23 lines
1.1 KiB
TeX
23 lines
1.1 KiB
TeX
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\begin{proof}
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The inclusion holds because $\iota_n (\cR^+) \subseteq \dB_\dR^+$ for all $n$ , and equality holds because more exactly
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\[
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\cap_n \iota_n^{-1} \dB_\dR^+ =
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\]
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and therefore, as by assumption the lowest filtration jump of $\rD_\cris(V)$ is $-h$, for all $n$ in $\N$,
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\[
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\cR_{\Q_{1/p^n}}^{+, r_0} \ox_{\varphi^{-n}_\rst{\cE^+}} \rN(V) \subseteq \Fil^0 ( t^{-h} \cdot \cR_{\Q_{1/p^n}}^{+, r_0} \ox_{\Q_p} \rD_\cris(V) ).
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\]
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where the left-hand side tensor product $\ox_{\varphi_{\rst{\cE^+}}}$ is given by the restriction
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\[
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\varphi^{-n} \from \cE^+ \to \cR^+_{\Q_{1/p^n}}
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\]
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of $\varphi^{-n} \from \cR^{+, r_n}_{\Q_{1/p^n}} \to \cR^{+, r_0}_{\Q_{1/p^n}}$ to $\cE^+$.
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Because the left-hand side and right-hand side module have the same rank and the same elementary divisors as submodule of $ t^{-h} \cdot \cR_{\Q_{1/p^n}}^{r_0} \ox_{\Q_p} \rD_\cris(V) $, the modules are equal for all $n$ in $\N$.
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By \cref{prop:NVoxRpInDcrisVoxRp},
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\[
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\cR^+ \ox_{\cE^+} \rN(V) \subseteq (t/X)^{-h} \cdot \cR^+ \ox_{\Q_p} \rD_\cris(V)
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\]
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\end{proof}
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