The Stacks project

Proof. Since $\text{Gal}(k'/k) = \text{Aut}(k'/k)$ it acts (from the right) on $\mathop{\mathrm{Spec}}(k')$, hence it acts (from the right) on $X_{k'} = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$, and since $\mathop{\mathrm{Pic}}\nolimits (-)$ is a contravariant functor, it acts (from the left) on $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$. If $k'/k$ is an infinite Galois extension, then we write $k' = \mathop{\mathrm{colim}}\nolimits k'_\lambda $ as a filtered colimit of finite Galois extensions, see Fields, Lemma 9.22.3. Then $X_{k'} = \mathop{\mathrm{lim}}\nolimits X_{k_\lambda }$ (as in Limits, Section 32.2) and we obtain

by Limits, Lemma 32.10.3. Moreover, the transition maps in this system of abelian groups are injective by Varieties, Lemma 33.30.3. It follows that every element of $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is fixed by one of the open subgroups $\text{Gal}(k'/k_\lambda )$, which exactly means that the action is continuous. Injectivity of the transition maps implies that it suffices to prove the statement on fixed points in the case that $k'/k$ is finite Galois.

Assume $k'/k$ is finite Galois with Galois group $G = \text{Gal}(k'/k)$. Let $\mathcal{L}$ be an element of $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ fixed by $G$. We will use Galois descent (Descent, Lemma 35.6.1) to prove that $\mathcal{L}$ is the pullback of an invertible sheaf on $X$. Recall that $f_\sigma = \text{id}_ X \times \mathop{\mathrm{Spec}}(\sigma ) : X_{k'} \to X_{k'}$ and that $\sigma $ acts on $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ by pulling back by $f_\sigma $. Hence for each $\sigma \in G$ we can choose an isomorphism $\varphi _\sigma : \mathcal{L} \to f_\sigma ^*\mathcal{L}$ because $\mathcal{L}$ is a fixed by the $G$-action. The trouble is that we don't know if we can choose $\varphi _\sigma $ such that the cocycle condition $\varphi _{\sigma \tau } = f_\sigma ^*\varphi _\tau \circ \varphi _\sigma $ holds. To see that this is possible we use that $X$ has a $k$-rational point $x \in X(k)$. Of course, $x$ similarly determines a $k'$-rational point $x' \in X_{k'}$ which is fixed by $f_\sigma $ for all $\sigma $. Pick a nonzero element $s$ in the fibre of $\mathcal{L}$ at $x'$; the fibre is the $1$-dimensional $k' = \kappa (x')$-vector space

Then $f_\sigma ^*s$ is a nonzero element of the fibre of $f_\sigma ^*\mathcal{L}$ at $x'$. Since we can multiply $\varphi _\sigma $ by an element of $(k')^*$ we may assume that $\varphi _\sigma $ sends $s$ to $f_\sigma ^*s$. Then we see that both $\varphi _{\sigma \tau }$ and $f_\sigma ^*\varphi _\tau \circ \varphi _\sigma $ send $s$ to $f_{\sigma \tau }^*s = f_\tau ^*f_\sigma ^*s$. Since $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ these two isomorphisms have to be the same (as one is a global unit times the other and they agree in $x'$) and the proof is complete. $\square$