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Maybe the answer to my question is obvious.

Let $p$ be a prime $\geq 3$. Let $D$ be an étale $(\varphi, \Gamma)$-module over $A_{\mathbb{Q}_p} = \{ \sum_{n \in \mathbb{Z}} a_n X^n \, \vert \, a_n \in \mathbb{Z}_p, a_n \to 0 \mbox{ as } n\to -\infty \}$ of finite rank.

Recall that the action of $\varphi$ is given by $\varphi(X) = (1+X)^p -1$ and that of $\gamma \in \Gamma$ by $\gamma(X) = (1+X)^{\omega(\gamma)} -1$ where $\omega$ is the $p$-adic cyclotomic character.

Denote by $A^+$ the subring of $A_{\mathbb{Q}_p}$ of power series without denominators, that is $A^+ = \{ \sum_{n \in \mathbb{N}} a_n X^n \, \vert \, a_n \in \mathbb{Z}_p \}$.

Say that $D$ is of finite height if $D= D^+ \otimes_{A^+} A_{\mathbb{Q}_p}$ where $D^+$ is an étale $(\varphi, \Gamma)$-module over $A^+$.

Is there an étale $(\varphi, \Gamma)$-module $D$ over $A_{\mathbb{Q}_p}$ (of finite rank) which is not of finite height and such that for all étale $(\varphi, \Gamma)$-module of rank $1$ $\delta$ over $A_{\mathbb{Q}_p}$, $D \otimes \delta$ is not of finite height ?

Obviously if such an object exists, it should be of rank $\geq 2$.

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1 Answer 1

up vote 8 down vote accepted

Proposition: If $D$ is not of finite height, then nor is $D \otimes \delta$ for any $\delta$ of rank 1.

Proof: It suffices to show the contrapositive: if $D$ is of finite height so is $D \otimes \delta$ for any rank 1 $\delta$. This follows easily from the fact that every continuous character of $G_{\mathbf{Q}_p}$ (thus any etale rank 1 $(\varphi, \Gamma)$-module) is the product of an unramified character and a character factoring through $\Gamma$; twisting by a character of either type will just change the actions of $\varphi$ and of a generator of $\Gamma$ on $D$ by elements of $\mathbf{Z}_p^\times$, and thus sends a $(\varphi, \Gamma)$-stable $\mathbf{A}^+_{\mathbf{Q}_p}$-sublattice in $D$ to one in $D(\delta)$.

Since representations that aren't of finite height do exist, the answer to your question is thus "yes" -- any non-finite-height $D$ will do.

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This was indeed not so hard. Thanks for your anwser ! –  user33624 Sep 19 '13 at 10:34
    
Hmmmm, David, it seems to me that you are proving the converse, not the contrapositive... –  Chris Wuthrich Sep 19 '13 at 18:34
    
It's the contrapositive, but with $D$ and $\delta$ relabelled as $D \otimes \delta$ and $\delta^{-1}$, so it is logically equivalent to the desired statement. –  David Loeffler Sep 19 '13 at 20:04
    
Oh, yes, of course, I see. Thanks –  Chris Wuthrich Sep 19 '13 at 21:02
    
"Since representations that aren't of finite height do exist": take a semistable noncrystalline representation. –  Laurent Berger Oct 1 '13 at 12:49

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