Let $K/\mathbb{Q}_p$ be a finite extension and let $q \in K^{\times}$ be such that $|q| <1$. Let $E_q:= \bar{K}^{\times}/q^{\mathbb{Z}}$ be the Tate curve where $q^{\mathbb{Z}}:= \{q^n| n\in \mathbb{Z}\}$. If we define $V_p(E_q):= T_p(E_q)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$ where $T_p(E_q):= \varprojlim_n E_q[p^n]$, with transition maps being multiplication by $p$. I was recently able to prove (with a lot of guidance) that $V_p(E_q)$ is a De Rham representation. Subsequently it was natural to ask if it is crystalline or not. And while in Berger's article (An introduction to the theory of p-adic representations https://arxiv.org/pdf/math/0210184.pdf) we see a very beautifully worked out proof of $V_p(E_q)$ being De Rham, I am finiding it hard to prove(or find a source of) that it is not crystalline. While I understand why the tactics used to prove that it is De Rham fails to give me a $G_K$ invariant basis of $D_{cris}(V_p(E_q))$ but that does not mean there can not be other ones. I will be glad to see a hands-on, computational proof of this.

A couple of comments: a) I read somewhere it is not crystalline and also was told that morally the purpose of Crystalline representations was to detect good reduction, so Tate module of the Tate curve should not be Crystalline. I would also like some elaboration on this comment

b) A possible strategy that was suggested to me was to prove that it is semi-stable with something about the monodromy operator (that it is non-zero?) would give me the desired result. But if possible I want to avoid using the machinery of semi-stable representations.

I apologize for being vague because of my lack of understanding,especially with my comments. And I appreciate any helpful comments or answers.

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    $\begingroup$ The Tate module of the Tate elliptic curve is absolutely the canonical example of a semistable non-crystalline representation, so it seems positively perverse not to go via a computation of the monodromy operator as in (b). $\endgroup$ – David Loeffler Mar 19 '17 at 8:38
  • $\begingroup$ Ah I see! Thanks for pointing this out. $\endgroup$ – Aranya Lahiri Mar 19 '17 at 15:14

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