Lemma mc T = mc T --> T = T
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@ -154,12 +154,14 @@ the $t$th upper (resp. lower) ramification jump of $X \to Y$ at $P$.
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Suppose that $\pi : X \to Y$ is a $\ZZ/p^n$-cover. Let $\langle G_P : P \in X(k) \rangle = \ZZ/p^m = G_{P_0}$ for $P_0 \in X(k)$. Then, as $k[\ZZ/p^n]$-modules:
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%
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\[
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H^1_{dR}(X) \cong J_{p^n}^{2 (g_Y - 1)} \oplus J_{p^n - p^{n-m} + 1}^2 \oplus \bigoplus_{P \neq P_0} J_{p^n - \frac{p^n}{e_{X/Y, P}}}^2
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\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} J_{p^n - p^t}^{u_{X/Y, P}^{(t+1)} - u_{X/Y, P}^{(t)}}.
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H^1_{dR}(X) \cong J_{p^n}^{2 (g_Y - 1)} \oplus J_{p^n - p^{n-m} + 1}^2 \oplus \bigoplus_{P \neq P_0} J_{p^n - p^n/e_P}^2
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\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} J_{p^n - p^t}^{u_{P}^{(t+1)} - u_{P}^{(t)}},
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\]
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where $e_P := e_{X/Y, P}$ and $u_P^{(t)} := u_{X/Y, P}^{(t)}$.
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\end{Theorem}
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Write $H := \ZZ/p^n = \langle \sigma \rangle$.
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Write $H := \langle \sigma \rangle \cong \ZZ/p^n$.
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For any $k[H]$-module $M$ denote:
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\begin{align*}
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@ -168,9 +170,9 @@ For any $k[H]$-module $M$ denote:
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\end{align*}
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Recall that $\dim_k T^i M$ determines the structure of $M$ completely (cf. ????).
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In the inductive step we use also the group $\ZZ/p^{n-1}$. In this case
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we denote the irreducible $k[\ZZ/p^{n-1}]$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
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and $\mc T^i M := T^i_{\ZZ/p^{n-1}} M$ for any $k[\ZZ/p^{n-1}]$-module $M$.
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In the inductive step we use also the group $H' := \ZZ/p^{n-1}$. In this case
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we denote the irreducible $k[H']$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
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and $\mc T^i M := T^i_{H'} M$ for any $k[H']$-module $M$.
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Note also that for $j \ge 1$:
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@ -187,7 +189,7 @@ Note also that for $j \ge 1$:
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\end{itemize}
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\begin{Lemma} \label{lem:G_invariants_etale}
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\begin{Lemma} \label{lem:G_invariants_\'{e}tale}
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If the $G$-cover $X \to Y$ is \'{e}tale, then the natural map
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\[
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@ -224,31 +226,72 @@ Note also that for $j \ge 1$:
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\end{proof}
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\begin{Lemma}
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Let $M$ be a $k[H]$-module. Let $T^i M$ be as above and
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$\mc T^i M := T^i_{H'} M$ for $H' \le H$, $H' \cong \ZZ/p^{n-1}$.
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If $\mc T^i M \cong \mc T^{i+1} M$ for some $i$ then:
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\begin{Lemma} \label{lem:lemma_mcT_and_T}
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Let $M$ be a $k[H]$-module. Let $T^i M$ and $\mc T^i M$ be as above.
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If $\dim_k \mc T^i M = \dim_k \mc T^{i+1} M$ for some $i$ then:
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\[
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T^{pi + p} M \cong T^{pi + p - 1} M \cong \ldots \cong T^{pi - p + 1} M.
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\dim_k T^{pi + p} M = \dim_k T^{pi + p - 1} M = \ldots = \dim_k T^{pi - p + 1} M.
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\]
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\end{Lemma}
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\begin{proof}
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??
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By Lemma~\ref{lem:TiM_isomorphism}:
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\begin{align*}
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\dim_k \mc T^i M &= \dim_k T^{pi} M + \ldots + \dim_k T^{pi - p + 1} M\\
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&\ge \dim_k T^{p^n - p^{n-1}} M + \ldots + \dim_k T^{p^n - p^{n-1}} M
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= \dim_k \mc T^{p^{n-1} - p^{n-2}} M.
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\end{align*}
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Since the left-hand side and right hand side are equal, we conclude by Lemma~\ref{lem:TiM_isomorphism}
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\end{proof}
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\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
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We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
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$H'' := H/\langle \sigma^{p^{n-1}} \rangle \cong \ZZ/p^{n-1}$, $Y' := X/H'$, $X'' := X/H''$.
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Write also $\mc M := H^1_{dR}(X)$.
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We consider now two cases. If the cover $X \to Y$ is \'{e}tale, then by Lemma~\ref{lem:G_invariants_\'{e}tale} we have
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\[
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\dim_k T^1 \mc M = 2 g_Y
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\]
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Moreover, by induction assumption, since $2(g_{Y'} - 1) = p \cdot 2 \cdot (g_Y - 1)$:
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\[
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\mc M \cong \mc J_{p^{n-1}}^{2 p \cdot (g_Y - 1)} \oplus k^{\oplus 2}.
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\]
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Therefore $\dim_k \mc T^2 \mc M = \ldots = \dim_k \mc T^{p^{n-1}} \mc M = 2 p (g_Y - 1)$,
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which implies that
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\[
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\dim_k T^p \mc M = \ldots = \dim_k T^{p^n} \mc M = 2(g_Y - 1).
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\]
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by Lemma~\ref{lem:lemma_mcT_and_T}. Thus, for $i = 2, \ldots, p$:
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\[
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\dim_k T^i \mc M \ge 2(g_Y - 1) = \dim_k T^{p+1} \mc M
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\]
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On the other hand:
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\begin{align*}
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\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).
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\end{align*}
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Thus $\dim_k T^i \mc M = 2(g_Y - 1)$ for every $i \ge 2$, which ends the proof in this case.
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Assume now that $X \to Y$ is not \'{e}tale. Therefore $X \to X''$ is also not \'{e}tale.
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By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
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\[
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\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - 1 -m'} + 1}^2 \oplus \bigoplus_{P \neq P_0} \mc J_{p^n - \frac{p^{n-1}}{e_{X/Y', P}}}^2
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\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - 1 -m'} + 1}^2 \oplus \bigoplus_{P \neq P_0} \mc J_{p^{n-1} - p^{n-1}/e'_P}^2
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\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} \mc J_{p^n - p^t}^{u_{X/Y', P}^{(t+1)} - u_{X/Y', P}^{(t)}}
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\]
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where
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where $e'_P := e_{X/Y', P}$ and
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\[
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m' :=
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@ -257,7 +300,7 @@ Note also that for $j \ge 1$:
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m, & \textrm{ otherwise.}
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\end{cases}
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\]
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Therefore, for $i \le p^n - p^{n-1}$
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\begin{align*}
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@ -268,30 +311,20 @@ Note also that for $j \ge 1$:
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\end{align*}
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In particular, $\dim_k \mc T^1 \mc M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M$.
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On the other hand, by Lemma~\ref{lem:TiM_isomorphism}:
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\begin{align*}
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\dim_k \mc T^1 \mc M &= \dim_k T^1 \mc M + \ldots + \dim_k T^p \mc M\\
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&\ge \dim_k T^{p^n - p^{n-1}} \mc M + \ldots + \dim_k T^{p^n - p^{n-1}} \mc M
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= \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M.
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\end{align*}
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Since the left-hand side and right hand side are equal, we conclude by Lemma~\ref{lem:TiM_isomorphism}
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that
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Thus by Lemma~\ref{lem:lemma_mcT_and_T}
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\[
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\dim_k T^1 \mc M = \ldots = \dim_k T^{p^n - p^{n-1}} \mc M = \frac{1}{p} \dim_k \mc T^1 \mc M.
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\]
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We consider now two cases. If the cover $X \to Y$ is \'{e}tale, then by Lemma~\ref{lem:G_invariants_etale} we have
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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
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in $\FF_p[x]$ we have the identity:
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\[
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\dim_k T^1 \mc M = 2 g_{X''}
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1 + x + \ldots + x^{p-1} = (x - 1)^{p-1}.
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\]
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then the cover $X \to Y$ must be also \'{e}tale.
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Thus the proof follows in this case by~\cite{Nakajima??Inventiones}. Suppose now that
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$X \to X''$ is not \'{e}tale. Then, by Lemma~\ref{lem:trace_surjective}, the map $\tr_{X/X''} : H^1_{dR}(X) \to H^1_{dR}(X'')$ is surjective. Moreover, note that in the group ring $k[H]$ we have:
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Therefore in the group ring $k[H]$ we have:
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\[
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\tr_{X/X''} = \sum_{j = 0}^{p-1} (\sigma^{p^{n-1}})^j = (\sigma^{p^{n-1}} - 1)^{p-1} =
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@ -304,7 +337,7 @@ Note also that for $j \ge 1$:
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\ker(\tr_{X/X''} : \mc M \to \mc M'') = \mc M^{(p^n - p^{n-1})}
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\]
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and that $\tr_{X/X''}$ induces a $k$-linear isomorphism $T^{i + p^n - p^{n-1}} M \to \mc T^i M''$ for any $i \ge 1$. Thus:
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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:
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\[
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\dim_k T^{i + p^n - p^{n-1}} \mc M = \dim_k \mc T^i \mc M'' = ....
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@ -315,7 +348,7 @@ Note also that for $j \ge 1$:
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\section{Hypoelementary covers}
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Assume now that $G = H \rtimes_{\chi} \ZZ/c$.
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Assume now that $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$.
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\begin{Proposition} \label{prop:main_thm_for_hypoelementary}
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Main Theorem holds for a hypoelementary $G$ as above and $k = \ol k$.
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@ -377,7 +410,7 @@ Assume now that $G = H \rtimes_{\chi} \ZZ/c$.
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is an isomorphism of $k$-vector spaces between $T^{i+1} \mc M$ and $T^i \mc M$ for
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$i = 2, \ldots, p^n$. This yields an isomorphism of $k[C]$-modules for $i \ge 2$:
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\begin{equation} \label{eqn:TiM=T1M_chi_etale}
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\begin{equation} \label{eqn:TiM=T1M_chi_\'{e}tale}
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T^i \mc M \cong (T^2 \mc M)^{\chi^{-i+2}}
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\end{equation}
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@ -394,10 +427,10 @@ Assume now that $G = H \rtimes_{\chi} \ZZ/c$.
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Thus, since by induction hypothesis $\mc T^i \mc M$ is determined by ramification data,
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we have by Lemma~\ref{lem:N+Nchi+...} that $T^2 \mc M$ is determined by ramification data.
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Moreover, by Lemma~\ref{lem:G_invariants_etale}, $T^1 \mc M \cong H^1_{dR}(X'')$
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Moreover, by Lemma~\ref{lem:G_invariants_\'{e}tale}, $T^1 \mc M \cong H^1_{dR}(X'')$
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is also determined by ramification data (???).
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Assume now that $X \to Y$ is not etale. Analogously as in the previous case, Lemma~\ref{lem:TiM_isomorphism_hypoelementary} and proof of Theorem~\ref{thm:cyclic_de_rham}
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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}
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yield an isomorphism of $k[C]$-modules:
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\begin{equation} \label{eqn:TiM=T1M_chi}
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