diff --git a/article_de_rham_cyclic.synctex.gz b/article_de_rham_cyclic.synctex.gz
index 4728aca..7d3580c 100644
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diff --git a/article_de_rham_cyclic.tex b/article_de_rham_cyclic.tex
index 8359d2c..659682f 100644
--- a/article_de_rham_cyclic.tex
+++ b/article_de_rham_cyclic.tex
@@ -615,19 +615,15 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
 %
 \begin{Lemma} \label{lem:N+Nchi+...}
 	Keep the above notation. Let $M$, $N$ be $k[C]$-modules. 
-	\begin{enumerate}[(1)]
-		\item If $N \cong M \oplus M^{\chi} \oplus \ldots \oplus M^{\chi^{p-1}},$
-		then $N$ is uniquely determined by $M$.
-		
-		\item If $N^{\oplus (p-1)} \cong M \oplus M^{\chi} \oplus \ldots \oplus M^{\chi^{p-2}}$,
-		then $M \cong N \cong N^{\chi}$.
-	\end{enumerate}
+	If
 	%
-	
-	%If $p-1 | j$, then $N_1 \cong N_2^{\chi^i}$ for some $i$.
+	\[
+		N \cong M \oplus M^{\chi} \oplus \ldots \oplus M^{\chi^{p-1}},
+	\]
+	then $N$ is uniquely determined by $M$.
 \end{Lemma}
 \begin{proof}
-	(1) Note that
+	Note that
 	%
 	\[
 	N \cong M^{\oplus 2} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}}.
@@ -657,8 +653,7 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
 		\cong N^{\oplus (p-1)}.
 	\end{align*}
 	%
-	The isomorphism~\eqref{eqn:M+N=N} clearly proves the thesis.\\
-	(2) 
+	The isomorphism~\eqref{eqn:M+N=N} clearly proves the thesis.
 \end{proof}
 %
 \begin{Lemma} \label{lem:TiM_isomorphism_hypoelementary}
@@ -696,42 +691,17 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
 
 \begin{proof}[Proof of Main Theorem]
 	As explained at the beginning of this section, it suffices to show this in the case when $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$ and $k = \ol k$ by Lemma~\ref{lem:reductions}.
-	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.
-	Note that the proof of Lemma~\ref{lem:G_invariants_\'{e}tale} implies
-	that there exists an exact sequence of $k[C]$-modules:
+	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. Then the results of \cite{Garnek_equivariant} imply that the Hodge--de Rham exact sequence splits and
 	%
-	\[
-		0 \to k \to H^1_{dR}(Y) \to T^1 \mc M \to k \to 0,
-	\]
-	%
-	where $k$ denotes the trivial $k[C]$-module. Therefore, since the category of $k[C]$-modules is semisimple, $T^1 \mc M \cong H^1_{dR}(Y)$. By induction hypothesis we conclude that
-	$T^1 \mc M$ is determined by higher ramification data.
-	Recall now 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}}
+	\begin{equation}
+		H^1_{dR}(X) \cong H^0(X, \Omega_X) \oplus H^1(X, \mc O_X).
 	\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:
-	%
-	\begin{align}
-		\mc T^1 \mc M &\cong 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)}} \label{eqn:decomposition_of_mc_T1}\\
-		\mc T^i \mc M &\cong T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p-1)}} \quad \textrm{ for } 2 \le i \le p^{n-1}. \label{eqn:decomposition_of_mc_Ti}
-	\end{align}
-	%
-	If $n \ge 2$, then by induction hypothesis, Lemma~\ref{lem:N+Nchi+...} and by~\eqref{eqn:decomposition_of_mc_Ti} the $k[C]$-module $T^2 \mc M$ is determined by higher ramification data. If $n = 1$, then the $k[C]$-module
-	%
-	\[
-		\mc T^1 \mc M/T^1 \mc M = H^1_{dR}(X)/H^1_{dR}(Y)
-	\]
-	%
-	is of the form $N^{\oplus (p-1)}$ for a $k[C]$-module $N$
-	by Lemma~????. Therefore by Lemma~????? we have $T^2 \mc M \cong N$.
-	This ends the proof, as $N$ is determined by the fundamental characters of the group action of $C$ on $X$.\\
+	By Lemma~\ref{lem:G_invariants_\'{e}tale} and \cite[Corollary~2.4]{Garnek_equivariant} we have $\dim_k H^1_{dR}(X)^H = 2g_Y = \dim_k H^0(X, \Omega_X)^H + \dim_k H^1(X, \mc O_X)^H$. Therefore the Hodge--de Rham exact sequence splits by \cite[Lemma~5.3]{Garnek_equivariant}.
+	Recall that by~\cite[Theorem~1.1]{Bleher_Chinburg_Kontogeorgis_Galois_structure}
+	the $k[G]$-module structure of $H^0(X, \Omega_X)$ is determined by the higher ramification data. This ends the proof in this case, 
+	as $H^1(X, \mc O_X) \cong H^0(X, \Omega_X)^{\vee}$ by Serre's duality (cf. ???).\\
 	
 	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: