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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