
> 
> Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map
>  \begin{equation}
>  \mathrm{Pic}(X) \to (\mathrm{Pic}(X\_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1)
> \end{equation}
>  surjective?
> 
> 
> 


Remarks:


1. If $X$ is a proper, then the map (1) is in fact an isomorphism. This is usually proved using the Hochshild--Serre spectral sequence.
2. The map (1) need not be injective in general. Take $X$ to be the complement of a closed point of degree two in $\mathbb{P}^1\_k$. Then $\mathrm{Pic}(X) = \mathbb{Z}/2\mathbb{Z}$ but $\mathrm{Pic}(X\_{\bar{k}}) = 0$.



Galois invariant Picard group elements
**Updated. The example by @Lucifer is completely correct.** Thanks to @Count Dracula who explained the example proposed by @Lucifer. 
That example is fine. I am keeping the counterexamples below, since they arise in a different way: as Severi-Brauer schemes over multiplicative group schemes over a field (and schemes mapping to such Severi-Brauer schemes). The counterexamples are given after a general statement that can (sometimes) be used to prove surjectivity of the morphism (1). 


**Setup.** Denote by $S$ the scheme $\text{Spec}\ k$ (to save typing). As in Borovoi-- Colliot-Thélène -- Skorobogatov, denote by $\mathfrak{g}$ the Galois group $\text{Gal}(\overline{k}/k)$. Let $\pi:X\to S$ be a smooth, separated, quasi-compact morphism with geometrically irreducible fiber. Let $\sigma:S\to X$ be a section of $\pi$. Denote by $G\_X$, resp. $G\_{X,\sigma}$, the group $S$-scheme that represents the functor sending every smooth $S$-scheme $T$ to $\mathbb{G}\_{m,X}(X\times\_S T)$, resp. to the kernel of $$\sigma^\*(T):\mathbb{G}\_{m,X}(X\times\_S T) \to \mathbb{G}\_{m,S}(T).$$ Pullback by $\sigma$ defines a group homomorphism, $$\sigma^\*:\text{Br}(X)\to \text{Br}(S).$$ Denote the kernel by $\text{Br}(X)\_\sigma$. Pullback by $\pi$ defines a group homomorphism, $$\pi^\*:H^2\_{\mathfrak{g}}(G\_{X,\sigma}(\overline{k})) \to \text{Br}(X)\_\sigma.$$


**Lemma.** The cokernel of the natural restriction morphism (1) equals the kernel of $H^2\_{\mathfrak{g}}(G\_{X,\sigma}(\overline{k})) \to \text{Br}(X)\_\sigma$. The group $G\_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. In particular, the cokernel of the natural restriction morphism is a torsion group.


**Proof.** The relevant terms in the Hochschild-Serre spectral sequence give the following long exact sequence, $$0 \to H^1\_{\mathfrak{g}}(G\_X(\overline{k})) \to \text{Pic}(X)\xrightarrow{\text{res}} \text{Pic}(X\_{\overline{k}})^{\mathfrak{g}} \xrightarrow{\delta} H^2\_{\mathfrak{g}}(G\_X(\overline{k})) \to \text{Br}(X).$$ By Hilbert's Theorem 90, the first term is $H^1\_{\mathfrak{g}}(G\_{X,\sigma}(\overline{k}))$. Pullback by the section $\sigma$ defines a splitting of the map $$H^2\_{\mathfrak{g}}(\mathbb{G}\_{m,S}(\overline{k}))\to \text{Br}(X).$$ Thus, the cokernel of $\text{res}$ equals the kernel of $\pi^\*$.


The group $G\_{X,\sigma}(\overline{k})$ is finitely generated by Rosenlicht. Every torsion-element satisfies an integral equation over $\overline{k}$. Since $X\_{\overline{k}}$ is integral, $\overline{k}$ is integrally closed in $H^0(X\_{\overline{k}},\mathcal{O}\_X)$. Thus, $G\_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. The profinite Galois cohomology group $H^2\_{\mathfrak{g}}(G\_\sigma(\overline{k}))$ is torsion. Thus, $H^2\_{\mathfrak{g}}(G\_\sigma(\overline{k}))$ is a torsion group. **QED**


The contravariant functor $G$ that sends a pointed $S$-scheme, $(X,\sigma)$, to the group scheme $G\_{X,\sigma}$ has an adjoint. For every smooth group $S$-scheme $M$ with $M\times\_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ isomorphic to $\mathbb{G}\_{m,\overline{k}}^\rho$, the group scheme $G\_{M,e}$ is the Cartier dual group scheme $M^D$. For every morphism of group $S$-schemes, $$\phi:M^D\to G\_{X,\sigma},$$ there is a unique $S$-morphism, $$f:X\to M,$$ sending $\sigma$ to the identity section $e$ and such that the pullback map on unit group $S$-schemes is $\phi$.


**Proposition.** The Galois cohomology group $H^2\_{\mathfrak{g}}(M^D)$ is canonically isomorphic to the kernel, $\text{Br}\_1(M)$, of the pullback map $$\text{pr}\_M^\*:\text{Br}(M)\to \text{Br}(M\times\_{\text{Spec}\ k} \text{Spec}\ \overline{k}).$$ 


**Proof.** Since $\text{Pic}(\mathbb{G}\_{m,\overline{k}}^\rho)$ is the trivial group, this follows from the previous lemma. **QED**


The Cartier dual group $S$-scheme to $G\_\sigma$ is a smooth group $S$-scheme $M$ of multiplicative type. There is a unique $S$-morphism $$f:X\to M,$$ mapping $\sigma$ to the identity section $e$ and such that the induced homomorphism of groups of units is an isomorphism of $S$-group schemes. 


**Proposition.** The cokernel of $\text{res}$ is identified with the kernel of the group homomorphism $$f^\*:\text{Br}\_1(M) \to \text{Br}(X).$$
In particular, if there exists a degree $1$ zero-cycle in the generic fiber of $f$, then $\text{res}$ is surjective.


**Proof.** This follows from the previous proposition and functoriality of the Hochschild-Serre spectral sequence with respect to $f$. If the fiber of $f$ over $\eta=\text{Spec}\ k(M)$ has a degree $1$ zero-cycle, then the pullback map $\text{Br}(k(M))\to \text{Br}(X\_\eta)$ is injective by the lemma. Since $M$ is smooth, it follows that also $f^\*$ is injective. **QED**


**Counterexamples.** This suggests how to construct a counterexample, namely as a Severi-Brauer scheme over a multiplicative group scheme $M$ over $k$. 


Let $k$ be a finite field $\mathbb{F}\_q$. Let $M$ be $\mathbb{G}\_{m,k}= \text{Spec}\ k[t,t^{-1}]$. The Cartier dual group scheme $M^D$ has $M^D(k) = M^D(\overline{k}) \cong \mathbb{Z}$, generated by the element $t$. 


For every integer $\ell$, let $k\_\ell=\mathbb{F}\_{q^\ell}$ be a degree $\ell$ extension of $k$ with cyclic Galois group generated by a Frobenius map. This extension together with the generator of the Galois group is unique up to unique isomorphism. Associated to the cyclic extension $k\_\ell/k$ and the element $t\in M^D(k)$, there is a cyclic algebra $A\_{\ell,t}$ that is a locally free $\mathcal{O}\_M$-module of rank $\ell^2$ and that is an Azumaya algebra over $\mathcal{O}\_M$. The Brauer class of $A\_{\ell,t}$ in $\text{Br}(M)$ is an element of order $\ell$ generating the full $\ell$-torsion subgroup of $\text{Br}(M)$. By Tsen's Theorem, the Brauer group of $M\times\_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is trivial. Thus, the class of $A\_{\ell,t}$ generates a cyclic subgroup of order $\ell$ in $\text{Br}(M)\_k$. 


Now let $f:X\to M$ be the Severi-Brauer scheme parameterizing left ideals in $A\_{\ell,t}$ of rank $\ell$. Since the Brauer group of $k$ is trivial (by Wedderburn or Chevalley-Warning), the restriction of $f$ over the $k$-rational point $e\in M(k)$ is a trivial Severi-Brauer scheme. Thus, there exists a section $$\sigma:\text{Spec}\ k \to X,$$ with image in the fiber of $f$ over $e$. The Picard group of $X\times\_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is $\mathbb{Z}$, and the generator restricts on each projective space (geometric) fiber of $f$ to a generator $\mathcal{O}(1)$ of the projective space. However, there cannot be such an invertible sheaf on $X$, or else $\mathcal{A}\_{\ell,t}$ would have zero Brauer class. 


