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Let $X$ be an algebraic scheme over an algebraically closed field of characteristic $p$, let $U$ be an open dense subset and let $\mathcal F$ be a lisse $\overline{\mathbb Q_l}$ sheaf on $U$.

It is known that $\mathcal F$ corresponds to a finite dimensional $\overline{\mathbb Q_l}$ vector space $V$ together with a continuous $\pi_1(U, \bar x)$-action (for an $x \in |U|$).

Here I wonder what is $(j_*\mathcal F)_{\bar s}$, where $s \in X-U$ and $j : U \rightarrow X$ is the open immersion.

The textbook I read says it is $V^{I_s}$, where $I_s \subseteq \pi_1(U, \bar x)$ is the "ramification group" at $s$, which seems to be the absolute Galois group of the quotient field of the strict henselization of the local ring at $s$ in $\bar s$.

I don't see why the absolute Galois group is contained in $\pi_1(U, \bar x)$ and $(j_*\mathcal F)_{\bar s} = V^{I_s}$.

Thanks in advance.

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  • $\begingroup$ Is $X$ a curve ? $\endgroup$
    – Roland
    Oct 12, 2022 at 13:13
  • $\begingroup$ @Roland Yes, an irreducible curve. $\endgroup$
    – Liu Y
    Oct 12, 2022 at 13:16
  • $\begingroup$ What textbook are you reading ? This is pretty standard (but not obvious) and should be covered in most textbook. Aren't there any explanations or references ? Or is there something in the explanation that you don't understand ? Do you know the topological analogy ? $\endgroup$
    – Roland
    Oct 12, 2022 at 13:25
  • $\begingroup$ @Roland I am reading Kiehl and Weissauer's book, and I could not find any explanations or relevant things. I don't know either what the topological analogy is. $\endgroup$
    – Liu Y
    Oct 12, 2022 at 14:31
  • $\begingroup$ @Roland Where can one find a textbook account? Though this can be proved via judicious usage of base change theorems, I recall how I got stuck here and there by such "small" gaps when learning etale cohomology. $\endgroup$ Oct 14, 2022 at 6:17

1 Answer 1

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Someone brought this to my attention so hopefully I can resolve this satisfactorily.

Let $X$ be a noetherian normal $1$-dimensional scheme over a field $k$. Let $S$ be a finite set of closed points of $X$ and let $U=X-S$. Let $\mathcal{F}$ be a local system on $U$ and so choosing the geometric point $\bar{\eta}$ lying above the generic point $\eta\in U$ we can equivalently think of $\mathcal{F}$ as the $\pi_1(U,\bar{\eta})$ representation $\mathcal{F}_{\bar{\eta}}$.

Now let $s\in S$ and consider the strict localisation $\mathcal{O}_{X,\bar{s}}^{sh}$ at a geometric point $\bar{s}$ lying above $s$. This is a dvr and note that its spectrum $Z=\operatorname{Spec}(\mathcal{O}_{X,\bar{s}}^{sh})$ consists of precisely two points, the closed point $\bar{s}$ and the generic point which we again denote by $\eta$ by abuse of notation. We let $K_{\bar{s}}^{sh}$ denote $\operatorname{Frac}(\mathcal{O}_{X,\bar{s}}^{sh})$. Notice that $Z^0:=\operatorname{Spec}(\mathcal{O}_{X,\bar{s}})-\{\bar{s}\}=\operatorname{Spec}(K_{\bar{s}}^{sh}).$ So in fact $\pi_1(Z^0,\bar{\eta})=\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh})$ maps into $\pi_1(U,\bar{\eta})$. The image of this morphism is the inertia group at $s$ denoted $I_s$.

Let $j\colon U\to X$ be the canonical immersion; we want to compute $(j_*\mathcal{F})_{\bar{s}}.$ To do this we compute that $$(j_*\mathcal{F})_{\bar{s}}=H^0(\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})\times_X U, p^*\mathcal{F})$$ where $p\colon \mathcal{O}^{sh}_{\bar{s}}\times_X U\to U$ is the projection.

So it is enough to understand the scheme $\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})\times_X U$. This is the fiber of the canonical morphism $j\colon U\to X$ above the image of $\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})$ in $X$. The fiber above the closed point $\bar{s}$ is empty since $s\in S=X-U$ and the fiber above the generic point is precisely the generic point of $U$. So the fiber is canonically isomorphic to $\operatorname{Spec}(K^{sh}_{\bar{s}})\times_U X$. By replacing $U$ by an affine cover containing $\eta$ and similarly for $X$ we see that $\operatorname{Spec}(K^{sh}_{\bar{s}})\times_U X=\operatorname{Spec}(K^{sh}_{\bar{s}}).$

Thus we have $$(j_*\mathcal{F})_{\bar{s}}=H^0(\operatorname{Spec}(K^{sh}_{\bar{s}}), \mathcal{F}_{\bar{\eta}}).$$

Since étale cohomology over a point is just Galois cohomology we have the right side is equal to

$$H^0(\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh}),\mathcal{F}_{\bar{\eta}})=\mathcal{F}_{\bar{\eta}}^{\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh})}=\mathcal{F}_{\bar{\eta}}^{I_s}.$$

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  • $\begingroup$ Thanks. The equation $(j_*\mathcal{F})_{\bar{s}}=H^0(\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})\times_X U, p^*\mathcal{F})$ seems to use some sort of base change theorem, but I can't see what it is. $\endgroup$
    – Liu Y
    Nov 1, 2022 at 8:11
  • $\begingroup$ The fact you mention uses only the definition of the stalk and the fact that étale cohomology commutes with colimits under mild separation conditions. See here stacks.math.columbia.edu/tag/03Q9 $\endgroup$
    – shubhankar
    Nov 1, 2022 at 11:22

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