# Understanding the proof of arithmetic local monodromy theorem

I am reading the proof of Grothendieck's $$\ell$$-adic local monodromy theorem in the book Theory of $$p$$-adic representations by Fontaine and and Ouyang. More concretely, let $$K$$ be a $$p$$-adic field, $$I_K=\mathrm{Gal}(K^{unr}/K)$$ the inertia group, $$P_{K,\ell} = \mathrm{Gal}(K^{tr,\ell}/K)$$, $$G_{K,\ell} = \mathrm{Gal}(K^{tr,\ell}/K)$$, where $$K^{unr},K^{tr,\ell}$$ denote the maximal unramified extension and the $$\ell$$-part of the maximal tamely ramified extension, respectively. It is known that $$I_K/P_{K,\ell}\simeq \mathbb{Z}_{\ell}(1)$$ and there is a natural injection $$I_K/P_{K,\ell} \hookrightarrow G_{K,\ell}$$.

Let $$t$$ be a topological generator of $$\mathbb{Z}_{\ell}(1)$$, then I do not see why (as the authors claimed)

• For any $$g \in G_{K,\ell}$$, $$gtg^{-1}=t^{\chi_{\ell}(g)}$$ with $$\chi_{\ell} \colon G_{K,\ell} \longrightarrow \mathbb{Z}_{\ell}^{\times}$$ is some character. Since $$K^{tr,\ell}=\bigcup K^{unr}(\sqrt[\ell^n]{\pi})$$ ($$\pi$$ a uniformizer), I assume that we can compute $$(gtg^{-1})$$ on each $$\sqrt[\ell^n]{\pi}$$. As $$g$$ permutes the roots, $$g\sqrt[\ell^n]{\pi}=\zeta_{\ell^n}\sqrt[\ell^n]{\pi}$$ for $$\zeta$$ some root of unity. This may reduce the question to finding some element $$\bullet$$ s.t. $$t\sqrt[\ell^n]{\pi} = \zeta_{\ell^n}^{\bullet}\sqrt[\ell^n]{\pi}$$.
• One more thing is I am not sure how to make sense of the element $$t^{\chi_{\ell}(g)}$$ as $$\chi_{\ell}(g)$$ lies in $$\mathbb{Z}_{\ell}^{\times}$$, not $$\mathbb{Z}$$. I assume that $$t^{\chi_{\ell}(g)}$$ is a limit of the form $$\varprojlim_{m \in \mathbb{Z},m \to \chi_{\ell}(g)}t^m$$.
• Suppose that no finite extension of the residue field contains all roots of unity of order a power of $$\ell$$, then $$\mathrm{Im}(\chi_{\ell})$$ is an open subgroup of $$\mathbb{Z}_{\ell}^{\times}$$ and hence $$\chi_{\ell}$$ must take infinitely many integral values. I do not see why, an online note here seems to face to same problem with me.

For the first and second point, the key is that $$\mathbb{Z}_{\ell}(1)$$ is a free $$\mathbb{Z}_{\ell}$$-module of rank one. In particular, its automorphisms are given by multiplication by a unit in $$\mathbb{Z}_{\ell}$$.
Concretely, an element of $$\mathbb{Z}_{\ell}(1)$$ is a collection $$(\zeta_n)_{n \geq 1}$$ of roots of unity such that $$\zeta_1^{\ell}=1$$ and $$\zeta_{n+1}^{\ell}=\zeta_n$$.
$$\mathbb{Z}_{\ell}$$ acts on $$\mathbb{Z}_{\ell}(1)$$ as follows: if $$x \in \mathbb{Z}_{\ell}$$ and $$(\zeta_n)_{n \geq 1} \in \mathbb{Z}_{\ell}(1)$$, then $$x \cdot (\zeta_n)_{n \geq 1} = (\zeta_n^{x \text{ mod }\ell^n})_n$$.
For your third question, the image $$I$$ of $$\chi_{\ell}$$ is a compact (hence closed) subgroup of $$\mathbb{Z}_{\ell}^{\times} \simeq D_{\ell} \oplus \mathbb{Z}_{\ell}$$, where $$D_{\ell}$$ is a finite cyclic group of order $$\max(2,\ell-1)$$.
If the projection of $$I$$ to $$\mathbb{Z}_{\ell}$$ (which is again a closed subgroup) is zero, then $$I$$ is finite, hence (by definition of $$\chi_{\ell}$$) some finite extension of $$K$$ contains all the $$\ell^n$$-th roots of unity.
If not, since $$D_{\ell}$$ is finite, it implies that $$I$$ has finite index: since it is the complement of the reunion of finitely many (compact hence closed) $$I$$-cosets, $$I$$ is open.