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Let $K$ be a discrete valuation field with valuation $v:K\rightarrow \mathbb Z\cup \{\infty\}$ which is normalized by $v(\pi)=1$ for a prime element $\pi$. Let $v:\overline K\rightarrow \mathbb Q\cup\{\infty\}$ be an extension of the normalized discrete valuation $v$ to a separable closure of $K$. For a rational number $r$, we put $\mathfrak m^r=\{a\in\overline K|v(a)\geq r\}$ and $\mathfrak m^{r+}=\{a\in\overline K| v(a)>r\}$. Let $I$ be the inertia subgroup of $G$.

Question: Is there a canonical action of $I$ on the vector space $\mathfrak m^r/\mathfrak m^{r+}$ for any rational number $r>0$ ? Moreover,Is this action can be given explicitly?

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up vote 3 down vote accepted

Yes, there is. Inertia is a subgroup of the decomposition group which by definition preserves the extended valuation. Consequently, the action of $I$ on $\overline{K}$ preserves both $\mathfrak{m}^r$ and $\mathfrak{m}^{r+}$, hence induces the action on the quotient. The induced action on the quotient preserves the $k = \mathcal{O}_{\overline{K}}/\mathfrak{m}$-vector space structure because by definition the action induced by $I$ on the residue field $k$ is trivial.

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Thank you very much. Moreover,Is this action can be given more explicitly? – Int Apr 24 '13 at 12:58
Yes, it can. See Serre "Proprietes galoissienes..." Prop. 7 in section 1.8. – Kestutis Cesnavicius Apr 24 '13 at 13:10
Wonderful~You really do me a great favor! Thanks~~ – Int Apr 24 '13 at 13:16

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