1
$\begingroup$

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.
$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .