Let $k$ be a perfect field. For a geometrically integral smooth $k$-variety $X$, let $Z^{1}(X)$ be the group of Weil divisors ( free $\mathbb{Z}$-module generated by irreducible closed subsets of codimension 1 of $X$ ), $P^{1}(X)$ be its subgroup of principal divisors and $Cl(X) := Z^{1}(X) / P^{1}(X)$ be the divisor class gruop.

Let $\pi : X_{\overline{k}} \rightarrow X $ be the projection from the base change $k \rightarrow \overline{k}$. We have that the pullback $\pi^* : Z^{1}(X) \rightarrow Z^{1}(X_{\overline{k}})$ is injective and via this identification, $Z^{1}(X) = Z^{1}(X_{ \overline{k} })^G$, here $G := Gal( \overline{k} / k)$, and if we assume that $X/k$ is complete, $ Z^{1}(X) \cap P^{1}(X_{\overline{k}}) = P^1(X) $ and hence $ \pi^* $ induces an injective homomorphsim $ \pi^{*} : Cl(X) \rightarrow Cl(X_{\overline{k}}) $. The Galois group still acts on $ Cl(X_{\overline{k}}) $ and the image of $ \pi^* $ is in $ Cl(X_{\overline{k}})^{G} $.

In general, we don't have $ Cl(X) = Cl(X_{\overline{k}})^{G} $. In Milne's note ( abelian variety ), he mentioned that this is true if $X$ is a curve and we have a $k$-rational point on it. But I don't know the reason, and do we have this equality for any complete geometrically integral smooth $k$-variety which has an irreducible closed subset $Z$ of codimension 1 of $X$ such that $ Z_{red} \times_k \overline{k} $ is still an integral closed subvariety of $X_\overline{k}$ ?