The Stacks project

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. $\square$