etale case new pt 1

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jgarnek 2024-12-02 21:01:06 +01:00
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@ -667,9 +667,20 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
\end{proof}
\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}.
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:
%
\[
0 \to k \to H^1_{dR}(Y) \to T^1 \mc M \to k \to 0.
\]
%
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 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}: