Morphism of admissible $1$-dimensional $(\varphi,N)$-module

Question

In these notes in section 7.3 admissible one-dimensional $(\varphi,N)$-modules are classified. Let $K$ be a $p$-adic field. Then for any $\lambda\in K_0$, they give an element of $MF_K^{ad}(\varphi,N)$ called $D_\lambda$, whose Frobenius is $\varphi=\lambda \sigma$, the monodromy operator is $N=0$, and $$Fil^r(D_{\lambda}\otimes_{K_0} K)=\begin{cases} D_{\lambda}\otimes_{K_0} K&, r\leq v_p(\lambda)\\ 0&, r>v_p(\lambda)\end{cases}.$$ Then a morphism $D_\lambda\to D_\mu$ is given by a $K_0$-linear morphism given by multypling $u\in K_0^\times$ such that $$\lambda=\frac{\sigma(u)}{u}\mu.$$ However, in the notes it is claimed that $u$ actually has to lie in $\mathcal{O}_{K_0}^{\times}$ and I don't understand why. Clearly I'm missing something and I'd appreciate any help.

Answer 1

I admit the way it is phrased in the notes seems a bit ambiguous / confusing, but the actual statement there is that there is an isomorphism $D_\lambda \simeq D_\mu$ if and only if there exists a $u \in \mathcal{O}_{K_0}^\times$ such that $\mu = \dfrac{\sigma(u)}{u} \lambda$.

Now if $D_\lambda \simeq D_\mu$, then indeed a priori there is just a $\tilde u \in K_0^\times$ such that the iso is given by multiplication with $\tilde u$. But now note that $p$ is a prime element of $\mathcal{O}_{K_0}$ (which is totally unramified), i.e. $\tilde u = u \cdot p^n$ for some $n \in \mathbb Z$ and $u \in \mathcal{O}_{K_0}^\times$, and because $p \in \mathbb Z$ is fixed by $\sigma$ we have$\dfrac{\sigma( \tilde u)}{\tilde u} = \dfrac{\sigma(u)}{u}$.

So indeed there exists a $u \in \mathcal{O}_{K_0}^\times$ such that ..., it's just that the iso is not necessarily given by multiplication with that $u$, but by multiplication with some $u \cdot p^n$.