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. $\square$

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