Let $\mathbb{Q}\_p$ be the field of $p$-adic numbers. Consider an unramified representation $\rho : Gal(\bar{\mathbb{Q}}\_p / \mathbb{Q}\_p) \to \mathbb{F}\_p^{\times}$ which sends the arithmetic Frobenius to an element $\mu \in \mathbb{F}\_p^{\times}$.


I know that the corresponding $(\varphi, \Gamma)$-module is of rank 1 over $\mathbb{F}\_p((X))$ and that it admits a basis $e$ in which $\varphi(e) = 1 / \mu$ and the action of $\Gamma$ is trivial on $e$.


I realized that I am not able to prove it and it feels that it should be trivial. How can one obtain such a description ?



Rank one (phi,Gamma)-modules
Nice question! I remember doing this exercise myself once. This can be extracted from Fontaine's article in the Grothendieck Festschrift, but it takes a little bit of work. The key observation is that since your representation is unramified, it factors through $\operatorname{Gal}(\overline{\mathbb{F}}\_p / \mathbb{F}\_p)$, so it suffices to work with the much simpler period ring $\overline{\mathbb{F}}\_p$, which is contained in Fontaine's big ring $\mathbb{E}$. 


The idea is to show that $\overline{\mathbb{F}}\_p^\times$ contains an element $y$ such that $\varphi(y) = \mu^{-1} y$, where $\varphi$ is the arithmetic Frobenius. This you can see directly using Hilbert's theorem 90 applied to $\overline{\mathbb{F}}\_p / \mathbb{F}\_p$. Then $y \otimes 1$ is an element of $$(\overline{\mathbb{F}}\_p \otimes \rho)^{G\_{\mathbb{Q}\_p}} \subseteq (\mathbb{A} \otimes \rho)^{H\_{\mathbb{Q}\_p}}$$
and this is your basis vector $x$. Since it's non-zero, it's an $\mathbb{E}\_{\mathbb{Q}\_p}$-basis, and $\varphi$ visibly acts as $\mu^{-1}$.


