The Étale Cohomology from the Variety to its Generic Point

Question

Let $n\in \mathbb{Z}_{>0}$ and $X$ be a smooth projective variety over $k$, where $k$ contains all $n$-th roots of unity and $\operatorname{char}(k)=p$. Here $p$ and $n$ are coprime. I wonder whether the following map is injective:

$$f: H^{1}_{\text{ét}}(X, \mu_{n})\to H^{1}_{\text{ét}}(K(X), \mu_{n})$$ where $K(X)$ is the function field of $X$, $\mu_{n}$ is the $n$-th cyclic group and the morphism is induced by $\operatorname{Spec}K(X)\to X$.

By applying the $0\to \mu_{n}\to \mathbb{G}_{m}\to \mathbb{G}_{m}\to 0$ to $X$ and $K(X)$, we can get $0\to \Gamma(X,\mathcal{O}_{X})^{*}/\Gamma(X,\mathcal{O}_{X})^{*n}\to H^{1}_{\text{ét}}(X, \mu_{n})\to \operatorname{Pic}(X)[n]\to 0$ and $H^{1}_{\text{ét}}(K(X),\mu_{n})\cong K(X)^{*}/K(X)^{*n}$.

By snake lemma, $f$ is injective if and only if $\operatorname{Pic}(X)[n]\to K(X)^{*}/K(X)^{*n}/(\Gamma(X,\mathcal{O}_{X})^{*}/\Gamma(X,\mathcal{O}_{X})^{*n})$ is injective.

I think the map should be this: Given a $n$-torsion line bundle, we can construct a $n$-th cyclic cover of $X$, it will induce a $n$-th cyclic extension of $K(X)$. So if line bundle map to $1$, it should be trivial line bundle. Thus $f$ should be injective.

Am I correct? It seems this is not a rigorous. Can you give a rigorous proof?

Also, if this is true, it seems this is also true when $X$ is just a locally factorial variety. Am I right?

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I asked the question at Math Stack Exchange yesterday. But I did not get an answer. So I ask here again. Hope this is OK.

Answer 1

Let $j : \eta \to X$ denote the inclusion of the generic point into $X$. From the low degree terms of the Leray spectral sequence, we have an exact sequence $$0 \to H^1(X, j_\ast \mu_n) \to H^1(\eta, \mu_n) $$ and thus it remains to prove that $j_\ast \mu_n = \mu_n$. Let $U \to X$ be an affine etale open. Then since normality is local in the etale topology, $U$ is also normal. In particular, connectedness is equivalent to irreducibility and thus we may assume that $U$ itself is the spectrum of a normal domain $A$ with fraction field $K$.

Let $f\in K$ satisfy $f^n = 1$. Then $f$ is integral over $A$. But $A$ is integrally closed in its field of fractions and hence $f \in A$. This proves that $j_\ast\mu_n = \mu_n$.