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Int
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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 more explicitly?

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 more explicitly?

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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Int
  • 93
  • 4

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 more explicitly?

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 $r>0$ ?

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 more explicitly?

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Int
  • 93
  • 4

ramification of discrete valuation field

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 $r>0$ ?