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$\newcommand{\ur}{\mathrm{ur}}\newcommand{\cris}{\mathrm{cris}}$Let $K$ be a finite extension of $\mathbb{Q}_p$, $G_K=\operatorname{Gal}(\overline{K}/K)$ and $I_K \subset G_K$ its inertial subgroup. Let $V$ be a finite-dimensional representation of $G_K$. Assume that $V$ is crystalline as $G_K$-representation. Is it true that it is crystalline as $I_K$-representation?

By definition, we need to prove that $$\dim_{\widehat{K^{\ur}}}(V \otimes B_{\cris})^{I_K}=\dim_{K_0}(V \otimes B_{\cris})^{G_K}=\dim_{\mathbb{Q}_p} V$$ where $\widehat{K^{\ur}}$ and $K_0$ are the maximal complete unramified extension of $\mathbb{Q}_p$ in $\overline{K}$ and $K$ respectively.

Edit: As remarked by David, $K^{\ur}$ need to be replaced by its completion.

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This is purely formal. If $V$ is crystalline, then $V \otimes \mathbf{B}_{\mathrm{cris}}$ has a basis as a $\mathbf{B}_{\mathrm{cris}}$-module in which the action of $G_K$ is trivial. Hence a fortiori it has a basis in which the action of $I_K$ is trivial.

What is much less obvious, but also true, is that the converse holds: if $V$ is crystalline as an $I_K$-representation, then it's actually crystalline as a $G_K$-representation. This is because $(B_{\mathrm{cris}})^{I_K} = \widehat{K^{\mathrm{nr}}}$ contains the periods of all unramified representations. (Incidentally, the assertion in your question that $(B_{\mathrm{cris}})^{I_K} = K^{\mathrm{nr}}$ is incorrect; it is genuinely the completion that you get here.)

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  • $\begingroup$ Thanks David for your remarks. However, for the converse part, in Fontaine-Ouyang's they only state this for de Rham and semi-stable cases not for crystalline. (prop. 7.15, 7.16 page 154). imo.universite-paris-saclay.fr/~fontaine/galoisrep.pdf $\endgroup$
    – Desunkid
    Jul 24, 2022 at 10:28
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    $\begingroup$ Are you trying to suggest that if it isn't in Fontaine--Ouyang then it must be wrong? (The context of that section is that they're trying to show "de Rham implies potentially semistable" and restricting from $G_K$ to $G_{K^{nr}}$ simplifies matters. So they don't deal with the crystalline case because it isn't relevant for what they're trying to prove, not because it isn't true.) $\endgroup$ Jul 24, 2022 at 11:49

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