Why does having an integral model make étale cohomology unramified?

Question

Let $K$ be a number field, and $\mathcal O$ its ring of integers. Fix a finite set $S$ of rational primes, and let $\mathcal O_S$ be the ring of integers with these primes inverted.

Let $X$ be a smooth proper scheme over $K$. Suppose that $X$ arises as the base change of a smooth proper $\mathfrak X$ defined over $\mathcal O_S$. Fix a non-archimedean place $v$ of $K$ not lying over any prime in $S$, and write $K_v$ for the completion of $K$ at $v$.

Let us consider the étale cohomology groups $\mathrm{H}^i(X_v \times_{K_v} \overline K_v, \mathbf{Q}_\ell)$ as Galois representations of $\mathrm{Gal}(\overline K_v/K_v)$. It seems that the existence of $\mathfrak X$ implies that these representations are unramified at $v$ (i.e. the inertia group $I_v$ acts trivially). Why is this the case?

Answer 1

This is really a local question: you may as well assume $X$ is defined over a local field $L$, it doesn't matter whether it comes from a number field.

The point is that if $S = \operatorname{Spec} O_L$ and $\pi: \mathfrak{X} \to S$ is the (smooth proper) structure map, then for any locally constant torsion sheaf $\mathcal{F}$ on $\mathfrak{X}$ whose order is invertible on $S$, there is a locally constant sheaf $R^i\pi_\star\mathcal{F}$ on $S$ whose fibre at any geometric point $\bar{x}$ is $H^i(\mathfrak{X}_{\bar{x}}, \mathcal{F}_{\bar{x}})$; see Corollary VI.4.2 of Milne's "Lectures on Etale Cohomology" for the details. Since $S$ is connected, we get an isomorphism between the fibres at the closed point of $S$ and the generic point. That is, $$H^i(X, \mathcal{F}) = H^i(X_0, \mathcal{F}_0)$$ where $X_0$ is the special fibre of $\mathfrak{X}$. This isomorphism is Galois-equivariant; but the inertia subgroup acts trivially on $X_0$. Hence $H^i(X, \mathcal{F})$ is unramified.