before changing ramification jumps
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@ -181,13 +181,13 @@ 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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Note also that for $j \ge 1$:
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\[
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\[
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l_{X/Y, P}^{(j)} - l_{X/Y, P}^{(j-1)} = \frac{1}{p^{j-1}} (u_{X/Y, P}^{(j)} - u^{(j-1)}_{X/Y, P})
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u_{X/Y, P}^{(j)} - u_{X/Y, P}^{(j-1)} = \frac{1}{p^{j-1}} (l_{X/Y, P}^{(j)} - l^{(j-1)}_{X/Y, P})
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\]
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\]
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(in particular, $u_{X/Y, P}^{(1)} = l_{X/Y, P}^{(1)}$). Moreover, if $X' \to Y$ is the $\ZZ/p^N$-subcover of $X \to Y$ for $N \le n$ then:
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(in particular, $u_{X/Y, P}^{(1)} = l_{X/Y, P}^{(1)}$). Moreover, if $X' \to Y$ is the $\ZZ/p^N$-subcover of $X \to Y$ for $N \le n$ then:
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\begin{itemize}
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\begin{itemize}
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\item $u_{X'/Y, P}^{(t)} = u_{X/Y, P}^{(t)}$ for $t \le N$,
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\item $u_{X'/Y, P'}^{(t)} = u_{X/Y, P}^{(t)}$ for $t \le N$ (here $P'$ denotes the image of~$P$ on~$X'$),
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\item $l_{X/X', P}^{(t)} = l_{X/Y, P}^{(t + N)}$ for $t \le n-N$.
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\item $l_{X/X', P}^{(t)} = l_{X/Y, P}^{(t + N)}$ for $t \le n-N$.
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\end{itemize}
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\end{itemize}
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@ -290,6 +290,37 @@ shows that $m_{\sigma - 1}$ is well-defined and injective.
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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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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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\end{proof}
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\begin{Lemma}
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For any $P \in X(k)$:
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\[
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p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
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\]
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\end{Lemma}
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\begin{proof}
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Note that:
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\begin{align*}
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u^{(n)}_{X/Y, P} - u^{(1)}_{X/Y, P} &= \sum_{j = 1}^{n-1} (u^{(j+1)}_{X/Y, P} - u^{(j)}_{X/Y, P}) = \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j+1)}_{X/Y, P} - l^{(j)}_{X/Y, P}).
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\end{align*}
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Similarly:
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\[
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u^{(n-1)}_{X/Y', P} - u^{(1)}_{X/Y', P} = \sum_{j = 1}^{n-2} \frac{1}{p^j} (l^{(j+1)}_{X/Y', P} - l^{(j)}_{X/Y', P}).
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\]
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We have:
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\begin{align*}
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p \cdot u^{(n)}_{X/Y, P} &= p \cdot (u^{(n)}_{X/Y, P} - u^{(1)}_{X/Y, P}) + u^{(1)}_{X/Y, Q}\\
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&=p \cdot \sum_{j = 1}^{n-1} (u^{(j+1)}_{X/Y, P} - u^{(j)}_{X/Y, P}) + u^{(1)}_{X/Y, P}\\
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&=p \cdot \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j+1)}_{X/Y, P} - l^{(j)}_{X/Y, P}) + u^{(1)}_{X/Y, P}\\
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&=p \cdot \sum_{j = 1}^{n-1} \frac{1}{p^j} (l^{(j)}_{X/Y', P} - l^{(j-1)}_{X/Y', P}) + u^{(1)}_{X/Y, P}\\
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\end{align*}
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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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We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
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We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
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@ -340,7 +371,7 @@ shows that $m_{\sigma - 1}$ is well-defined and injective.
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p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
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p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
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\]
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\]
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where $Q$ denotes the image of~$P$ in~$Y'$.
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where $Q$ denotes the image of~$P$ in~$Y'$. Indeed, ?????
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Therefore, for $i \le p^{n-1} - p^{n-2}$, using the Riemann--Hurwitz formula (cf. ????):
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Therefore, for $i \le p^{n-1} - p^{n-2}$, using the Riemann--Hurwitz formula (cf. ????):
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\begin{align*}
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\begin{align*}
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