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. $\square$

Proof. First we observe that the map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is injective by Varieties, Lemma 33.30.3. Hence we have to show the map in the lemma is surjective. Let $\mathcal{L}$ be an invertible $\mathcal{O}_{X_{k'}}$-module which has order dividing $n$ in $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$. Choose an isomorphism $\alpha : \mathcal{L}^{\otimes n} \to \mathcal{O}_{X_{k'}}$ of invertible modules. We will prove that we can descend the pair $(\mathcal{L}, \alpha )$ to $X$.

Set $A = k' \otimes _ k k'$. Since $k'/k$ is purely inseparable, the kernel of the multiplication map $A \to k'$ is a locally nilpotent ideal $I$ of $A$. Observe that

comes with two projections $\text{pr}_ i : X_ A \to X_{k'}$, $i = 0, 1$ which agree over $A/I$. Hence the invertible modules $\mathcal{L}_ i = \text{pr}_ i^*\mathcal{L}$ agree over the closed subscheme $X_{A/I} = X_{k'}$. Since $X_{A/I} \to X_ A$ is a thickening and since $\mathcal{L}_ i$ are $n$-torsion, we see that there exists an isomorphism $\varphi : \mathcal{L}_0 \to \mathcal{L}_1$ by More on Morphisms, Lemma 37.4.2. We may pick $\varphi $ to reduce to the identity modulo $I$. Namely, $H^0(X, \mathcal{O}_ X) = k$ implies $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ by Cohomology of Schemes, Lemma 30.5.2 and $A \to k'$ is surjective hence we can adjust $\varphi $ by multiplying by a suitable element of $A$. Consider the map

We can view $\lambda $ as an element of $A$ because $H^0(X_ A, \mathcal{O}_{X_ A}) = A$ (same reference as above). Since $\varphi $ reduces to the identity modulo $I$ we see that $\lambda = 1 \bmod I$. Then there is a unique $n$th root of $\lambda $ in $1 + I$ (Algebra, Lemma 10.32.8) and after multiplying $\varphi $ by its inverse we get $\lambda = 1$. We claim that $(\mathcal{L}, \varphi )$ is a descent datum for the fpqc covering $\{ X_{k'} \to X\} $ (Descent, Definition 35.2.1). If true, then $\mathcal{L}$ is the pullback of an invertible $\mathcal{O}_ X$-module $\mathcal{N}$ by Descent, Proposition 35.5.2. Injectivity of the map on Picard groups shows that $\mathcal{N}$ is a torsion element of $\mathop{\mathrm{Pic}}\nolimits (X)$ of the same order as $\mathcal{L}$.

Proof of the claim. To see this we have to verify that

As before the diagonal morphism $\Delta : X_{k'} \to X_{k' \otimes _ k k' \otimes _ k k'}$ is a thickening. The left and right hand sides of the equality signs are maps $a, b : p_0^*\mathcal{L} \to p_2^*\mathcal{L}$ compatible with $p_0^*\alpha $ and $p_2^*\alpha $ where $p_ i : X_{k' \otimes _ k k' \otimes _ k k'} \to X_{k'}$ are the projection morphisms. Finally, $a, b$ pull back to the same map under $\Delta $. Affine locally (in local trivializations) this means that $a, b$ are given by multiplication by invertible functions which reduce to the same function modulo a locally nilpotent ideal and which have the same $n$th powers. Then it follows from Algebra, Lemma 10.32.8 that these functions are the same. $\square$