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Let $X$ be a quasi-projective variety over a number field $E$. Let $v$ be a place in $E$ and let $E_v$ denote the $v$-adic completion. Let $p$ be the prime number determined by $v$, and let $\Lambda = \mathbb Z_{\ell}/\ell^k\mathbb Z_{\ell}$, where $\ell$ is a prime number different from $p$ and $k\geq 1$.

Under what conditions do we have an isomorphism between the etale cohomology of $X/E$ and the etale cohomology of $X_v/E_v$, both with coefficients in $\Lambda$?

In particular, I am interested in the case where $v$ is non-archimedean.

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You shouldn't expect such an isomorphism in general, even if $X$ is projective. When $X=\mathrm{Spec}(E)$, you are asking whether $H^i(E,\mathbb Z/\ell^k)\cong H^i(E_v,\mathbb Z/\ell^k)$.

For example when $E$ contains a $\ell$-th root of unity, then $\mathbb Z/\ell\cong\mu_\ell$ as Galois modules, and by Hilbert's 90, the short exact sequence $$1\to\mu_\ell\to\mathbb G_m\xrightarrow{\ell}\mathbb G_m\to 1,$$ so that $$1\to \mu_\ell\to E^\times\xrightarrow\ell E^\times\to H^1(E,\mu_\ell)\to 1,$$ so that $H^1(E,\mu_\ell)\cong E^\times/(E^\times)^\ell$ while $H^1(E_v,\mu_\ell)\cong E_v^\times/(E_v^\times)^\ell$. These are far from isomorphic in general: $E^\times/(E^\times)^\ell$ is infinitely-generated, while $E_v^\times/(E_v^\times)^\ell$ is certainly finite.

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  • $\begingroup$ Thank you, you are absolutely right! $\endgroup$
    – Suzet
    Jul 19 at 6:08

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