notation; m_{X/Y,P}; pt 2
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@ -171,6 +171,7 @@ Throughout the paper we will use the following notation for any $P \in X(\ol k)$
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at $P$,
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at $P$,
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\item $m_{X/Y, P} := \ord_p(e_{X/Y, P})$ is the maximal power of~$p$
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\item $m_{X/Y, P} := \ord_p(e_{X/Y, P})$ is the maximal power of~$p$
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dividing the ramification index,
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dividing the ramification index,
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\item $u_{X/Y, P} := u_{X/Y, P}^{(m_{X/Y, P})}$ is the last ramification jump,
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\item $m_{X/Y} := \max \{ m_{X/Y, P} : P \in X(k) \}$.
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\item $m_{X/Y} := \max \{ m_{X/Y, P} : P \in X(k) \}$.
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\end{itemize}
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\end{itemize}
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@ -219,7 +220,8 @@ For any $k[H]$-module $M$ denote:
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T^i M &= T^i_H M := M^{(i)}/M^{(i-1)} \quad \textrm{ for } i = 1, \ldots, p^n.
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T^i M &= T^i_H M := M^{(i)}/M^{(i-1)} \quad \textrm{ for } i = 1, \ldots, p^n.
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\end{align*}
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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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Recall that $\dim_k T^i M$ determines the structure of $M$ completely (see \cite[p. 108]{Valentini_Madan_Automorphisms} -- they give the argument for $M := H^0(X, \Omega_X)$,
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but it works for an arbitrary module).
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Moreover, for $i > 0$:
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Moreover, for $i > 0$:
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\begin{equation} \label{eqn:dim_of_Ti_Jl}
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\begin{equation} \label{eqn:dim_of_Ti_Jl}
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@ -239,29 +241,7 @@ where we use the Iverson bracket notation:
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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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In the inductive step we use also the group $H' := \ZZ/p^{n-1}$. In this case
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we denote the indecomposable $k[H']$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
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we denote the indecomposable $k[H']$-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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and $\mc T^i M := T^i_{H'} M$ for any $k[H']$-module $M$.
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\noindent Recall also that by \cite[???]{Serre1979} there exist integers $i_{X/Y, P}^{(0)}, i_{X/Y, P}^{(1)}, \ldots$ such that:
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\begin{align*}
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u_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} + \ldots + i_{X/Y, P}^{(t-1)}\\
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l_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p + \ldots + i_{X/Y, P}^{(t-1)} \cdot p^{t-1}.
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\end{align*}
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Assume now that $X' \to Y$ is the $\ZZ/p^N$-subcover of $X \to Y$ for $N \le n$. Let $P' \in X'(k)$ be the image of $P \in X(k)$. Then, if
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$e_{X/Y, P} = p^n$, we have:
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\begin{align*}
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i_{X/X', P}^{(t)} &=
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\begin{cases}
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i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p + \ldots + i_{X/Y, P}^{(N)} \cdot p^N, & t = 0\\
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p^N \cdot i_{X/Y, P}^{(N+t)}, & t = 1, \ldots, n-N-1.
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\end{cases}\\
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i_{X'/Y, P'}^{(t)} &= i_{X/Y, P}^{(t)} \qquad \textrm{ for } t < N.
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\end{align*}
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If $e_{X/Y, P} \le p^{n - N}$ then $i_{X/Y, P}^{(t)} = i_{X/X', P}^{(t)}$
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for all $t$.
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\begin{Lemma} \label{lem:G_invariants_\'{e}tale}
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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
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If the $G$-cover $X \to Y$ is \'{e}tale, then
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@ -406,7 +386,28 @@ shows that $m_{\sigma - 1}$ is well-defined and injective.
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\end{equation}
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\end{equation}
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Assume now that $Q \in B_{Y'/Y}$. Then there exists a unique point $Q' \in Y'(k)$
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Assume now that $Q \in B_{Y'/Y}$. Then there exists a unique point $Q' \in Y'(k)$
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in the preimage of $Q$ through $Y' \to Y$. Moreover, $m_{X/Y, Q} = m_{X/Y', Q'}$. By using ????above formulas???:
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in the preimage of $Q$ through $Y' \to Y$. Moreover, $m_{X/Y, Q} = n$, $m_{X/Y', Q'} = n-1$.
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Recall also that by \cite[???]{Serre1979} there exist integers $i_{X/Y, P}^{(0)}, i_{X/Y, P}^{(1)}, \ldots$ such that for every $t \ge 0$:
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\begin{align*}
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u_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} + \ldots + i_{X/Y, P}^{(t-1)}\\
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l_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p + \ldots + i_{X/Y, P}^{(t-1)} \cdot p^{t-1}.
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\end{align*}
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Observe that:
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\begin{align*}
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i_{X/X', P}^{(0)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p,\\
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i_{X/X', P}^{(t)} &= p \cdot (i_{X/Y, P}^{(t + 1)} + \ldots + i_{X/Y, P}^{(n-1)}) \quad \textrm{ for } t > 0.
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\end{align*}
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This implies that
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\begin{equation} \label{eqn:Q_in_B'}
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p \cdot (u_{X/Y, Q} - 1) = (u_{X/Y', Q'} - 1) + (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1).
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\end{equation}
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Indeed, by using the above formulas:
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\begin{align*}
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\begin{align*}
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p \cdot (u_{X/Y, Q} - 1) &=
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p \cdot (u_{X/Y, Q} - 1) &=
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@ -416,6 +417,7 @@ shows that $m_{\sigma - 1}$ is well-defined and injective.
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&= (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1) + (u_{X/Y', Q'} - 1).
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&= (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1) + (u_{X/Y', Q'} - 1).
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\end{align*}
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\end{align*}
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The proof follows by summing~\eqref{eqn:Q_not_in_B'} and~\eqref{eqn:Q_in_B'} over all $Q \in B_{X/Y}$.
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\end{proof}
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\end{proof}
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\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
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\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
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@ -468,13 +470,9 @@ shows that $m_{\sigma - 1}$ is well-defined and injective.
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2(g_{Y'} - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
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2(g_{Y'} - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
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&= 2 p (g_Y - 1) + \sum_{Q' \in Y'(k)} (p-1) \cdot (l_{Y'/Y, Q'}^{(1)} + 1)\\
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&= 2 p (g_Y - 1) + \sum_{Q' \in Y'(k)} (p-1) \cdot (l_{Y'/Y, Q'}^{(1)} + 1)\\
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&+ 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
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&+ 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
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&= p \cdot \left( 2(g_Y - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y(k)} (u_{X/Y, Q'} - 1) \right),\\
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&= p \cdot \left( 2(g_Y - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y(k)} (u_{X/Y, Q'} - 1) \right).
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\end{align*}
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\end{align*}
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where
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\[ B := \{ Q \in Y(k) : e_Q > 1 \} = \{ Q \in Y(k) : e'_Q > 1 \}. \]
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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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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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Thus by Lemma~\ref{lem:lemma_mcT_and_T} for any $1 \le i \le p^n - p^{n-1}$:
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Thus by Lemma~\ref{lem:lemma_mcT_and_T} for any $1 \le i \le p^n - p^{n-1}$:
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