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Let $p$ be a prime number $\geq 3$. Let $V$ be a representation of $Gal(\bar{\mathbb{Q}}_p/ \mathbb{Q}_p)$ with coefficients in $\mathbb{F}_p$. Assume $V$ is a non-split extension of two characters $\delta_1$ and $\delta_2$.

Let $D$ be the $(\varphi, \Gamma)$-module associated to $V$. It is a non-split extension of two $(\varphi, \Gamma)$ of rank $1$. Denote by $\Phi$ the matrix of $\varphi$ and by $G$ the matrix of a generator $\gamma$ of $\mathbb{Z}_p^*$. In a suitable basis, we have $\Phi = \begin{pmatrix} f_1 & f \\ 0 & f_2 \end{pmatrix}$ and $G = \begin{pmatrix} g_1 & g \\ 0 & g_2 \end{pmatrix}$ where $f_1, f_2, g_1$ and $g_2$ are elements of $\mathbb{F}_p^{\times}$ (see One dimensional (phi,Gamma)-modules in char p) and where $f,g \in \mathbb{F}_p ((X))$.

Can we find a basis where $f$ and $g$ are in $\mathbb{F}_p[[X]]$ ? If not, what is the "nicest" form possible for $f$ and $g$ ? (i.e. can we kill some of the denominators and which one ?)

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up vote 4 down vote accepted

The answer is "no" in general. If $\delta_2=1$ and $\delta_1$ is the mod $p$ cyclotomic character, then there are both peu and très ramifiées extensions. Let $res(g)$ denote the coefficient of $1/X$ in $g$. Theorem: in your notation, the extension is peu ramifiée iff $res(g)=0$. See for instance proposition 3.7.5 of Mathieu Vienney's PhD

As to the "nicest" possible form: you can construct elements of $H^1(G_{Q_p},V)$ from elements of $D(V)^{\psi=1}$, see lemma I.5.5 of Cherbonnier and Colmez' JAMS paper. If you make this explicit, you'll have strong restrictions on $f$ and $g$.

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Thanks for the reference. I am a little bit confused though. If $V$ is an extension of $1$ by $\chi_p$, then it is always the reduction of a crystalline representation (if the extension is très ramifié one can choose the weights to be 0 and p). But a crystalline representation is of finite height so the matrices of the action of $\varphi$ and $\gamma$ should not have denominators. What am I missing here ? – user33624 Aug 26 '13 at 10:10
"Finite height" does mean that in some basis, there are no denominators. This does not imply the same property holds in your favorite basis :) – Laurent Berger Aug 27 '13 at 6:31
Oh I see. Thanks for the insight. – user33624 Aug 27 '13 at 10:27

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