notation; m_{X/Y,P}; pt 2

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jgarnek 2024-11-25 13:30:42 +01:00
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commit 3d12edc1c8
2 changed files with 28 additions and 30 deletions

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