$\newcommand{\ur}{\mathrm{ur}}\newcommand{\cris}{\mathrm{cris}}$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.



Crystalline when restricted to inertial subgroup
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.)


