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% !TeX spellcheck = en_GB
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\newtheorem{Conjecture}[Theorem]{Conjecture}
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\begin{document}
\title[The de Rham...]{?? The de Rham cohomology of covers\\ with cyclic $p$-Sylow subgroup}
\author[A. Kontogeorgis and J. Garnek]{Aristides Kontogeorgis and J\k{e}drzej Garnek}
\address{???}
\email{jgarnek@amu.edu.pl}
\subjclass[2020]{Primary 14G17, Secondary 14H30, 20C20}
\keywords{de~Rham cohomology, algebraic curves, group actions,
characteristic~$p$}
\urladdr{http://jgarnek.faculty.wmi.amu.edu.pl/}
\date{}
\begin{abstract}
????
\end{abstract}
\maketitle
\bibliographystyle{plain}
%
\section{Introduction}
%
\begin{mainthm}
Suppose that $G$ is a group with a $p$-cyclic Sylow subgroup.
Let $X$ be a curve with an action of~$G$ over a field $k$ of characteristic $p$.
The $k[G]$-module structure of $H^1_{dR}(X)$ is uniquely determined by the lower ramification groups and the fundamental characters of closed
points $x$ of $X$ that are ramified in the cover $X \to X/G$.
\end{mainthm}
%
Note that if $p > 2$ and the $p$-Sylow subgroup of $G$ is not cyclic, the structure
of $H^1_{dR}(X)$ isn't determined uniquely by the ramification data, see \cite{??Garnek_indecomposables}.
\section{Cyclic covers}
%
Let for any $\ZZ/p^n$-cover $X \to Y$
%
\begin{align*}
u_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_P^{(t)} \cong \ZZ/p^{n-t} \},\\
l_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_{P, t} \cong \ZZ/p^{n-t} \}.
\end{align*}
%
Note that if $G_P = \ZZ/p^n$, this coincides with the standard definition of
the $t$th upper (resp. lower) ramification jump of $X \to Y$ at $P$.
If $G_P = \ZZ/p^m$, then (??relation with usual jumps??). By Hasse--Arf theorem (cf. ???),
the numbers $u_{X/Y, P}^{(t)}$ are integers.
%
\begin{Theorem} \label{thm:cyclic_de_rham}
Suppose that $\pi : X \to Y$ is a $\ZZ/p^n$-cover. Let $\langle G_P : P \in X(k) \rangle = \ZZ/p^m$. Pick arbitrary $P_0 \in X(k)$ such that $G_{P_0} \cong \ZZ/p^m$. Then, as $k[\ZZ/p^n]$-modules:
%
\[
H^1_{dR}(X) \cong J_{p^n}^{2 (g_Y - 1)} \oplus J_{p^n - p^{n-m} + 1}^2 \oplus \bigoplus_{\substack{P \in X(k)\\ P \neq P_0}} J_{p^n - p^n/e_P}^2
\oplus \bigoplus_{P \in X(k)} \bigoplus_{t = 0}^{n-1} J_{p^n - p^t}^{u_{P}^{(t+1)} - u_{P}^{(t)}},
\]
%
where $e_P := e_{X/Y, P}$ and $u_P^{(t)} := u_{X/Y, P}^{(t)}$.
\end{Theorem}
%
Write $H := \langle \sigma \rangle \cong \ZZ/p^n$.
For any $k[H]$-module $M$ denote:
%
\begin{align*}
M^{(i)} &:= \ker ((\sigma - 1)^i : M \to M),\\
T^i M &= T^i_H M := M^{(i)}/M^{(i-1)} \quad \textrm{ for } i = 1, \ldots, p^n.
\end{align*}
%
Recall that $\dim_k T^i M$ determines the structure of $M$ completely (cf. ????).
In the inductive step we use also the group $H' := \ZZ/p^{n-1}$. In this case
we denote the indecomposable $k[H']$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
and $\mc T^i M := T^i_{H'} M$ for any $k[H']$-module $M$.\\
Note also that for $j \ge 1$:
%
\[
u_{X/Y, P}^{(j)} - u_{X/Y, P}^{(j-1)} = \frac{1}{p^{j-1}} (l_{X/Y, P}^{(j)} - l^{(j-1)}_{X/Y, P})
\]
%
(in particular, $u_{X/Y, P}^{(1)} = l_{X/Y, P}^{(1)}$). Moreover, if $X' \to Y$ is the $\ZZ/p^N$-subcover of $X \to Y$ for $N \le n$ then:
%
\begin{itemize}
\item $u_{X'/Y, P'}^{(t)} = u_{X/Y, P}^{(t)}$ for $t \le N$ (here $P'$ denotes the image of~$P$ on~$X'$),
\item $l_{X/X', P}^{(t)} = l_{X/Y, P}^{(t + N)}$ for $t \le n-N$.
\end{itemize}
\begin{Lemma} \label{lem:G_invariants_\'{e}tale}
If the $G$-cover $X \to Y$ is \'{e}tale, then the natural map
%
\[
H^1_{dR}(Y) \to H^1_{dR}(X)^G
\]
%
is an isomorphism.
\end{Lemma}
\begin{proof}
????
\end{proof}
%
\begin{Lemma} \label{lem:trace_surjective}
Suppose that $G$ is a $p$-group.
If the $G$-cover $X \to Y$ is totally ramified, then the map
%
\[
\tr_{X/Y} : H^1_{dR}(X) \to H^1_{dR}(Y)
\]
%
is an epimorphism.
\end{Lemma}
\begin{proof}
%
By induction, it suffices to prove this in the case when $G = \ZZ/p$.
Consider the following commutative diagram:
%
\begin{center}
% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKMgEYqtRizYduvbHgJFh5MfWatEIABIA9YgAoAGqQAEAHVMB5ALYwA5nQD6BgJRceIDHIGLSo6qskNHX0ATRNzaztHENcZDz55QWQAJmV-CXUtbWEHYCgAJU5DWPdPfgUUVL9xNTYdHLzCvRi3WXKkgGY0msCs4UNw0ysAY2MLJxK2xKIu6oDM+ubBkbGHFriy6ZQAFm75qVb4rwrkXbmMg84xGChbeCJQADMAJwgrJDIQHAgkZLiXt6-ajfJDbf6vd6IXZfH6IABs4MB8OBsIAHIjIQB2FFIACcGKQAFYcYhMQTEF0YUTyUoqRTyak6ZT9hpzDhnrkDAB6EKcQ4AyHQkGIYk9TJsjnAbm8-kQpBwknYsVsCWcnl8q6cIA
\begin{tikzcd}
0 \arrow[r] & {H^0(X, \Omega_X)} \arrow[r] \arrow[d, "\tr_{X/Y}"] & H^1_{dR}(X) \arrow[r] \arrow[d, "\tr_{X/Y}"] & {H^1(X, \mc O_X)} \arrow[r] \arrow[d, "\tr_{X/Y}"] & 0 \\
0 \arrow[r] & {H^0(Y, \Omega_Y)} \arrow[r] & H^1_{dR}(Y) \arrow[r] & {H^1(Y, \mc O_Y)} \arrow[r] & 0
\end{tikzcd}
\end{center}
%
where the rows are Hodge--de Rham exact sequences. Recall that by~\cite[Theorem~1]{Valentini_Madan_Automorphisms}, in this case $H^0(X, \Omega_X)$ contains
a copy of $k[G]^{\oplus g_Y}$ as a direct summand. Thus, since trace is injective on $k[G]^{\oplus g_Y}$, the dimension
of the image of
%
\begin{equation} \label{eqn:trace_H0_Omega}
\tr_{X/Y} : H^0(X, \Omega_X) \to H^0(Y, \Omega_Y)
\end{equation}
%
is $g_Y$. Therefore the map~\eqref{eqn:trace_H0_Omega} is surjective.
Similarly, by Serre's duality, also $H^1(X, \mc O_X)$ contains $k[G]^{\oplus g_Y}$ as a direct summand
and one shows similarly that the trace map
%
\begin{equation*} %\label{eqn:trace_H0_Omega}
\tr_{X/Y} : H^1(X, \mc O_X) \to H^1(Y, \mc O_Y)
\end{equation*}
%
is surjective. Therefore, since the outer vertical maps in the diagram are surjective,
the trace map on the de Rham cohomology must be surjective as well.
%
\end{proof}
%
\begin{Lemma} \label{lem:TiM_isomorphism}
For any $i \le p^n - 1$ we have the following $k$-linear monomorphism:
%
\[
m_{\sigma - 1} : T^{i+1} M \hookrightarrow T^i M.
\]
\end{Lemma}
\begin{proof}
%
We define $m_{\sigma - 1}$ as follows:
%
\[
m_{\sigma - 1}(\ol x) := (\sigma - 1) \cdot x,
\]
%
where for $\ol x \in T^{i+1} M$ we picked any representative $x \in M^{(i+1)}$.
Indeed, if $x \in M^{(i+1)}$ then clearly $(\sigma - 1) \cdot x \in M^{(i)}$.
Moreover $(\sigma - 1) \cdot x \in M^{(i-1)}$ holds if and only if $x \in M^{(i)}$. This
shows that $m_{\sigma - 1}$ is well-defined and injective.
\end{proof}
%
\begin{Lemma} \label{lem:lemma_mcT_and_T}
Let $M$ be a $k[H]$-module. Let $T^i M$ and $\mc T^i M$ be as above.
If $\dim_k \mc T^i M = \dim_k \mc T^{i+1} M$ for some $i$ then:
%
\[
\dim_k T^{pi + p} M = \dim_k T^{pi + p - 1} M = \ldots = \dim_k T^{pi - p + 1} M.
\]
\end{Lemma}
\begin{proof}
Note that $\mc T^i M = M^{(pi)}/M^{(pi - p)}$. This easily implies that:
%
\begin{align*}
\dim_k \mc T^i M &= \dim_k T^{pi} M + \ldots + \dim_k T^{pi - p + 1} M\\
&\ge \dim_k T^{pi+p} M + \ldots + \dim_k T^{pi+1} M
= \dim_k \mc T^{i+1} M.
\end{align*}
%
Since the left-hand side and right hand side are equal, we conclude by Lemma~\ref{lem:TiM_isomorphism}
\end{proof}
%
\begin{Lemma}
For any $P \in X(k)$:
%
\[
p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
\]
%
\end{Lemma}
\begin{proof}
Note that:
%
\begin{align*}
u^{(n)}_{X/Y, P} - u^{(1)}_{X/Y, P} &= \sum_{j = 1}^{n-1} (u^{(j+1)}_{X/Y, P} - u^{(j)}_{X/Y, P}) = \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j+1)}_{X/Y, P} - l^{(j)}_{X/Y, P}).
\end{align*}
%
Similarly:
%
\[
u^{(n-1)}_{X/Y', P} - u^{(1)}_{X/Y', P} = \sum_{j = 1}^{n-2} \frac{1}{p^j} (l^{(j+1)}_{X/Y', P} - l^{(j)}_{X/Y', P}).
\]
We have:
%
\begin{align*}
p \cdot u^{(n)}_{X/Y, P} &= p \cdot (u^{(n)}_{X/Y, P} - u^{(1)}_{X/Y, P}) + u^{(1)}_{X/Y, Q}\\
&=p \cdot \sum_{j = 1}^{n-1} (u^{(j+1)}_{X/Y, P} - u^{(j)}_{X/Y, P}) + u^{(1)}_{X/Y, P}\\
&=p \cdot \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j+1)}_{X/Y, P} - l^{(j)}_{X/Y, P}) + u^{(1)}_{X/Y, P}\\
&=p \cdot \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j)}_{X/Y', P} - l^{(j-1)}_{X/Y', P}) + u^{(1)}_{X/Y, P}\\
\end{align*}
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
$H'' := H/\langle \sigma^{p^{n-1}} \rangle \cong \ZZ/p^{n-1}$, $Y' := X/H'$, $X'' := X/\langle \sigma^{p^{n-1}} \rangle$. Note that $H''$ naturally acts on $X''$.
Write also $\mc M := H^1_{dR}(X)$.
We consider now two cases. If the cover $X \to Y$ is \'{e}tale, then by induction assumption, since $2(g_{Y'} - 1) = p \cdot 2 \cdot (g_Y - 1)$:
%
\[
\mc M \cong \mc J_{p^{n-1}}^{2 p \cdot (g_Y - 1)} \oplus k^{\oplus 2}.
\]
%
Therefore $\dim_k \mc T^2 \mc M = \ldots = \dim_k \mc T^{p^{n-1}} \mc M = 2 p (g_Y - 1)$,
which by Lemma~\ref{lem:lemma_mcT_and_T} implies that
%
\[
\dim_k T^p \mc M = \ldots = \dim_k T^{p^n} \mc M = 2(g_Y - 1).
\]
%
Thus, for $i = 2, \ldots, p$:
%
\[
\dim_k T^i \mc M \ge 2(g_Y - 1) = \dim_k T^{p+1} \mc M.
\]
%
On the other hand, by Lemma~\ref{lem:G_invariants_\'{e}tale} we have
%
$
\dim_k T^1 \mc M = 2 g_Y
$. Thus:
%
\begin{align*}
\sum_{i = 2}^p \dim_k T^i \mc M = 2g_X - \dim_k T^1 \mc M - \sum_{i = p+1}^{p^n} \dim_k T^i \mc M = (p-1) \cdot 2(g_Y - 1).
\end{align*}
%
Thus $\dim_k T^i \mc M = 2(g_Y - 1)$ for every $i \ge 2$, which ends the proof in this case.
Assume now that $X \to Y$ is not \'{e}tale. Therefore $X \to X''$ is also not \'{e}tale.
By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
%
\[
\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - m} + 1}^2 \oplus \bigoplus_{\substack{P \in X(k)\\P \neq P_0}} \mc J_{p^{n-1} - p^{n-1}/e'_P}^2
\oplus \bigoplus_{P \in X(k)} \bigoplus_{t = 0}^{n-2} \mc J_{p^n - p^t}^{u_{X/Y', P}^{(t+1)} - u_{X/Y', P}^{(t)}}
\]
%
where $e'_P := e_{X/Y', P}$. Note that for any $P \in X(k)$:
%
\[
p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
\]
%
where $Q$ denotes the image of~$P$ in~$Y'$. Indeed, ?????
Therefore, for $i \le p^{n-1} - p^{n-2}$, using the Riemann--Hurwitz formula (cf. ????):
%
\begin{align*}
\dim_k \mc T^i \mc M &=
2(g_{Y'} - 1) + 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y', P}^{(n-1)} - 1)\\
&= 2 p (g_Y - 1) + \sum_{Q \in Y'(k)} (p-1) \cdot (l_{Y'/Y, Q}^{(1)} + 1)\\
&+ 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y', P}^{(n-1)} - 1)\\
&= p \cdot \left( 2(g_Y - 1) + 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y, P}^{(n)} - 1) \right)
\end{align*}
%
where
%
\[ R := \{ P \in X(k) : e_P > 1 \} = \{ P \in X(k) : e'_P > 1 \}. \]
%
In particular, $\dim_k \mc T^1 \mc M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M$.
Thus by Lemma~\ref{lem:lemma_mcT_and_T}
%
\begin{align*}
\dim_k T^1 \mc M &= \ldots = \dim_k T^{p^n - p^{n-1}} \mc M = \frac{1}{p} \dim_k \mc T^1 \mc M\\
&= 2(g_Y - 1) + 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y, P}^{(n)} - 1).
\end{align*}
%
By Lemma~\ref{lem:trace_surjective} since $X \to X''$ is not \'{e}tale, the map $\tr_{X/X''} : H^1_{dR}(X) \to H^1_{dR}(X'')$ is surjective. Recall that
in $\FF_p[x]$ we have the identity:
%
\[
1 + x + \ldots + x^{p-1} = (x - 1)^{p-1}.
\]
%
Therefore in the group ring $k[H]$ we have:
%
\[
\tr_{X/X''} = \sum_{j = 0}^{p-1} (\sigma^{p^{n-1}})^j = (\sigma^{p^{n-1}} - 1)^{p-1} =
(\sigma - 1)^{p^n - p^{n-1}}.
\]
%
This implies that:
%
\[
\ker(\tr_{X/X''} : \mc M \to \mc M'') = \mc M^{(p^n - p^{n-1})}
\]
%
and that $\tr_{X/X''}$ induces a $k$-linear isomorphism $T^{i + p^n - p^{n-1}} \mc M \to \mc T^i \mc M''$ for any $i \ge 1$. Thus:
%
\[
\dim_k T^{i + p^n - p^{n-1}} \mc M = \dim_k \mc T^i \mc M'' = ....
\]
%
This ends the proof.
\end{proof}
\section{Hypoelementary covers}
%
Assume now that $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$.
Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-module $M$ and any character $\psi$ of $H$ we write $M^{\psi} := M \otimes_{k[C]} \psi$.
%
\begin{Proposition} \label{prop:main_thm_for_hypoelementary}
Main Theorem holds for a hypoelementary $G$ as above and $k = \ol k$.
\end{Proposition}
%
\begin{Lemma}
Let $M$ be a $k[G]$-module of finite dimension. The $k[G]$-structure of $M$
is uniquely determined by the $k[C]$-structure of $T^1 M, \ldots, T^{p^n} M$.
\end{Lemma}
\begin{proof}
See \cite[????]{Bleher_Chinburg_Kontogeorgis_Galois_structure} for a proof.
\end{proof}
%
\begin{Lemma} \label{lem:N+Nchi+...}
Keep the above notation. Let $M$, $N$ be $k[C]$-modules. Assume that
%
\[
M \cong N \oplus N^{\chi} \oplus \ldots \oplus N^{\chi^{p-1}}.
\]
%
Then $N$ is uniquely determined by $M$.
%If $p-1 | j$, then $N_1 \cong N_2^{\chi^i}$ for some $i$.
\end{Lemma}
\begin{proof}
Note that
%
\[
M \cong N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}}.
\]
%
By tensoring this isomorphism by $\chi^i$ we obtain:
%
\begin{align*}
M^{\chi^i} \cong (N^{\chi^i})^{\oplus 2} \oplus N^{\chi^{i+1}} \oplus N^{\chi^{i+2}} \oplus \ldots \oplus N^{\chi^{i + p-2}}
\cong (N^{\chi^i})^{\oplus 2} \oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j}
\end{align*}
%
for $i = 0, \ldots, p-2$. Therefore:
%
\begin{equation} \label{eqn:N+M=M}
N^{\oplus p} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}} \oplus
\cong M^{\oplus (p-1)}.
\end{equation}
%
Indeed, for the proof of~\eqref{eqn:N+M=M} note that
%
\begin{align*}
N^{\oplus p} &\oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}}
\cong N^{\oplus p} \oplus \bigoplus_{i = 1}^{p-2} \left((N^{\chi^i})^{\oplus 2}
\oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j} \right)\\
&\cong \left( N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}} \right)^{\oplus (p-1)}
\cong M^{\oplus (p-1)}.
\end{align*}
%
The isomorphism~\eqref{eqn:N+M=M} clearly proves the thesis.
\end{proof}
%
\begin{Lemma} \label{lem:TiM_isomorphism_hypoelementary}
For any $i \le p^n - 1$ the map~$m_{\sigma - 1}$ from Lemma~\ref{lem:TiM_isomorphism}
yields a $k[C]$-equivariant monomorphism:
%
\[
m_{\sigma - 1} : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}.
\]
\end{Lemma}
\begin{proof}
By Lemma~\ref{lem:TiM_isomorphism} this map is injective. Thus it suffices to check that it is $k[C]$-equivariant.
Note that we have the following identity in the ring~$k[C]$:
%
\[
(\sigma - 1) \cdot \rho = \rho \cdot (\sigma^{\chi(\rho)^{-1}} - 1)
= \rho \cdot (\sigma - 1) \cdot (1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1})
\]
%
Note that $\sigma$ acts trivially on $T^i M$, so that for any $\ol x \in T^i M$:
%
\[
(1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1}) \cdot \ol x = \chi(\rho)^{-1} \cdot \ol x.
\]
%
This easily shows that
%
\[
m_{\sigma - 1}(\rho \cdot \ol x) = \chi(\rho)^{-1} \cdot \rho \cdot m_{\sigma - 1}(\ol x),
\]
%
which ends the proof.
%
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:main_thm_for_hypoelementary}]
We prove this by induction on $n$. If $n = 0$, then it follows by Chevalley--Weil theorem.
Consider now two cases. Firstly, we assume that $X \to Y$ is \'{e}tale.
Recall that by proof of Theorem~\ref{thm:cyclic_de_rham}, the map $(\sigma - 1)$
is an isomorphism of $k$-vector spaces between $T^{i+1} \mc M$ and $T^i \mc M$ for
$i = 2, \ldots, p^n$. This yields an isomorphism of $k[C]$-modules for $i \ge 2$ by Lemma~\ref{lem:TiM_isomorphism_hypoelementary}:
%
\begin{equation} \label{eqn:TiM=T1M_chi_\'{e}tale}
T^i \mc M \cong (T^2 \mc M)^{\chi^{-i+2}}
\end{equation}
%
Observe that $\mc T^i \mc M$ has the filtration $\mc M^{(pi)} \supset \mc M^{(pi - 1)} \supset \ldots \supset \mc M^{(pi - p)}$ with subquotients $T^{pi} \mc M, \ldots, T^{pi - p} \mc M$.
Thus, since the category of $k[C]$-modules is semisimple, for $i \le p^n - p^{n-1}$:
%
\begin{align*}
\mc T^i \mc M &\cong
\begin{cases}
T^1 \mc M \oplus T^2 \mc M \oplus (T^2 \mc M)^{\chi^{-1}} \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p - 2)}}, & i = 1\\
T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p-1)}}, & i > 1.
\end{cases}
\end{align*}
%
Thus, since by induction hypothesis $\mc T^2 \mc M$ is determined by ramification data,
we have by Lemma~\ref{lem:N+Nchi+...} that $T^2 \mc M$ is determined by ramification data.
Moreover, by Lemma~\ref{lem:G_invariants_\'{e}tale} and induction hypothesis, $T^1 \mc M \cong H^1_{dR}(X'')$
is also determined by ramification data.
Assume now that $X \to Y$ is not \'{e}tale. Analogously as in the previous case, Lemma~\ref{lem:TiM_isomorphism_hypoelementary} and proof of Theorem~\ref{thm:cyclic_de_rham}
yield an isomorphism of $k[C]$-modules:
%
\begin{equation} \label{eqn:TiM=T1M_chi}
T^{i+1} \mc M \cong (T^1 \mc M)^{\chi^{-i}}
\end{equation}
%
for $i \le p^n - p^{n-1}$. Observe that $\mc T^i M$ has the filtration $\mc M^{(pi)} \supset \mc M^{(pi - 1)} \supset \ldots \supset \mc M^{(pi - p)}$ with subquotients $T^{pi} \mc M, \ldots, T^{pi - p + 1} \mc M$.
Thus, since the category of $k[C]$-modules is semisimple, for $i \le p^n - p^{n-1}$:
%
\begin{align*}
\mc T^i \mc M &\cong T^{pi - p + 1} \mc M \oplus \ldots \oplus T^{pi} \mc M\\
&\cong T^1 \mc M \oplus (T^1 \mc M)^{\chi^{-1}} \oplus \ldots \oplus
(T^1 \mc M)^{\chi^{-p}}.
\end{align*}
%
By induction assumption, the $k[C]$-module structure of $\mc T^i \mc M$ is uniquely determined by the ramification data. Thus, by Lemma~\ref{lem:N+Nchi+...} for $N := T^1 \mc M$ and by~\eqref{eqn:TiM=T1M_chi} the $k[C]$-structure of the modules $T^i \mc M$ is uniquely determined by the ramification data for $i \le p^n - p^{n-1}$.
By similar reasoning, $\tr_{X/X'}$ yields an isomorphism:
%
\[
T^{i + p^n - p^{n-1}} \mc M \cong (\mc T^i \mc M'')^{\chi^{-1??}}.
\]
%
Thus, by induction hypothesis for $\mc M''$, the $k[C]$-structure of $T^{i + p^n - p^{n-1}} \mc M$
is determined by ramification data as well.
\end{proof}
\section{Proof of Main Theorem}
%
(Conlon induction ???) (algebraic closure ???)
\bibliography{bibliografia}
\end{document}
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