The Stacks project

Lemma 44.7.2. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a quasi-compact and quasi-separated scheme over $k$ with $H^0(X, \mathcal{O}_ X) = k$. Let $n$ be an integer prime to $p$. Then the map

\[ \mathop{\mathrm{Pic}}\nolimits (X)[n] \longrightarrow \mathop{\mathrm{Pic}}\nolimits (X_{k'})[n] \]

is bijective for any purely inseparable extension $k'/k$.

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

\[ X_ A = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(A) = X_{k'} \times _ X X_{k'} \]

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

\[ \lambda : \mathcal{O}_{X_ A} \xrightarrow {\text{pr}_0^*\alpha ^{-1}} \mathcal{L}_0^{\otimes n} \xrightarrow {\varphi ^{\otimes n}} \mathcal{L}_1^{\otimes n} \xrightarrow {\text{pr}_0^*\alpha } \mathcal{O}_{X_ A} \]

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

\[ \text{pr}_{12}^*\varphi \circ \text{pr}_{01}^*\varphi = \text{pr}_{02}^*\varphi \quad \text{on}\quad X_{k'} \times _ X X_{k'} \times _ X X_{k'} = X_{k' \otimes _ k k' \otimes _ k k'} \]

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CD5. Beware of the difference between the letter 'O' and the digit '0'.