# Why does having an integral model make étale cohomology unramified?

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?

• How detailed an answer would you like? I’m confident the answer starts with the proper base change theorem – but it’s not enough, since any abelian variety has a proper model as a variety but does not always have unramified $\ell$-adic cohomology (by Néron-Ogg-Shafarevich). Jan 21, 2022 at 8:21
• I'm happy to see as much detail as you're willing to add. Nothing about this subject is too basic or boring for me to review again.
– Bun
Jan 21, 2022 at 8:22

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.