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Let $X$ be a connected noetherian scheme and $\ell$ a prime invertible on $X$. Let $D \subset X$ be a regular effective Cartier divisor (or more generally a normal crossings divisor, I suppose). Write $U = X \setminus D$ and let $\mathcal{F}$ be a lisse $\overline{\mathbb{Q}}_{\ell}$-sheaf on $U$. For any $\zeta \in D$ a generic point of $D$, write $K_{\zeta}$ for the field of fractions of $\mathcal{O}^h_{X,\zeta}$, and let $\overline{K_{\zeta}}$ be a separable closure of $K_{\zeta}$. By pulling back $\mathcal{F}$ along $\text{Spec} K_{\zeta} \rightarrow U$ we obtain a lisse sheaf on $\text{Spec} K_{\zeta}$, which corresponds to a continuous $\overline{\mathbb{Q}}_{\ell}$-linear representation $$ \rho_{\zeta} \colon \text{Gal}(\overline{K_{\zeta}}/K_{\zeta}) \rightarrow \text{GL}(\mathcal{F}_{\text{Spec} \overline{K_{\zeta}}}).$$ I will say that $\mathcal{F}$ is unramified along $D$ if for all such $\zeta$, $\rho_{\zeta}$ vanishes on the inertia group $I_{\zeta} \subset \text{Gal}(\overline{K_{\zeta}}/K_{\zeta})$. Now, is it true that if $\mathcal{F}$ is unramified along $D$, that it can then be extended to a lisse $\overline{\mathbb{Q}}_{\ell}$-sheaf on $X$?

A very similar statement for finite étale covers (or finite locally constant sheaves) $Y \rightarrow U$ is discussed in Tag0BSE. Can I somehow prove this statement by representing $\mathcal{F}$ by a lisse $\mathcal{O}_E$-sheaf $\mathcal{F}_{\mathcal{O}_E}$ for some $\ell$-adic field $\mathbb{Q}_{\ell} \subset E \subset \overline{\mathbb{Q}}_{\ell}$, and then extend each of the finite locally constant sheaves $\mathcal{F}_{\mathcal{O}_E} \otimes \mathcal{O}_E/\mathfrak{m}^n$ in a compatible way?

Thank you!

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    $\begingroup$ Is $X$ assumed to be regular (at least at every point of $D$)? $\endgroup$ Commented Apr 22 at 20:16
  • $\begingroup$ If $x$ is a point of $D$, then $D$ is cut out by a nonzero divisor $d \in \mathfrak{m}_x$ at $x$. By assumption, $\mathcal{O}_{D,x} = \mathcal{O}_{X,x}/d\mathcal{O}_{X,x}$ is regular. Then by Tag00NU, $\mathcal{O}_{X,x}$ is itself regular. I am fine with assuming that $X$ is regular everywhere, should this be necessary. @PiotrAchinger $\endgroup$
    – Hugo Zock
    Commented Apr 23 at 10:00

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