Lemma 44.7.1. Let k be a field. Let X be a quasi-compact and quasi-separated scheme over k with H^0(X, \mathcal{O}_ X) = k. If X has a k-rational point, then for any Galois extension k'/k we have
\mathop{\mathrm{Pic}}\nolimits (X) = \mathop{\mathrm{Pic}}\nolimits (X_{k'})^{\text{Gal}(k'/k)}
Moreover the action of \text{Gal}(k'/k) on \mathop{\mathrm{Pic}}\nolimits (X_{k'}) is continuous.
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
\mathop{\mathrm{Pic}}\nolimits (X_{k'}) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Pic}}\nolimits (X_{k_\lambda })
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
\mathcal{L}_{x'} \otimes _{\mathcal{O}_{X_{k'}, x'}} \kappa (x').
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
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