=a$ as in Lemma 1.9(b). Set $s=0,\;t=i\mathbf{v}(\alpha)$,
and replace $(a,b)$ by $(a,b-i\mathbf{v}(a))$, so that we have $m_{a}>m_{1}$.
This is valid since we have

$-\mathbf{A}(a)\mathbf{v}(\alpha)+F(\mathbf{v}(\alpha))=\mathbf{v}(a)$ (3.12)

by (3.10).

Define $\alpha,\gamma_{j}\in M_{n}\;(1\leq j\leq n)$ as in Lemma 1.9(b). Since
$M_{n}$ is totally ramified, $K(\alpha),K(\gamma_{1})$ are ramified. By
Proposition 2.5(b), we have $m_{a},m_{1}>0$.

Performing (I) and (II) successively, conditions (i) and (ii) are satisfied.

Suppose $n\leq p-1$, $m_{a}=m_{1}$ and
$\bar{a}^{-1}\bar{b}_{1}\in\mathbb{F}_{p}$. Performing (III), condition (iii)
is satisfied. However, since (III) may change $m_{j}\;(1\leq j\leq n)$, we
have to perform (II) again to ensure conditions (i) and (ii) are satisfied.
Since (II) does not make $m_{1}$ larger, condition (iii) remains satisfied. ∎

Take $a\in K,\;b\in K^{n}$ satisfying the conditions of Lemma 3.2. Define
$r_{j}\;(1\leq j\leq n)$ as in Lemma 3.1.

Let

$\omega=\begin{pmatrix}{\omega_{n}}\\\ \vdots\\\
{\omega_{1}}\end{pmatrix}=\mathbf{A}(-a)db\in(\Omega_{K}^{1})^{n},$ (3.13)

where $db$ denotes the component-wise derivation.

We will now state our main theorem.

######  Theorem 3.3.

Let $K$ be a complete discrete valuation field of equal characteristics $p>0$.
Assume that the residue field $k$ of $K$ is perfect. Let $v_{K}$ denote the
valuation of $K$ and $\Omega_{K}^{1}$ defined at the beginning of Section 2.
Take $K\subset M_{1}\subset\cdots\subset M_{n}$ as in Definition 1.8. Assume
that $M_{n}/K$ is totally ramified. Take $a\in K,\;b\in K^{n}$ satisfying the
conditions of Lemma 3.2. Define $b_{1},\ldots,b_{n}\in K$ as in (3.5). Let
$m_{a}=-v_{K}(a),\;m_{j}=-v_{K}(b_{j})\;(1\leq j\leq n)$. Define the sequence
$m_{a}\leq r_{1}<\cdots<r_{n}$ as in Lemma 3.1, and $\omega_{j}\;(1\leq j\leq
n)$ as in (3.13). Then we have

$r_{j}=\max\left(\max_{1\leq i\leq
j}\left(\frac{j-i}{p}m_{a}-v_{K}\left(\omega_{i}\right)\right),\frac{(j+p-2)m_{a}+m_{1}}{p}\right)$
(3.14)

for all $2\leq j\leq n$.

We will prove this theorem at the last of this paper.

Define $\alpha,\gamma_{j}\;(1\leq j\leq n)$ as in Lemma 1.9(b). Let

$c=\begin{pmatrix}c_{n}\\\ \vdots\\\
c_{1}\end{pmatrix}=\mathbf{A}(-\alpha^{p})b,$ (3.15)

and let $m^{\prime}_{j}\;(2\leq j\leq n)$ denote the smallest integer such
that $c_{j}\in L$ defines an element of $F_{m^{\prime}_{j}}H^{1}(L)$. Then we
have $P(\gamma_{j})=c_{j}$ and $M_{j}=K(\alpha,\gamma_{1},\ldots,\gamma_{j})$
for all $1\leq j\leq n$ by Lemma 1.9(b).

######  Lemma 3.4.

  1. (a)

Let $2\leq j\leq n$. We have

$-v_{L}(c_{j})\leq p\max_{1\leq i\leq j}((j-i)m_{a}+m_{i})$ (3.16)

and

$-v_{L}(dc_{j})=\max_{1\leq i\leq j}((j-i)m_{a}-v_{L}(\omega_{i})).$ (3.17)

Furthermore, there exists a unique $1\leq i\leq j$ satisfying

$-v_{L}(dc_{j})=(j-i)m_{a}-v_{L}(\omega_{i}).$ (3.18)

For this unique $i$, the image of $dc_{j}$ in
$\operatorname{Gr}_{-v_{L}(dc_{j})}\Omega_{L}^{1}$ equals that of
$\displaystyle{\binom{-\alpha}{j-i}\omega_{i}}$.

  2. (b)

We have $p\nmid m^{\prime}_{j}=\psi_{L/K}(r_{j})$ for $2\leq j\leq n$, and
$m_{a}<m^{\prime}_{2}<\cdots<m^{\prime}_{n}$.

###### Proof.

(a) Since $d(a+\alpha)=d(\alpha^{p})=0$ in $\Omega_{L}^{1}$, we have

$c=\mathbf{A}(-a-\alpha)b,\;dc=\mathbf{A}(-a-\alpha)db=\mathbf{A}(-\alpha)\omega.$
(3.19)

Thus we have

$c_{j}=\sum_{i=1}^{j}\binom{-a-\alpha}{j-i}b_{i}$ (3.20)

and

$dc_{j}=\sum_{i=1}^{j}\binom{-\alpha}{j-i}\omega_{i}.$ (3.21)

We get (3.16) from (3.20). By Lemma 2.2 and Proposition 2.5(a), we have

$-v_{L}\left(\binom{-\alpha}{j-i}\omega_{i}\right)=(j-i)m_{a}-v_{L}(\omega_{i})$
$=(j-i)m_{a}-pv_{K}(\omega_{i})-(p-1)m_{a}\equiv(j-i+1)m_{a}\mod p.$ (3.22)

Therefore, the valuations of the terms in the right-hand side of the equation
(3.21) do not coincide with each other. Thus we have (3.17) and the rest of
the claim.

(b) By Lemma 2.4, we have ${G_{M_{j}/L}}^{\psi_{L/K}(i)}={G_{M_{j}/K}}^{i}\cap
G_{M_{j}/L}$ for all $i\geq-1$. Thus by Lemma 3.1,
$U_{M_{j}/L}=\\{\psi_{L/K}(r_{j})\\}\sqcup U_{M_{j-1}/L}$. Note that
$\psi_{L/K}(r_{j})$ is larger than all of the elements of $U_{M_{j-1}/L}$ and
that $M_{j}=M_{j-1}(\gamma_{j})$. Hence $m^{\prime}_{j}=\psi_{L/K}(r_{j})$. We
get $r_{1}\geq m_{a}$ from $U_{M_{1}/K}\supset U_{L/K}=\\{m_{a}\\}$ and Lemma
3.1. Since we have $r_{1}<r_{2}<\cdots<r_{n}$ by Lemma 3.1, we have
$m_{a}<m^{\prime}_{2}<\cdots<m^{\prime}_{n}$. Since $c_{j}\notin P(L)$, by
definition of $c^{\prime}_{j}$ (3.23), we have $p\nmid m^{\prime}_{j}>0$. ∎

Take $c^{\prime}_{j},c^{\prime\prime}_{j}\in L\;(2\leq j\leq n)$ such that

$c_{j}-P(c^{\prime\prime}_{j})=c^{\prime}_{j},\;-v_{L}(c^{\prime}_{j})=m^{\prime}_{j}.$
(3.23)

Then we have $dc_{j}+dc^{\prime\prime}_{j}=dc^{\prime}_{j}$, since
$d({c^{\prime\prime}_{j}}^{p})=0$.

######  Lemma 3.5.

We have

$-v_{L}(dc^{\prime\prime}_{j})\leq\max_{1\leq i\leq j}((j-i)m_{a}+m_{i}).$
(3.24)

###### Proof.

By definition of $c^{\prime\prime}_{j}$ (3.23), we have

$-v_{L}(dc^{\prime\prime}_{j})\leq-
v_{L}(c^{\prime\prime}_{j})\leq-\frac{1}{p}v_{L}(c_{j}).$ (3.25)

By (3.16), we have (3.24). ∎

Let $\bar{a},\bar{b}_{j}\;(1\leq j\leq n)$ denote the image of $a,b_{j}$ in
$\operatorname{Gr}_{m_{a}}K,\operatorname{Gr}_{m_{j}}K$ respectively, when
$m_{j}\neq-\infty$.

######  Proposition 3.6.

Suppose we have

$-v_{L}(dc_{n})\leq\max_{1\leq i\leq n}((n-i)m_{a}+m_{i}).$ (3.26)

  1. (a)

For all $2\leq j\leq n$, We have

$-v_{K}(\omega_{j})<(j-1)m_{a}+m_{1}$ (3.27)

  2. (b)

For all $1\leq j\leq n$, we have

$p\nmid m_{j}=(j-1)m_{a}+m_{1}>0,$ (3.28)

and

$\bar{b}_{j}=\frac{m_{1}}{(j-1)!m_{j}}\bar{a}^{j-1}\bar{b}_{1}.$ (3.29)

  3. (c)

We have $n\leq p-1$.

  4. (d)

For all $1\leq j\leq n$, we have

$-v_{L}(c_{j})=pm_{j}=p(j-1)m_{a}+pm_{1},$ (3.30)

and the image of $c_{j}$ in $\operatorname{Gr}_{pm_{j}}L$ equals
$\displaystyle{\frac{(-m_{a}\bar{a})^{j-1}\bar{b}_{1}}{\prod_{i=2}^{j}m_{i}}\neq
0}$.

  5. (e)

For all $1\leq j\leq n$, we have

$-v_{L}(c^{\prime\prime}_{j})=-v_{L}(dc^{\prime\prime}_{j})=m_{j}=(j-1)m_{a}+m_{1}$
(3.31)

and the image of $c^{\prime\prime}_{j}$ in $\operatorname{Gr}_{m_{j}}L$ equals
$\displaystyle{\frac{\left(-m_{a}\sqrt[p]{\bar{a}}\right)^{j-1}\sqrt[p]{\bar{b}_{1}}}{\prod_{i=2}^{j}m_{i}}}$.

###### Proof.

(a) By Lemma 2.2 and Proposition 2.5(a), we have
$-v_{L}(\chi)=-pv_{K}(\chi)-(p-1)m_{a}$ for all $\chi\in\Omega_{K}^{1}$. Hence
it suffices to show that for all $2\leq j\leq n$, we have

$-v_{L}(\omega_{j})<p((j-1)m_{a}+m_{1})-(p-1)m_{a}.$ (3.32)

We will apply Lemma 1.10 to

$\lambda=m_{a},\;\lambda_{j}=\max_{1\leq i\leq
j}((j-i)m_{a}+m_{i}),\;l_{j}=-v_{L}(\omega_{j}).$ (3.33)

By definition of $\omega$ (3.13), we have

$\omega_{j}=\sum_{i=1}^{j}\binom{-a}{j-i}db_{j}.$ (3.34)

We have

$-v_{L}\left(\binom{-a}{j-i}db_{j}\right)=p((j-i)m_{a}+m_{i})-(p-1)m_{a}$
(3.35)

by Lemma 2.2 and Proposition 2.5(a). By (3.34) and the property of valuation,
condition (i) of Lemma 1.10 is satisfied.

By (3.17) and (3.26), we get

$(n-j)m_{a}-v_{L}(\omega_{j})\leq\max_{1\leq i\leq n}((n-i)m_{a}+m_{i}).$
(3.36)

Thus condition (ii) of Lemma 1.10 is satisfied.

Since $m_{a},m_{1}>0$, we have

$m_{a}<(j-1)m_{a}+m_{1}\leq\max_{1\leq i\leq j}((j-i)m_{a}+m_{i})$ (3.37)

for all $2\leq j\leq n$. Thus condition (iii) of Lemma 1.10 is satisfied.

By applying Lemma 1.10, we get (3.32) for all $2\leq j\leq n$.

(b) Since $db=\mathbf{A}(a)\omega$, we have

$-v_{K}(db_{j})=m_{j}\leq\max_{1\leq i\leq j}((j-i)m_{a}-v_{K}(\omega_{i})).$
(3.38)

By (3.27) for $2\leq i\leq j$, we have

$(j-i)m_{a}-v_{K}(\omega_{i})<(j-1)m_{a}+m_{1}$ (3.39)

for $2\leq i\leq j$. On the other hand, we have $\omega_{1}=db_{1}$. Thus, we
have

$(j-1)m_{a}-v_{K}(\omega_{1})=(j-1)m_{a}+m_{1}.$ (3.40)

Thus we have

$-v_{K}(db_{j})=m_{j}=\max_{1\leq i\leq
j}((j-i)m_{a}-v_{K}(\omega_{i}))=(j-1)m_{a}+m_{1}>0$ (3.41)

by the property of valuation. By conditions (i) and (ii) of Lemma 3.2, we have
(3.28) for all $1\leq j\leq n$. The image of $db_{j}$ equals the image of
$\displaystyle{\binom{a}{j-1}db_{1}}$ in
$\operatorname{Gr}_{m_{j}}\Omega_{K}^{1}$. Hence we have (3.29).

(c) By (b), we have (3.28) for all $2\leq j\leq n$. Since
$(j-1)m_{a}+m_{1}\mod p\;(1\leq j\leq n)$ are different from each other and
$m_{1},\ldots,m_{n}$ are all prime to $p$ by (3.28), we have $n\leq p-1$.

(d) By (3.16) and (3.28), we have $-v_{L}(c_{j})\leq pm_{j}$. By (3.20), Lemma
1.11 and (3.29), the image of $c_{j}$ in $\operatorname{Gr}_{pm_{j}}L$ equals

$\sum_{i=1}^{j}\frac{(-\bar{a})^{j-i}}{(j-i)!}\bar{b}_{i}=\sum_{i=1}^{j}\frac{(-\bar{a})^{j-i}m_{1}}{(j-i)!(i-1)!m_{i}}\bar{a}^{i-1}\bar{b}_{1}=\frac{(-m_{a}\bar{a})^{j-1}\bar{b}_{1}}{\prod_{i=2}^{j}m_{i}}\neq
0.$ (3.42)

Thus we have (3.30).

(e) Since we have (3.30) and $m_{j}$ is prime to $p$ by (3.28), we have
(3.31). Hence, the image of $c^{\prime\prime}_{j}$ in
$\operatorname{Gr}_{m_{j}}L$ equals
$\displaystyle{\frac{\left(-m_{a}\sqrt[p]{\bar{a}}\right)^{j-1}\sqrt[p]{\bar{b}_{1}}}{\prod_{i=2}^{j}m_{i}}}$,
since we have $\sqrt{x}=x$ for $x\in\mathbb{F}_{p}$. ∎

We will now express $\max(-v_{L}(dc_{n}),-v_{L}(dc^{\prime\prime}_{n}))$ in
terms of $a,b$.

######  Corollary 3.7.

  1. (a)

The following conditions are equivalent:

    1. (i)

$-v_{L}(dc_{n})<-v_{L}(dc^{\prime\prime}_{n})$.

    2. (ii)

$-v_{L}(dc_{n})<\max_{1\leq j\leq n}((n-j)m_{a}+m_{j})$.

    3. (iii)

$-v_{L}(dc_{n})<(n-1)m_{a}+m_{1}$.

  2. (b)

We have

$m^{\prime}_{n}=-v_{L}(dc^{\prime}_{n})\leq\max(-v_{L}(dc_{n}),-v_{L}(dc^{\prime\prime}_{n}))$
$=\max\left(\max_{1\leq i\leq
n}\left((n-i-p+1)m_{a}-pv_{K}\left(\omega_{i}\right)\right),(n-1)m_{a}+m_{1}\right).$
(3.43)

###### Proof.

(a) $\text{(\ref{ra-c-1})}\Rightarrow\text{(\ref{ra-c-2})}$: This follows from
(3.24).

$\text{(\ref{ra-c-3})}\Rightarrow\text{(\ref{ra-c-2})}$: This clearly holds.

$\text{(\ref{ra-c-2})}\Rightarrow\text{(\ref{ra-c-1}),(\ref{ra-c-3})}$: This
follows from (3.31).

(b) Since $dc_{n}+dc^{\prime\prime}_{n}=dc^{\prime}_{n}$, we have
$m^{\prime}_{n}\leq\max(-v_{L}(dc_{n}),-v_{L}(dc^{\prime\prime}_{n}))$. By
(a), (3.17), and (3.31), we have (3.43). ∎

We will now prove our main theorem.

###### Proof of Theorem 3.3.

It suffices to show the case where $j=n$, since the case where $j<n-1$ can be
reduced to the case $j=n$ by replacing $n$ by $j$.

Let

$s_{n}=\psi_{L/K}\left(\max\left(\max_{1\leq i\leq
n}\left(\frac{n-i}{p}m_{a}-v_{K}\left(\omega_{i}\right)\right),\frac{(n+p-2)m_{a}+m_{1}}{p}\right)\right).$
(3.44)

It suffices to show $m^{\prime}_{n}=\psi_{L/K}(r_{n})=s_{n}$ to complete the
proof, since $\psi_{L/K}$ is injective by Proposition 2.5(c).

Since $\displaystyle{\frac{(n+p-2)m_{a}+m_{1}}{p}>m_{a}}$, we have

$\max\left(\max_{1\leq i\leq
n}\left(\frac{n-i}{p}m_{a}-v_{K}\left(\omega_{i}\right)\right),\frac{(n+p-2)m_{a}+m_{1}}{p}\right)>m_{a}.$
(3.45)

By Proposition 2.5(c), we have $\psi_{L/K}(x)=px-(p-1)m_{a}$ if $x\geq m_{a}$.
Then by Corollary 3.7(b), we get

$s_{n}=\max(-v_{L}(dc_{n}),\;-v_{L}(dc^{\prime\prime}_{n}))\geq
m^{\prime}_{n}.$ (3.46)

We have $m^{\prime}_{n}=s_{n}$ when $-v_{L}(dc_{n})\neq-
v_{L}(dc^{\prime\prime}_{n})$. It suffices to show that we have
$m^{\prime}_{n}=s_{n}$ also when
$-v_{L}(dc_{n})=-v_{L}(dc^{\prime\prime}_{n})$.

Assume $-v_{L}(dc_{n})=-v_{L}(dc^{\prime\prime}_{n})$. Then by (3.24), the
hypothesis (3.26) of Proposition 3.6 is satisfied. By Proposition 3.6(c), we
have $n\leq p-1$. By Lemma 2.2 and Proposition 2.5(a), we have
$-v_{L}(\omega_{i})=-pv_{K}(\omega_{i})-(p-1)m_{a}\equiv m_{a}\mod p$ for all
$1\leq i\leq n$. Thus, by Lemma 3.4(a), there exists a unique integer $1\leq
j\leq n$ satisfying
$-v_{L}(dc_{n})=(n-j)m_{a}-v_{L}(\omega_{j})\equiv(n-j+1)m_{a}\mod p$.
Combining with (3.31), we get

$v_{L}(dc_{n})-v_{L}(dc^{\prime\prime}_{n})=0\equiv(j-2)m_{a}+m_{1}\mod p.$
(3.47)

Since $m_{i}=(i-1)m_{a}+m_{1}$ is prime to $p$ for all $1\leq i\leq n$ by
(3.28), we have $j\neq 2,\ldots,n+1$. Therefore, since $j$ is an integer
satisfying $1\leq j\leq n$, we have $j=1$. Hence, we have

$m_{n}=(n-1)m_{a}+m_{1}=-v_{L}(dc^{\prime\prime}_{n})=-v_{L}(dc_{n})=(n-1)m_{a}-v_{L}(\omega_{1}).$
(3.48)

Therefore, we have $m_{1}=-v_{L}(\omega_{1})$. By definition, we have
$\omega_{1}=db_{1}$. Thus we have $m_{1}=pm_{1}-(p-1)m_{a}$ by Lemma 2.2 and
Proposition 2.5(a). Hence we have $m_{a}=m_{1}$. By (3.28) we get
$m_{j}=jm_{a}$ for all $1\leq j\leq n$. Hence we have
$s_{n}=-v_{L}(dc_{n})=-v_{L}(dc^{\prime\prime}_{n})=m_{n}$.

We have only to check that the image of
$dc_{n}+dc^{\prime\prime}_{n}=dc^{\prime}_{n}$ in
$\operatorname{Gr}_{m_{n}}\Omega_{L}^{1}$ does not vanish. Define the
$k$-linear isomorphism
$\mu:\operatorname{Gr}L\to\operatorname{Gr}\Omega_{L}^{1}$ as in Lemma 2.1(a)
and $\theta\in\operatorname{Gr}_{-(p-1)m_{a}}L$ as in Lemma 2.2. Since we have
(3.48), the image of $dc_{n}$ in $\operatorname{Gr}_{m_{n}}\Omega_{L}^{1}$
equals that of $\displaystyle{\binom{-\alpha}{n-1}\omega_{1}}$ by Lemma
3.4(a). Thus, by Lemma 2.1(b), the image of $dc_{n}$ in
$\operatorname{Gr}_{m_{n}}\Omega_{L}^{1}$ equals
$\displaystyle{-m_{1}\frac{(-\sqrt[p]{\bar{a}})^{n-1}}{(n-1)!}\mu(\theta\bar{b}_{1})}$.

Meanwhile, by Proposition 3.6(e), Lemma 2.1(b), and $m_{j}=jm_{a}=jm_{1}$ for
all $1\leq j\leq n$, the image of $dc^{\prime\prime}_{n}$ in
$\operatorname{Gr}_{m_{n}}\Omega_{L}^{1}$ equals
$\displaystyle{-m_{1}\frac{(-\sqrt[p]{\bar{a}})^{n-1}}{(n-1)!}\mu\left(\sqrt[p]{\bar{b}_{1}}\right)}$.

Define $u:\operatorname{Gr}_{pm_{a}}L\to\operatorname{Gr}_{m_{a}}L$ as in
Proposition 2.5(d). Then the image of
$dc_{n}+dc^{\prime\prime}_{n}=dc^{\prime}_{n}$ in
$\operatorname{Gr}_{m_{n}}\Omega_{L}^{1}$ equals
$\displaystyle{-m_{1}\frac{(-\sqrt[p]{\bar{a}})^{n-1}}{(n-1)!}}$ times

$\mu\left(\theta\bar{b}_{1}+\sqrt[p]{\bar{b}_{1}}\right)=\mu(u(\bar{b}_{1}))$
(3.49)

by Proposition 2.5(d). Since $n\leq p-1$, by condition (iii) of Lemma 3.2 and
Proposition 2.5(d), we have $u(\bar{b}_{1})\neq 0$. Thus the image of
$dc^{\prime}_{n}$ in $\operatorname{Gr}_{m_{n}}L$ does not vanish. ∎

######  Example 3.8.

  1. 1.

We give an example for $n=2$, the simplest case where Theorem 3.3 can be
applied.

Let $n=2$. Then $G$ is the algebraic group of upper triangular unipotent
matrices of order 3 over $\mathbb{F}_{p}$. Then $M_{2}/K$ is defined by

$\begin{pmatrix}1&x^{p}&{y_{2}}^{p}\\\ 0&1&{y_{1}}^{p}\\\
0&0&1\end{pmatrix}=\begin{pmatrix}1&a&b_{2}\\\ 0&1&b_{1}\\\
0&0&1\end{pmatrix}\begin{pmatrix}1&x&y_{2}\\\ 0&1&y_{1}\\\ 0&0&1\end{pmatrix}$
(3.50)

for some $a,b_{1},b_{2}\in K$ satisfying the conditions of Lemma 3.2. We have

$\omega_{1}=db_{1},\;\omega_{2}=db_{2}-adb_{1}.$ (3.51)

By Theorem 3.3, we have

$U_{M_{2}/K}=U_{M_{1}/K}\cup\\{r_{2}\\}$ (3.52)

where

$r_{2}=\max\left(-v_{K}(db_{2}-adb_{1}),\frac{m_{a}}{p}+m_{1},m_{a}+\frac{m_{1}}{p}\right).$
(3.53)

We can calculate $U_{M_{1}/K}$ by Proposition 2.5(b), since $M_{1}/K$ is an
abelian extension.

  2. 2.

We give an example where the maximum of the right-hand side of (3.14) is
achieved by the first term or the second term, depending on the parameters
$\eta,\eta^{\prime}$.

Let $p>2$, $n=p-1$, $\eta,\eta^{\prime}\in\mathbb{Z}_{\geq 0}$. For $1\leq
j\leq p-1$, define $f_{j}(x)\in\mathbb{Z}\left[\frac{1}{j!},x^{-1}\right]$ as
a polynomial of $x^{-1}$ satisfying

$\frac{df_{j}}{dx}=-\binom{x^{-\eta p-1}}{j-1}x^{-\eta^{\prime}p-2}.$ (3.54)

Let $M_{n}/K$ be the Galois extension defined by

$\begin{pmatrix}\mathbf{A}(x^{p})&F(y)\\\
0&1\end{pmatrix}=\begin{pmatrix}\mathbf{A}(a)&b\\\
0&1\end{pmatrix}\begin{pmatrix}\mathbf{A}(x)&y\\\ 0&1\end{pmatrix},$ (3.55)

with $a=t^{-\eta p-1},\;b_{j}=\epsilon f_{j}(t)\;(1\leq j\leq n)$, where $t$
is a uniformizer of $K$ and $\epsilon\in k-\mathbb{F}_{p}$. These $a,b$
satisfy the conditions of Lemma 3.2. Then by Proposition 2.5(b), we have
$U_{M_{1}/K}=\\{\eta p+1,\eta^{\prime}p+1\\}$.

We have $db=-\epsilon\mathbf{A}(t^{-\eta
p-1})\mathbf{v}(1)t^{-\eta^{\prime}p-2}dt$. Thus
$\omega=-\epsilon\mathbf{v}(1)t^{-2}dt$. Thus we have
$-v_{K}(\omega_{1})=\eta^{\prime}p+1$ and $-v_{K}(\omega_{j})=-\infty$ for
$2\leq j\leq p-1$. Hence, by Theorem 3.3, we have

$r_{j}=\max\left(\frac{(j-1)(\eta
p+1)+p(\eta^{\prime}p+1)}{p},\frac{(j+p-2)(\eta
p+1)+\eta^{\prime}p+1}{p}\right)$
$=(j-1)\left(\eta+\frac{1}{p}\right)+\eta^{\prime}+1+(p-1)\max(\eta,\eta^{\prime}).$
(3.56)

Thus, if we have $\eta\geq\eta^{\prime}$ (resp. $\eta\leq\eta^{\prime}$), then
the maximum of the right-hand side of (3.14) is achieved by the first (resp.
second) term.

## References

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  * [2] Viktor Aleksandrovich Abrashkin  “A ramification filtration of the Galois group of a local field”  In _Proceedings of the St. Petersburg Mathematical Society Volume III American Mathematical Society Translations: Series 2_ , 1995, pp. 35â100  DOI: 10.1090/trans2/166/02
  * [3] Armand Borel  “Linear algebraic groups” 126, Graduate Texts in Mathematics  Springer, 1991
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# Risks of Using Non-verified Open Data: A case study on using Machine
Learning techniques for predicting Pregnancy Outcomes in India

Anusua Trivedi∗, Sumit Mukherjee∗, Edmund Tse∗, Anne Ewing, Juan Lavista
Ferres
Microsoft, Redmond, USA.
∗ These authors contributed equally

###### Abstract

Artificial intelligence (AI) has evolved considerably in the last few years.
While applications of AI is now becoming more common in fields like retail and
marketing, application of AI in solving problems related to developing
countries is still an emerging topic. Specially, AI applications in resource-
poor settings remains relatively nascent. There is a huge scope of AI being
used in such settings. For example, researchers have started exploring AI
applications to reduce poverty and deliver a broad range of critical public
services. However, despite many promising use cases, there are many dataset
related challenges that one has to overcome in such projects. These challenges
often take the form of missing data, incorrectly collected data and improperly
labeled variables, among other factors. As a result, we can often end up using
data that is not representative of the problem we are trying to solve. In this
case study, we explore the challenges of using such an open dataset from
India, to predict an important health outcome. We highlight how the use of AI
without proper understanding of reporting metrics can lead to erroneous
conclusions.

## Introduction

An estimated 2.6 million still births and 2.7 million neo-natal deaths occur
globally each year [8]. A disproportionate percentage (approximately 99%) of
these deaths occur in low-income and middle-income developing countries [9].
India accounts for approximately 19 births/1,000 population (2017 est.) and
39.1 deaths/1,000 live births (2017 est.)[16]. While India has made laudable
progress in reducing overall child mortality over the past 25 years, similar
reductions in neonatal mortality have lagged. As a result, India fell short of
the Millennium Development Goal 4 of a two-thirds reduction in under-five
mortality from 1990 to 2015 [10]. Previous research on evidence-based
approaches to reducing infant mortality have included efforts to increase
access to preventive and curative antenatal, perinatal and postnatal care
[11]. Pregnancy outcomes is a strong indicator of such cares [12].
Unfortunately there is a big divide between the developed and developing world
on this pregnancy outcome. This is not because of a lack of clarity in
defining a pregnancy outcome, but because of an inherent weakness in the
health information and recording systems [17]. According to Indian Council of
Medical Research (ICMR) [13], data on health and demographics in India is
plagued by incomplete information, overestimation, and over-reporting that
lead to hindrance in policy planning. The National Data Quality Forum (NDQF)
[14], formulated by ICMR’s National Institute for Medical Statistics (ICMR -
NIMS), in partnership with Population Council has identified gaps in data
compilation. The NDQF attempted to identify issues in data quality, and found
several factors effecting the data quality:

  * •

lack of comparability

  * •

poor usability of national level data sources

  * •

discordance between system and survey level estimates

  * •

increased questionnaire length

  * •

capturing wrong subsections for questions in questionnaire

  * •

questions on socially restricted conversation topics

  * •

age-reporting errors

  * •

non-response and intentional skipping of questions

  * •

under reporting due to subjective question interpretation

  * •

incompleteness of data

All these factors translate to poor data quality. For our case study we
explore open reporting data sources in India and try to show how use of AI
without understanding these reporting can lead to wrong estimates. unexplored
in our experiment.

## Data

### Pregnancy Outcome Data

For calculating pregnancy outcomes in India, we started exploring two major
data sources: Open Government Data (OGD) platform [2] and Kaggle [3]. OGD is a
platform for supporting Open Data initiative of Government of India. The
portal is used by Govt. of India to publish datasets, documents, services,
tools and applications collected for public use. It intends to increase
transparency in the functioning of Government and also open avenues for many
more innovative uses of Government Data to give different perspective. The
base Open Government Data Platform India is a joint initiative of Government
of India and US Government. Kaggle is a crowd-sourced platform to attract,
nurture, train and challenge data scientists from all around the world to
solve data science, machine learning and predictive analytics problems. Kaggle
enables data scientists and other developers to engage in running machine
learning contests, write and share code, and to host datasets. Kaggle users
have shared over one thousand datasets and is the number one go-to place for
AI practitioners to look for data [4]. For our case study we looked at the
Annual Health Survey (AHS) on Woman Health for 9 states in India [5].
Specifically, we focused on the Woman dataset and WPS dataset in [5]. The
Woman dataset is also represented in Kaggle [6] with acknowledgement to the
source being OGD Platform. The data curator of the Kaggle pregnancy outcome
dataset states - "These nine states, which account for about 48 percent of the
total population, 59 percent of Births, 70 percent of Infant Deaths, 75
percent of Under 5 Deaths and 62 percent of Maternal Deaths in the country,
are the high focus States in view of their relatively higher fertility and
mortality". These numbers did not make any sense to us, so we started to
explore this dataset further.

### Data Gap Analysis

We came across the AHS questionnaire form used by the Census Office in India
[7]. This questionnaire was used to collect the pregnancy outcome dataset in
[5]. Further exploration of this questionnaire form showed us some interesting
facts which helped us identify some data gaps.

Figure 1: Snapshot of the AHS questionnaire for the state of Rajasthan in
India

Figure 1 shows a snapshot of the AHS questionnaire for the state of Rajasthan
in India. In Figure 1, if the answer to the Question 7 is Yes, that answer
leads to Question 8, and if the answer to the question 7 is No, that answer
leads to Question 9. But this is not correctly represented in the Woman
dataset being used by Kaggle in [6]. In [6], the Question 8 data is completely
missing. So the pregnancy outcome column in Kaggle dataset [6] comprises of
just binary pregnancy outcome labels coming from Question 7 (Yes, No), instead
of 4 pregnancy outcome labels coming from Question 8 (Live Birth, Still Birth,
Induced Abortion, Spontaneous Abortion). The pregnancy outcome column in the
WPS dataset on the other hand comprises of 4 pregnancy outcome labels coming
from Question 8 (Live Birth, Still Birth, Induced Abortion, Spontaneous
Abortion). This data gap leads us to believe that the Woman dataset should not
be used to train AI models for pregnancy outcome prediction. That makes the
Kaggle dataset in [6] unfit for training such a model. We should rather be
using WPS dataset to train AI models for pregnancy outcome prediction. To
evaluate our hypothesis, we run some experiments on both Woman and WPS
dataset. unexplored in our experiment.

## Experiment

Figure 2: a) Cumulative density function of fraction of null entries of
features in the WPS and Woman datasets. b) Comparison of models trained on
Woman and WPS datasets - we see that women trains with the Women dataset leads
to poor performance on the WPS data in comparison to models trained on WPS
data.

To demonstrate the problems with using incorrect labels, we sought to
demonstrate how models trained with the incorrect and correct labels are
different from each other. To this end, we first sought to identify usable
features from each dataset. The Women dataset had 202 features, while WPS had
355 features. The features in the Woman dataset were a sub-set of those in the
WPS dataset. Furthermore, the WPS dataset is supposed to only contain
individuals who responded Yes to the pregnancy-outcome question. To keep the
models comparable, we only consider the features that are available in both
datasets. We further down the features to discard all features that were not
related to social, economic, health or demographic indicators e.g. all id
related fields were discarded. Since most confounding features which captured
birth outcome information (e.g. is-receive-birth-certification, baby-checkup-
after birth-in-days etc.). Since the outcome variable for the WPS dataset was
not a binary variable we converted it into a binary survival prediction
problem by treating Live birth surviving as one category and everything else
as the Non-surviving category. For both datasets, we discarded entries for
which the outcome variable was null. As a further step in the data cleaning
process, we discarded all features that had over 10 percent entries as null.
We noticed an extremely high number of null entries in both datasets as can be
seen in Figure 2a, which highlights a significant problem in the data
collection process. We further proceeded to one-hot-encode all categorical
variables present in the datasets. Having performed the data
cleaning/processing, we split each processed dataset into a 2:1 training/test
split. At this stage both datasets have the same features and we train
separate logistic regression models on both training sets using the scikit-
learn python package. We then evaluate the learned models on: i) the test set
corresponding to their training set, ii) the test set corresponding to the
other dataset (Figure 2b). We see a large drop in precision/recall/accuracy
for the model trained on the Woman dataset and tested on the WPS dataset in
comparison to the model trained on the WPS dataset. Moreover, the model
trained on the Woman performs extremely well on the test data for Woman.
Despite the extremely good performance on many of the metrics, this model is
actually prediction whether or not the woman was pregnant based on the
available feature information and not what the outcome of the pregnancy was.
This model was in fact a rather poorly performing model for the task of
predicting pregnancy outcomes compared to a model actually trained to predict
pregnancy outcomes. This highlights a one of the major issues with using such
datasets, which is the lack of clarity in naming convention which can lead to
misinterpretation of features and outcomes. While a long term solution to this
problem is adoption of clear and consistent naming conventions for variables,
a shorter term solution is for users of such datasets to rigorously explore
the data dictionaries and survey forms. It must be noted that there may have
been many other types of reporting errors which were unexplored in our
experiment.

## Discussion

Due to the complicated textual nature of the AHS questionnaire, we end up
capturing wrong subsections for questions in questionnaire for our case study.
By taking steps to simplify current AHS forms and making sure to correctly
transition to online data, we will have a much more comprehensive data set to
draw meaningful conclusions from. By doing so, we can ensure that we are
applying AI in a more efficient manner to predict pregnancy outcomes. Despite
the highlighted issues in the data collection and processing of such datasets
from developing countries, there is still a lot of value in analyzing such
datasets. While this is beyond the scope of this paper, the WPS dataset, once
appropriately processed, could be used to build an accurate pregnancy outcome
prediction model. In our simple experiment, we already saw decent recall and
accuracy for the model trained and tested on WPS. Such a model, along with
analysis of the important predictors would help identify women whose
pregnancies are at risk as well as the socio-economic or healthcare factors
that are leading to adverse pregnancy outcome in such developing nations.
unexplored in our experiment.

## References

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WHO, UNICEF, UNFPA. Geneva: WHO, 2003

[2] Open Data Government Platform India: https://data.gov.in

[3] Kaggle: https://www.kaggle.com

[4] Kaggle Getting Started: https://www.kaggle.com/getting-started/44916

[5] Annual Health Survey - Woman Schedule: https://data.gov.in/catalog/annual-
health-survey-woman-schedule

[6] Kaggle Predict Outcome of Pregnancy: https://www.kaggle.com/rajanand/ahs-
woman-1

[7] Ministry of Home Affairs, Government of India:
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[8] UNICEF Data: monitoring the situation of children and women:
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[13] Indian Council of Medical Research: https://www.icmr.nic.in/

[14] National Data Quality Forum:
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[15] National Institute for Medical Statistics: http://nims-
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[16] India Demographics Profile 2018:
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[17] The Challenges Confronting Public Hospitals in India, Their Origins, and
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