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Denote by $G_p$ a choice of an absolute Galois group of $Q_p$, the field of $p$-adic numbers. Consider a continuous representations of $G_p$ on a $3$-dimensional $Q_p$ vector space that is a successive extension of the trivial character, the cyclotomic character ($\chi_p$) and the square of the cyclotomic character, i.e. there are exact sequences of $G_p$-modules :

$$0 \to Q_p(2) \to V \to W \to 0 $$ and $$0 \to U \to V \to Q_p \to 0$$

where $W$ is an extension $0 \to Q_p(1) \to W \to Q_p \to 0$ and $U$ is an extension $0 \to Q_p(2) \to U \to Q_p(1) \to 0$. Here $Q_p(n)$ denote a 1 dimensional $Q_p$ vector space endowed with an action of $G_p$ via the $n$-th power of the cyclotomic character.

Assume $U$ and $W$ to be crystalline, that is the $Q_p$-vector spaces $D_{crys}(U) = (B_{crys} \otimes_{Q_p} U)^{G_p}$ and $D_{crys}(W) = (B_{crys} \otimes_{Q_p} W)^{G_p}$ are both of dimension 2 ($B_{crys}$ is one of the ring of $p$-adic periods introduced by Fontaine).

In this case, is $V$ crystalline ?

It is the case if and only if the cocycle $[V] \in H^1(G_p, U)$ determined by (the extension determined by) $V$ is in the kernel of the map $H^1(G_p, U) \to H^1(G_p, B_{crys}\otimes_{Q_p}U)$, which is usually denoted by $H^1_{crys}(G_p, U)$.

One can show that in our case $H^{1}_{crys}(G_p,U)$ is 2 dimensional (there is a formula involving the dimension of $U$, the dimension of its invariants under $G_p$ and the Hodge-Tate weights). Now $H^{1}_{crys}(G_p,Q_p(2))$ and $H^{1}_{crys}(G_p,Q_p)$ are both 1 dimensional over $Q_p$ so we have an exact sequence of $Q_p$ vector spaces $$0 \to H^{1}_{crys}(G_p,Q_p(2)) \to H^{1}_{crys}(G_p,U) \to H^{1}_{crys}(G_p,Q_p) \to 0.$$

From there I don't know if there is more one can say...

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  • $\begingroup$ You don't say anything about having tried to just do the calculation on the semi-linear algebra side. (As you know, part of the power of these notions is to turn Galois-theoretic calculations into (semi-)linear algebra ones that can be easier or more tractable, all the more so over a ground field like $\mathbf{Q}_p$ for which the Frobenius operator is the identity and all semi-linearity is just usual linearity.) Have you tried to do the work on the linear algebra side rather than by looking at Galois cohomology? $\endgroup$ – user74230 Feb 8 '15 at 19:06
  • $\begingroup$ This is a good remark. Actually I didn't think of trying to do the calculations because I had in mind the case where the base field is not necessarily $Q_p$... $\endgroup$ – user65490 Feb 8 '15 at 19:34
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One has a natural exact sequence: $$0 \rightarrow H^1(G_p, \mathbb{Q}_p(2)) \rightarrow H^1(G_p,U) \xrightarrow{s} H^1(G_p,\mathbb{Q}_p(1)) \rightarrow 0,$$ where $ H^1(G_p, \mathbb{Q}_p(2)) \cong H^1_{crys}(G_p,\mathbb{Q}_p(2))$, $\dim H^1(G_p,U)=\dim H^1_{crys}(G_p,U)+1$, $\dim H^1(G_p, \mathbb{Q}_p(1))=\dim H^1_{crys}(G_p, \mathbb{Q}_p(1))+1$. So $s^{-1}\big(H^1_{crys}(G_p, \mathbb{Q}_p(1))\big)=H^1_{crys}(G_p,U)$. In other words, if $U$ is crystalline, then $V$ is crystalline $\Leftrightarrow$ $W$ is crystalline.

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