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Let $k \subset K$ be an extension of algebraically closed fields of characteristic $0$ (e.g. $\overline{\mathbb{Q}} \subset \mathbb{C}$).

Let X be a smooth variety over $k$. Then is the natural map $$\mathrm{Pic}(X)\,/\,n \to \mathrm{Pic}(X_K)\,/\,n$$ and isomorphism for all $n \in \mathbb{Z}$?

Here for an abelian group $A$ I denote by $A \, / \, n := A/nA$. Also $X_K$ denotes the base change of $X$ to $K$.

I can prove this if $X$ is proper using the theory of the Picard scheme. Here we have $\mathrm{Pic}^0(X)\,/\,n = 0$ (the set of points on an abelian variety over an algebraically closed field is divisible). The quotient $\mathrm{Pic}(X)/\mathrm{Pic}^0(X)$ is the Neron-Severi group, which is a finitely generated abelian group scheme hence its points don't change upon base change to $K$.

This proof breaks down for non-proper $X$ as the Picard functor is not representable in general.

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    $\begingroup$ Yes. By resolution of singularities we can realize $X$ as dense open in a smooth proper $k$-scheme $X'$, so by smoothness ${\rm{Pic}}(X')\to {\rm{Pic}}(X)$ is surjective. Hence, ${\rm{Pic}}(X')/(n)\to {\rm{Pic}}(X)/(n)$ is surjective and likewise over $K$, so the isomorphism property for $X'$ implies surjectivity of the map for $X$. Injectivity follows by the usual spreading-out arguments (descending a relation "$L_K \simeq N^{\otimes n}$" over $X_K$ to one over $X_A$ for a finitely generated $k$-subalgebra $A \subset K$, etc.). $\endgroup$ – nfdc23 Nov 28 '17 at 15:56
  • $\begingroup$ @Daniel: Do you mean "an extension of algebraically closed fields"? Do you mean $\overline{\Bbb Q}\subset \Bbb C$? $\endgroup$ – Mikhail Borovoi Nov 28 '17 at 16:05
  • $\begingroup$ @Borovoi: Yes of course; I've corrected the typo. $\endgroup$ – Daniel Loughran Nov 28 '17 at 16:08
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$\require{AMScd}$ The following result holds in much more generality than asked by the OP. There is no characteristic zero/smoothness assumption.

Proposition: Let $X/k$ be a qcqs scheme with $k$ separably closed. Let $K/k$ be a separable extension with $K$ separably closed. Let $n \in \mathbf{N}$ be a positive integer that is prime to the characteristic of $k$. Then the natural map $$\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} \to \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)}$$ is an isomorphism.

The proof goes as follows. The Kummer sequence gives the following commutative diagram with exact rows: $\require{AMScd}$ \begin{CD} 0 @>{}>>\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} @>{}>> H^2(X, \mu_n)@>{}>> H^2(X,\mathbf{G}_m) \\ {} @VVV @VVV @VVV {} \\ 0 @>{}>> \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)} @>{}>> H^2(X_K, \mu_n) @>{}>> H^2(X_K, \mathbf{G}_m) \end{CD}

The middle vertical arrow is an isomorphism by the smooth base change theorem in \'{e}tale cohomology (Corollary 1.3.1 of this):

Smooth Base Change: Suppose $K/k$ is an extension of separably closed fields, $X$ a qcqs $k$-scheme and $\mathscr{F}$ a torsion sheaf with torsion orders not divisible by the characteristic of $k$.Then, the base change map $$H^i(X,\mathscr{F}) \to H^i(X_K, \mathscr{F}_K)$$ is an isomorphism for all $i \geq 0$.

Therefore to prove the proposition, we reduce to proving the following injectivity statement concerning Brauer groups:

Proposition: Let $k \subseteq K$ be a separable extension of fields with $k$ separably closed. Let $X/k$ be a qcqs scheme. Then $H^2(X,\mathbf{G}_m)$ injects into $H^2(X_K, \mathbf{G}_m)$.

To prove this, write $K$ as a direct limit of smooth $k$-algebras. Since cohomology commutes with direct limits for qcqs schemes, if a (cohomological) Brauer class $\alpha$ in $H^2(X,\mathbf{G}_m)$ dies in $H^2(X_K,\mathbf{G}_m)$ it must die in $H^2(X_R, \mathbf{G}_m)$ for some smooth $k$-algebra $R$. By smoothness and using that $k$ is separably closed, $\operatorname{Spec} R$ has a $k$-rational point. Therefore the canonical map $f: X_{R} \to X$ has a section $g$ and so on cohomology the composition $$H^2(X,\mathbf{G}_m) \stackrel{f^\ast}{\to} H^2(X_R,\mathbf{G}_m) \stackrel{g^\ast}{\to} H^2(X,\mathbf{G}_m)$$ is the identity. We conclude that $\alpha = 0$ as desired.

Finally we remark that in the injectivity statement on Brauer groups, only the ground field is required to be separably closed. There is no requirement on $K$, only that $K/k$ be separable.

Edit: The proof also shows that for any \'{e}tale sheaf $\mathscr{F}$ on $X$ such that multiplication by $n$ is surjectve, then $$\frac{H^i(X,\mathscr{F})}{nH^i(X,\mathscr{F})} \to \frac{H^i(X_K,\mathscr{F})}{nH^i(X_K,\mathscr{F})} $$ is an isomorphism for all $i \geq 1$.

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  • $\begingroup$ Nice answer. I have a silly question though. Why do you say "By smoothness and [...]"? Does it not suffice that $R$ is a reduced $k$-finitely generated algebra for Spec $R$ to have a $k$-point? $\endgroup$ – Ariyan Javanpeykar Nov 29 '17 at 23:23
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    $\begingroup$ @AriyanJavanpeykar: What if $R$ is a purely inseparable finite extension of $k$? $\endgroup$ – Daniel Litt Nov 29 '17 at 23:37
  • $\begingroup$ @DanielLitt Thanks. That was bad of me. One needs that $R$ is geometrically reduced. $\endgroup$ – Ariyan Javanpeykar Nov 30 '17 at 0:18
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    $\begingroup$ @AriyanJavanpeykar Yes. See Lemma 1.7.4. here. $\endgroup$ – Ben Lim Nov 30 '17 at 0:42
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You may consult Florence Lecomte's paper: Rigidité des groupes de Chow, Duke Math. J. 53 (1986), no. 2, 405-426.

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There is also Uwe Jannsen's preprint http://www.mathematik.uni-regensburg.de/Jannsen/home/Preprints/RIGBAT.pdf (rigidity also for other functors)

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